ADVANCED ALGEBRA

Prof. Dr. B. Pareigis

Winter Semester 2001/02

Table of Contents

1. Tensor Products and Free Modules 3 1.1. Modules 3 1.2. Tensor products I 5 1.3. Free modules 6 1.4. Tensor products II 8 1.5. Bimodules 9 1.6. Complexes and exact sequences 12 2. Algebras and Coalgebras 15 2.1. Algebras 15 2.2. Tensor algebras 17 2.3. Symmetric algebras 19 2.4. Exterior algebras 21 2.5. Left A-modules 23 2.6. Coalgebras 23 2.7. Comodules 26 3. Projective Modules and Generators 30 3.1. Products and coproducts 30 3.2. Projective modules 34 3.3. Dual basis 36 3.4. Generators 39 4. Categories and Functors 40 4.1. Categories 40 4.2. Functors 42 4.3. Natural Transformations 43 5. Representable and Adjoint Functors, the Yoneda Lemma 46 5.1. Representable functors 46 5.2. The Yoneda Lemma 49 5.3. Adjoint functors 51 5.4. Universal problems 52 6. Limits and Colimits, Products and Equalizers 55 6.1. Limits of diagrams 55 6.2. Colimits of diagrams 57 6.3. Completeness 58 6.4. Adjoint functors and limits 59 7. The Morita Theorems 60 8. Simple and Semisimple rings and Modules 66 8.1. Simple and Semisimple rings 66 8.2. Injective Modules 67 8.3. Simple and Semisimple Modules 70 8.4. Noetherian Modules 73 9. Radical and Socle 76 10. Localization 81 10.1. Local rings 81 10.2. Localization 81 11. Monoidal Categories 87 12. Bialgebras and Hopf Algebras 92 12.1. Bialgebras 92 12.2. Hopf Algebras 94 13. Quickies in Advanced Algebra 101

1. Tensor Products and Free Modules

1.1. Modules.

Definition 1.1. Let R be a ring (always associative with unit element). A left R-module_{R}M is an Abelian group M (with composition written as addition) together with an operation

R × M ∋ (r, m) ↦ rm ∈ M

such that (1) (rs)m = r(sm),

(2) (r + s)m = rm + sm,

(3) r(m + m′) = rm + rm′,

(4) 1m = m for all r, s ∈ R, m, m′ ∈ M. If R is a field then a (left) R-module is a (called a) vector space over R. A homomorphism of left R-modules or simply an R-module homomorphism f : _{R}M ⟶ _{R}N is a homomorphism of groups with f(rm) = rf(m). Right R-modules and homomorphisms of right R-modules are defined analogously. We define

Hom_{R}(.M, .N) := {f : _{R}M ⟶ _{R}N|f is a homomorphism of left R-modules}.

Similarly Hom_{R}(M., N.) denotes the set of homomorphisms of right R-modules M_{R} and N_{R}. An R-module homomorphism f : _{R}M ⟶ _{R}N is

a monomorphism if f is injective, an epimorphism if f is surjective, an isomorphism if f is bijective, an endomorphism if M = N, an automorphism if f is an endomorphism and an isomorphism.

Problem 1.1. Let R be a ring and M be an Abelian group. Show that there is a one-to-one correspondence between maps f : R × M ⟶ M that make M into a left R-module and ring homomorphisms (always preserving the unit element) g : R ⟶ End(M).

Lemma 1.2. Hom_{R}(M, N) is an Abelian group by (f + g)(m) := f(m) + g(m).

Proof. Since N is an Abelian group the set of maps Map(M, N) is also an Abelian group. The set of group homomorphisms Hom(M, N) is a subgroup of Map(M, N) (observe that this holds only for Abelian groups). We show that Hom_{R}(M, N) is a subgroup of Hom(M, N). We must only show that f − g is an R-module homomorphism if f and g are. Obviously f − g is a group homomorphism. Furthermore we have (f − g)(rm) = f(rm) − g(rm) = rf(m) − rg(m) = r(f(m) − g(m)) = r(f − g)(m). □

Problem 1.2. Let f : M ⟶ N be an R-module homomorphism.

(1) f is an isomorphism if and only if (iff) there exists an R-module homomorphism g : N ⟶ M such that

fg = id_{N} and gf = id_{M} .

Furthermore g is uniquely determined by f.

(2) The following are equivalent:

(a) f is a monomorphism,

(b) for all R-modules P and all homomorphisms g, h : P ⟶ M

fg = fh ⟹ g = h,

(c) for all R-modules P the homomorphism of Abelian groups

Hom_{R}(P, f) : Hom_{R}(P, M) ∋ g ↦ fg ∈ Hom_{R}(P, N)

is a monomorphism.

(3) The following are equivalent:

(a) f is an epimorphism,

(b) for all R-modules P and all homomorphisms g, h : N ⟶ P

gf = hf ⟹ g = h,

(c) for all R-modules P the homomorphism of Abelian groups

Hom_{R}(f, P) : Hom_{R}(N, P) ∋ g ↦ gf ∈ Hom_{R}(M, P)

is a monomorphism.

Remark 1.3. Each Abelian group is a ℤ-module in a unique way. Each homomorphism of Abelian groups is a ℤ-module homomorphism.

Proof. By exercise 1.1 we have to find a unique ring homomorphism g : ℤ ⟶ End(M). This holds more generally. If S is a ring then there is a unique ring homomorphism g : ℤ ⟶ S. Since a ring homomorphism must preserve the unit we have g(1) = 1. Define g(n) := 1+…+1 (n-times) for n ≥ 0 and g(−n) := −(1+…+1) (n-times) for n > 0. Then it is easy to check that g is a ring homomorphism and it is obviously unique. This means that M is a ℤ-module by nm = m+…+m (n-times) for n ≥ 0 and (−n)m = −(m+…+m) (n-times) for n > 0. If f : M ⟶ N is a homomorphism of (Abelian) groups then f(nm) = f(m + … + m) = f(m) + … + f(m) = nf(m) for n ≥ 0 and f((−n)m) = f(−(m + … + m)) = −(f(m) + … + f(m)) = (−n)f(m) for n > 0. Hence f is a ℤ-module homomorphism. □

Problem 1.3. (1) Let R be a ring. Then _{R}R is a left R-module.

(2) Let M be a Abelian group and End(M) be the endomorphism ring of M. Then M is an End(M)-module.

(3) {(¯1, ¯0), (¯0, ¯1)} is a generating set for the ℤ-module ℤ/(2) × ℤ/(3).

(4) {(¯1, ¯1)} is a generating set for the ℤ-module ℤ/(2) × ℤ/(3).

(5) _{ℤ}ℤ/(n) has no basis as a module, i.e. this module is not free.

(6) Let V = ⨁_{i=0}^{∞} Kb_{i} be a countably infinite dimensional vector space over the field K. Let p, q, a, b ∈ Hom(V, V) be defined by

p(b_{i}) := b_{2i},
q(b_{i}) := b_{2i+1},
a(b_{i}) := \begin{cases}b_{i/2}, & if i is even, and \\ 0, & if i is odd. \end{cases}

b(b_{i}) := \begin{cases}b_{i−1/2}, & if i is odd, and \\ 0, & if i is even. \end{cases}

Show pa + qb = id_{V}, ap = bq = id, aq = bp = 0.

Show for R = End_{K}(V) that _{R}R = Ra ⊕ Rb and R_{R} = pR ⊕ qR holds.

(7) Are {(0, …, a, …, 0)|a ∈ K_{n}} and {(a, 0, …, 0)|a ∈ K_{n}} isomorphic as M_{n}(K)- modules?

(8) For each module P there is a module Q such that P ⊕ Q ≅ Q.

(9) Which of the following statements is correct?

(a) P_{1} ⊕ Q = P_{2} ⊕ Q ⟹ P_{1} = P_{2}?

(b) P_{1} ⊕ Q = P_{2} ⊕ Q ⟹ P_{1} ≅ P_{2}?

(c) P_{1} ⊕ Q ≅ P_{2} ⊕ Q ⟹ P_{1} ≅ P_{2}?

(10) ℤ/(2) ⊕ ℤ/(6) ⊕ ℤ/(6) ⊕ … ≅ ℤ/(6) ⊕ ℤ/(6) ⊕ ℤ/(6) ⊕ . . ..

(11) ℤ/(2) ⊕ ℤ/(4) ⊕ ℤ/(4) ⊕ … ≅ ℤ̸/(4) ⊕ ℤ/(4) ⊕ ℤ/(4) ⊕ . . ..

(12) Find two Abelian groups P and Q, such that P is isomorphic to a subgroup of Q and Q is isomorphic to a subgroup of P and P ≅ Q̸.

1.2. Tensor products I.

Definition and Remark 1.4. Let M_{R} and _{R}N be R-modules, and let A be an Abelian group. A map f : M × N ⟶ A is called R-bilinear if

(1) f(m + m′, n) = f(m, n) + f(m′, n),

(2) f(m, n + n′) = f(m, n) + f(m, n′),

(3) f(mr, n) = f(m, rn) for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N. Let Bil_{R}(M, N; A) denote the set of all R-bilinear maps f : M × N ⟶ A. Bil_{R}(M, N; A) is an Abelian group with (f + g)(m, n) := f(m, n) + g(m, n).

Definition 1.5. Let M_{R} and _{R}N be R-modules. An Abelian group M ⊗_{R} N together with an R-bilinear map

⊗ : M × N ∋ (m, n) ↦ m ⊗ n ∈ M ⊗_{R} N

is called a tensor product of M and N over R if for each Abelian group A and for each R-bilinear map f : M × N ⟶ A there exists a unique group homomorphism g : M ⊗_{R} N ⟶ A such that the diagram

[figure: M × N; M ⊗_{R} N; ⊗; f; A; g]

commutes. The elements of M ⊗_{R} N are called tensors, the elements of the form m ⊗ n are called decomposable tensors. Warning: If you want to define a homomorphism f : M ⊗_{R} N ⟶ A with a tensor product as domain you must define it by giving an R-bilinear map defined on M × N.

Proposition 1.6. A tensor product (M ⊗_{R} N, ⊗) defined by M_{R} and _{R}N is unique up to a unique isomorphism.

Proof. Let (M ⊗_{R} N, ⊗) and (M ⊠_{R} N, ⊠) be tensor products. Then

[figure: M × N; ⊗; ⊠; ⊗; ⊠; M ⊗_{R} N; M ⊠_{R} N; h; k; M ⊗_{R} N; M ⊠_{R} N; h]

implies k = h^{−1}. □

Because of this fact we will henceforth talk about the tensor product of M and N over R.

Proposition 1.7. (Rules of computation in a tensor product) Let (M ⊗_{R}N, ⊗) be the tensor product. Then we have for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N (1) M ⊗_{R} N = {∑_{i} m_{i} ⊗ n_{i} | m_{i} ∈ M, n_{i} ∈ N},

(2) (m + m′) ⊗ n = m ⊗ n + m′ ⊗ n,

(3) m ⊗ (n + n′) = m ⊗ n + m ⊗ n′,

(4) mr ⊗ n = m ⊗ rn (observe in particular, that ⊗ : M × N ⟶ M ⊗ N is not injective in general),

(5) if f : M × N ⟶ A is an R-bilinear map and g : M ⊗_{R} N ⟶ A is the induced homomorphism, then

g(m ⊗ n) = f(m, n).

Proof. (1) Let B := ⟨m ⊗ n⟩ ⊆ M ⊗_{R} N denote the subgroup of M ⊗_{R} N generated by the decomposable tensors m ⊗ n. Let j : B ⟶ M ⊗_{R} N be the embedding homomorphism. We get an induced map ⊗′ : M × N ⟶ B. The following diagram

[figure: M × N; B; ⊗′; M ⊗_{R} N; j; B; M ⊗_{R} N; j; ⊗′; jp; p]

induces a unique p with p ∘ j ∘ ⊗′ = p ∘ ⊗ = ⊗′ since ⊗′ is R-bilinear. Because of jp ∘ ⊗ = j ∘ ⊗′ = ⊗ = id_{M⊗_{R}N} ∘⊗ we get jp = id_{M⊗_{R}N}, hence the embedding j is surjective and thus the identity.

(2) (m + m′) ⊗ n = ⊗(m + m′, n) = ⊗(m, n) + ⊗(m′, n) = m ⊗ n + m′ ⊗ n.

(3) and (4) analogously.

(5) is precisely the definition of the induced homomorphism. □

To construct tensor products, we need the notion of a free module.

1.3. Free modules.

Definition 1.8. Let X be a set and R be a ring. An R-module RX together with a map ι : X ⟶ RX is called a free R-module generated by X (or an R-module freely generated by X), if for every R-module M and for every map f : X ⟶ M there exists a unique homomorphism of R-modules g : RX ⟶ M such that the diagram

[figure: X; RX; ι; f; M; g]

commutes. An R-module F is a free R-module if there is a set X and a map ι : X ⟶ F such that F is freely generated by X. Such a set X (or its image ι(X)) is called a free generating set for F.

Warning: If you want to define a homomorphism g : RX ⟶ M with a free module as domain you should define it by giving a map f : X ⟶ M.

Proposition 1.9. A free R-module ι : X ⟶ RX defined over a set X is unique up to a unique isomorphism of R-modules.

Proof. follows from the following diagram

[figure: X; ι; ι′; ι; ι′; RX; RX′; h; k; RX; RX′; h]
□

Proposition 1.10. (Rules of computation in a free R-module) Let ι : X ⟶ RX be a free R-module over X. Let x̃ := ι(x) ∈ RX for all x ∈ X. Then we have

(1) X̃ = {x̃| ∃x ∈ X : x̃ = ι(x)} is a generating set of RX, i.e. each element m ∈ RX is a linear combination m = ∑_{i=1}^{n} r_{i}x̃_{i} of the x̃.

(2) X̃ ⊆ RX is linearly independent and ι is injective, i.e. if ∑′_{x∈X} r_{x}x̃ = 0, then we have ∀x ∈ X : r_{x} = 0.

Proof. (1) Let M := ⟨x̃|x ∈ X⟩ ⊆ RX be the submodule generated by the x̃. Then the diagram

[figure: X; RX; ι; RX/M; 0; 0; ν]

commutes with both maps 0 and ν. Thus 0 = ν and RX/M = 0 and hence RX = M.

(2) Let ∑_{i=0}^{n} r_{i}x̃_{i} = 0 and r_{0} ≠ 0. Let j : X ⟶ R be the map given by j(x_{0}) = 1, j(x) = 0 for all x ≠ x_{0}. ⟹ ∃g : RX ⟶ R with

[figure: X; RX; ι; j; R; g]

commutative and 0 = g(0) = g(∑_{i=0}^{n} r_{i}x̃_{i}) = ∑_{i=0}^{n} r_{i}g(x̃_{i}) = ∑_{i=0}^{n} r_{i}j(x_{i}) = r_{0}. Contradiction. Hence the second statement. □

Notation 1.11. Since ι is injective we will identify X with it’s image in RX and we will write ∑_{x∈X} r_{x}x for an element ∑_{x∈X} r_{x}ι(x) ∈ RX. The coefficients r_{x} are uniquely determined.

Proposition 1.12. Let X be a set. Then there exists a free R-module ι : X ⟶ RX over X.

Proof. Obviously RX := {α : X ⟶ R| for almost all x ∈ X : α(x) = 0} is a submodule of Map(X, R) which is an R-module by componentwise addition and multiplication. Define ι : X ⟶ RX by ι(x)(y) := δ_{xy}. Let f : X ⟶ M be an arbitrary map. Let α ∈ RX. Define g(α) := ∑_{x∈X} α(x) · f(x). Then g is well defined, because we have α(x) ≠ 0 for only finitely many x ∈ X. Furthermore g is an R-module homomorphism: rg(α) + sg(β) = r ∑ α(x) · f(x) + s ∑ β(x) · f(x) = ∑(rα(x) + sβ(x)) · f(x) = ∑(rα + sβ)(x) · f(x) = g(rα + sβ). Furthermore we have gι = f : gι(x) = ∑_{y∈X} ι(x)(y) · f(y) = ∑ δ_{xy} · f(y) = f(x). For α ∈ RX we have α = ∑_{x∈X} α(x)ι(x) since α(y) = ∑ α(x)ι(x)(y). In order to show that g is uniquely determined by f, let h ∈ Hom_{R}(RX, M) be given with hι = f. Then h(α) = h(∑ α(x)ι(x)) = ∑ α(x)hι(x) = ∑ α(x)f(x) = g(α) hence h = g. □

Remark 1.13. If the base ring 𝕂 is a field then a 𝕂-module is a vector space. Each vector space V has a basis X (proof by Zorn’s lemma). V together with the embedding X ⟶ V is a free 𝕂-module (as one shows in Linear Algebra). Hence every vector space is free. This is why one always defines vector space homomorphisms only on the basis. For a vector space V any two bases have the same number of elements. This is not true for free modules over an arbitrary ring (see Exercise 1.4).

Problem 1.4. Show that for R := End_{𝕂}(V) for a vector space V of infinite countable dimension there is an isomorphism of left R-modules _{R}R ≅ _{R}R ⊕_{R}R. Conclude that R is a free module on a generating set {1} with one element and also free on a generating set with two elements.

Problem 1.5. Let ι : X ⟶ RX be a free module. Let f : X ⟶ M be a map and g : RX ⟶ M be the induced R-module homomorphism. Then

g(∑_{X} r_{x}x) = ∑_{X} r_{x}f(x).

1.4. Tensor products II.

Proposition 1.14. Given R-modules M_{R} and _{R}N. Then there exists a tensor product (M ⊗_{R} N, ⊗).

Proof. Define M ⊗_{R} N := ℤ(M × N)/U where ℤ(M × N) is a free ℤ-module over M × N (the free Abelian group) and U is generated by

ι(m + m′, n) − ι(m, n) − ι(m′, n)

ι(m, m + n′) − ι(m, n) − ι(m, n′)
ι(mr, n) − ι(m, rn)

for all r ∈ R, m, m′ ∈ M, n, n′ ∈ N. Consider

[figure: M × N; ℤ(M × N); ι; M ⊗_{R} N; ν; = ℤ(M × N)/U; A; ψ; ρ; g]

Let ψ be given. Then there is a unique ρ ∈ Hom(ℤ(M × N), A) such that ρι = ψ. Since ψ is R-bilinear we get ρ(ι(m+m′, n)−ι(m, n)−ι(m′n)) = ψ(m+m′, n)−ψ(m, n)−ψ(m′, n) = 0 and similarly ρ(ι(m, n + n′) − ι(m, n) − ι(m, n′)) = 0 and ρ(ι(mr, n) − ι(m, rn)) = 0. So we get ρ(U) = 0. This implies that there is a unique g ∈ Hom(M ⊗_{R} N, A) such that gν = ρ (homomorphism theorem). Let ⊗ := ν ∘ ι. Then ⊗ is bilinear since (m + m′) ⊗ n = ν ∘ι(m+m′, n) = ν(ι(m+m′, n)) = ν(ι(m+m′, n)−ι(m, n)−ι(m′, n)+ι(m, n)+ι(m′, n)) = ν(ι(m, n) + ι(m′, n)) = ν ∘ ι(m, n) + ν ∘ ι(m′, n) = m ⊗ n + m′ ⊗ n. The other two properties are obtained in an analogous way. We have to show that (M ⊗_{R} N, ⊗) is a tensor product. The above diagram shows that for each Abelian group A and for each R-bilinear map ψ : M × N ⟶ A there is a g ∈ Hom(M ⊗_{R} N, A) such that g ∘ ⊗ = ψ. Given h ∈ Hom(M ⊗_{R} N, A) with h ∘ ⊗ = ψ. Then h ∘ ν ∘ ι = ψ. This implies h ∘ ν = ρ = g ∘ ν hence g = h. □

Proposition and Definition 1.15. Given two homomorphisms

f ∈ Hom_{R}(M., M′.) and g ∈ Hom_{R}(.N, .N′).

Then there is a unique homomorphism

f ⊗_{R} g ∈ Hom(M ⊗_{R} N, M′ ⊗_{R} N′)

such that f ⊗_{R} g(m ⊗ n) = f(m) ⊗ g(n), i.e. the following diagram commutes

[figure: M′ × N′; M′ ⊗_{R} N′; ⊗; M × N; M ⊗_{R} N; ⊗; f × g; f ⊗_{R} g]

Proof. ⊗ ∘ (f × g) is bilinear. □

Notation 1.16. We often write f ⊗_{R} N := f ⊗_{R} 1_{N} and M ⊗_{R} g := 1_{M} ⊗_{R} g. We have the following rule of computation:

f ⊗_{R} g = (f ⊗_{R} N′) ∘ (M ⊗_{R} g) = (M′ ⊗_{R} g) ∘ (f ⊗_{R} N)

since f × g = (f × N′) ∘ (M × g) = (M′ × g) ∘ (f × N).

1.5. Bimodules.

Definition 1.17. Let R, S be rings and let M be a left R-module and a right S-module. M is called an R-S-bimodule if (rm)s = r(ms). We define Hom_{R-S}(.M., .N.) := Hom_{R}(.M, .N) ∩ Hom_{S}(M., N.).

Remark 1.18. Let M_{S} be a right S-module and let R × M ⟶ M be a map. M is an R-S-bimodule if and only if

(1) ∀r ∈ R : (M ∋ m ↦ rm ∈ M) ∈ Hom_{S}(M., M.),

(2) ∀r, r′ ∈ R, m ∈ M : (r + r′)m = rm + r′m,

(3) ∀r, r′ ∈ R, m ∈ M : (rr′)m = r(r′m),

(4) ∀m ∈ M : 1m = m.

Lemma 1.19. Let _{R}M_{S} and _{S}N_{T} be bimodules. Then _{R}(M ⊗_{S} N)_{T} is a bimodule by r(m ⊗ n) := rm ⊗ n and (m ⊗ n)t := m ⊗ nt.

Proof. Clearly we have that (r⊗_{S} id)(m⊗n) = rm⊗n = r(m⊗n) is a homomorphism. Then (2)-(4) hold. Thus M ⊗_{S} N is a left R-module. Similarly it is a right T -module. Finally we have r((m ⊗ n)t) = r(m ⊗ nt) = rm ⊗ nt = (rm ⊗ n)t = (r(m ⊗ n))t. □

Corollary 1.20. Given bimodules _{R}M_{S}, _{S}N_{T}, _{R}M′_{S}, _{S}N′_{T} and homomorphisms f ∈ Hom_{R-S}(.M., .M′.) and g ∈ Hom_{S-T} (.N., .N′.). Then we have f ⊗_{S} g ∈ Hom_{R-T} (.M ⊗_{S} N., .M′ ⊗_{S} N′.).

Proof. f ⊗_{S} g(rm ⊗ nt) = f(rm) ⊗ g(nt) = r(f ⊗_{S} g)(m ⊗ n)t. □

Remark 1.21. Unless otherwise defined 𝕂 will always be a commutative ring. Every module M over the commutative ring 𝕂 and in particular every vector space over a field 𝕂 is a 𝕂-𝕂-bimodule by λm = mλ. Observe that there are 𝕂-𝕂-bimodules that do not satisfy λm = mλ. Take for example an automorphism α : 𝕂 ⟶ 𝕂 and a left 𝕂-module M and define mλ := α(λ)m. Then M is such a 𝕂-𝕂-bimodule. The tensor product M ⊗_{𝕂} N of two 𝕂-𝕂-bimodules M and N is again a 𝕂-𝕂-bimodule. If we have, however, 𝕂-𝕂-bimodules M and N arising from 𝕂-modules as above, i.e. satisfying λm = mλ, then their tensor product M ⊗_{𝕂} N also satisfies this equation, so M ⊗_{𝕂} N comes from a (left) 𝕂-module. Indeed we have λm ⊗ n = mλ ⊗ n = m ⊗ λn = m ⊗ nλ. Thus we can also define a tensor product of two left 𝕂-modules. We often write the tensor product of two vector spaces or two left modules M and N over a commutative ring 𝕂 as M ⊗ N instead of M ⊗_{𝕂} N and the tensor product over 𝕂 of two 𝕂-module homomorphisms f and g as f ⊗ g instead of f ⊗_{𝕂} g. (Warning: Do not confuse this with a tensor f ⊗ g. See the following exercise.)

Problem 1.6. (1) Let M_{R}, _{R}N, M′_{R}, and _{R}N′ be R-modules. Show that the following is a homomorphism of Abelian groups:

μ : Hom_{R}(M, M′) ⊗_{ℤ} Hom_{R}(N, N′) ∋ f ⊗ g ↦ f ⊗_{R} g ∈ Hom(M ⊗_{R} N, M′ ⊗_{R} N′).

(2) Find examples where μ is not injective and where μ is not surjective.

(3) Explain why f ⊗ g is a decomposable tensor whereas f ⊗_{R} g is not a tensor.

Theorem 1.22. Let _{R}M_{S}, _{S}N_{T}, and _{T} P_{U} be bimodules. Then there are canonical isomorphisms of bimodules

(1) Associativity Law: α : (M ⊗_{S} N) ⊗_{T} P ≅ M ⊗_{S} (N ⊗_{T} P).

(2) Law of the Left Unit: λ : R ⊗_{R} M ≅ M.

(3) Law of the Right Unit: ρ : M ⊗_{S} S ≅ M.

(4) Symmetry Law: If M, N are 𝕂-modules then there is an isomorphism of 𝕂-modules τ : M ⊗ N ≅ N ⊗ M.

(5) Existence of Inner Hom-Functors: Let _{R}M_{T}, _{S}N_{T}, and _{S}P_{R} be bimodules. Then there are canonical isomorphisms of bimodules

Hom_{S-T} (.P ⊗_{R} M., .N.) ≅ Hom_{S-R}(.P., . Hom_{T} (M., N.).) and
Hom_{S-T} (.P ⊗_{R} M., .N.) ≅ Hom_{R-T} (.M., . Hom_{S}(.P, .N).).

Proof. We only describe the corresponding homomorphisms.

(1) Use 1.7 (5) to define α((m ⊗ n) ⊗ p) := m ⊗ (n ⊗ p).

(2) Define λ : R ⊗_{R} M ⟶ M by λ(r ⊗ m) := rm.

(3) Define ρ : M ⊗_{S} S ⟶ M by ρ(m ⊗ s) := ms.

(4) Define τ(m ⊗ n) := n ⊗ m.

(5) For f : P ⊗_{R} M ⟶ N define ϕ(f) : P ⟶ Hom_{T} (M, N) by ϕ(f)(p)(m) := f(p ⊗ m) and ψ(f) : M ⟶ Hom_{S}(P, N) by ψ(f)(m)(p) := f(p ⊗ m). □

Usually one identifies threefold tensor products along the map α so that we can use M ⊗_{S} N ⊗_{T} P := (M ⊗_{S}N)⊗_{T} P = M ⊗_{S}(N ⊗_{T} P). For the notion of a monoidal or tensor category, however, this canonical isomorphism (natural transformation) is of central importance and will be discussed later.

Problem 1.7.

(1) Give a complete proof of Theorem 1.22. In (5) show how Hom_{T} (M., N.) becomes an S-R-bimodule.

(2) Give an explicit proof of M ⊗_{R} (X ⊕ Y) ≅ M ⊗_{R} X ⊕ M ⊗_{R} Y.

(3) Show that for every finite dimensional vector space V there is a unique element ∑_{i=1}^{n} v_{i} ⊗ v_{i}^{∗} ∈ V ⊗ V^{∗} such that the following holds

∀v ∈ V : ∑_{i} v_{i}^{∗}(v)v_{i} = v.

(Hint: Use an isomorphism End(V) ≅ V ⊗ V^{∗} and dual bases {v_{i}} of V and {v_{i}^{∗}} of V^{∗}.)

(4) Show that the following diagrams (coherence diagrams or constraints) of 𝕂-modules commute:

[figure: ((A ⊗ B) ⊗ C) ⊗ D; (A ⊗ (B ⊗ C)) ⊗ D; α(A,B,C)⊗1; A ⊗ ((B ⊗ C) ⊗ D); α(A,B⊗C,D); α(A⊗B,C,D); 1⊗α(B,C,D); (A ⊗ B) ⊗ (C ⊗ D); A ⊗ (B ⊗ (C ⊗ D)); α(A,B,C⊗D)]

[figure: (A ⊗ 𝕂) ⊗ B; A ⊗ (𝕂 ⊗ B); α(A,𝕂,B); A ⊗ B; ρ(A)⊗1; 1⊗λ(B)]

(5) Write τ(A, B) : A ⊗ B ⟶ B ⊗ A for τ(A, B) : a ⊗ b ↦ b ⊗ a. Show that

[figure: (A ⊗ B) ⊗ C; (B ⊗ A) ⊗ C; τ(X,B)⊗1; B ⊗ (A ⊗ C); α; α; 1⊗τ(A,C); A ⊗ (B ⊗ C); (B ⊗ C) ⊗ A; τ(A,B⊗C); B ⊗ (C ⊗ A); α]

commutes for all 𝕂-modules A, B, C and that

τ(B, A)τ(A, B) = id_{A⊗B}

for all 𝕂-modules A and B. Let f : A ⟶ A′ and g : B ⟶ B′ be 𝕂-modules homomorphisms. Show that

[figure: A ⊗ B; B ⊗ A; τ(A,B); f⊗g; g⊗f; A′ ⊗ B′; B′ ⊗ A′; τ(A′,B′)]

commutes.

(6) Find an example of M, N ∈ 𝕂-Mod-𝕂 such that M ⊗_{𝕂} N ≅ N̸ ⊗_{𝕂} M.

Proposition 1.23. Let (RX, ι) be a free R-module and _{S}M_{R} be a bimodule. Then every element u ∈ M ⊗_{R} RX has a unique representation u = ∑_{x∈X} m_{x} ⊗ x.

Proof. By 1.10 ∑_{x∈X} r_{x}x is the general element of RX. Hence we have u = ∑ m_{i} ⊗ α_{i} = ∑ m_{i} ⊗ ∑ r_{x,i}x = ∑_{i} ∑_{x} m_{i}r_{x,i} ⊗ x = ∑_{x}(∑_{i} m_{i}r_{x,i}) ⊗ x. To show the uniqueness let ∑_{y∈X} m_{y} ⊗ y = 0. Let x ∈ X and f_{x} : RX ⟶ R be defined by f_{x}(ι(y)) = f_{x}(y) := δ_{xy}. Then (1_{M} ⊗_{R} f_{x})(∑ m_{y} ⊗ y) = ∑ m_{y} ⊗ f_{x}(y) = m_{x} ⊗ 1 = 0 for all x ∈ X. Now let

[figure: M × R; M ⊗_{R} R; ⊗; mult; M; ρ]

be given. Then ρ(m_{x} ⊗ 1) = m_{x} · 1 = m_{x} = 0 hence we have uniqueness. From 1.22 (3) we know that ρ is an isomorphism. □

Corollary 1.24. Let _{S}M_{R}, _{R}N be (bi-)modules. Let M be a free S-module over Y, and N be a free R-module over X. Then M ⊗_{R} N is a free S-module over Y × X.

Proof. Consider the diagram Let f be an arbitrary map. For all x ∈ X we define homomorphisms g(-, x) ∈ Hom_{S}(.M, .U) by the commutative diagram

[figure: Y × X; M × N; ι_{Y} × ι_{X}; _{S}M ⊗_{R} N; ⊗; _{S}U; f; g; h]

[figure: Y; _{S}M; ι_{Y}; f(-, x); _{S}U; g(-, x)]

Let g̃ ∈ Hom_{R}(.N, . Hom_{S}(.M_{R}, .U)) be defined by

[figure: X; _{R}N; ι_{x}; g(-, -); _{R} Hom_{S}(.M_{R}, .U); g̃]

with x ↦ g(-, x). Then we define g(m, n) := g̃(n)(m) =: h(m ⊗ n). Observe that g is additive in m and in n (because g̃ is additive in m and in n), and g is R-bilinear, because g(mr, n) = g̃(n)(mr) = (rg̃(n))(m) = g̃(rn)(m) = g(m, rn). Obviously g(y, x) = f(y, x), hence h ∘ ⊗ ∘ ι_{Y} × ι_{X} = f. Furthermore we have h(sm ⊗ n) = g̃(n)(sm) = s(g̃(n)(m)) = sh(m ⊗ n), hence h is an S-module homomorphism. Let k be an S-module homomorphism satisfying k∘⊗∘ι_{Y} ×ι_{X} = f, then k∘⊗(-, x) = g(-, x), since k ∘ ⊗ is S-linear in the first argument. Thus k ∘ ⊗(m, n) = g̃(n)(m) = h(m ⊗ n), and hence h = k. □

Problem 1.8. (Tensors in physics:) Let V be a finite dimensional vector space over the field 𝕂 and let V^{∗} be its dual space. Let t be a tensor in V ⊗…⊗V ⊗V^{∗}⊗…⊗V^{∗} = V^{⊗r}⊗(V^{∗})^{⊗s}.

(1) Show that for each basis B = (b_{1}, …, b_{n}) and dual basis B^{∗} = (b^{1}, …, b^{n}) there is a uniquely determined scheme (a family or an (r + s)-dimensional matrix) of coefficients (a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}}) with a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}} ∈ 𝕂 such that

(1)  t = ∑_{i_{1}=1}^{n} … ∑_{i_{r}=1}^{n} ∑_{j_{1}=1}^{n} … ∑_{j_{s}=1}^{n} a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}}b_{i_{1}} ⊗ … ⊗ b_{i_{r}} ⊗ b^{j_{1}} ⊗ … ⊗ b^{j_{s}}.

(2) Show that for each change of bases L : B ⟶ C with c_{j} = ∑ λ_{j}^{i}b_{i} (with inverse matrix (μ_{j}^{i})) the following transformation formula holds

(2)  a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}} = ∑_{k_{1}=1}^{n} … ∑_{k_{r}=1}^{n} ∑_{l_{1}=1}^{n} … ∑_{l_{s}=1}^{n} λ_{k_{1}}^{i_{1}} … λ_{k_{r}}^{i_{r}}μ_{j_{1}}^{l_{1}} … μ_{j_{s}}^{l_{s}}a(C)_{l_{1},...,l_{s}}^{k_{1},...,k_{r}}

(3) Show that every family of schemes of coefficients (a(B)|B basis of V) with a(B) = (a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}}) and a(B)_{j_{1},...,j_{s}}^{i_{1},...,i_{r}} ∈ K satisfying the transformation formula (2) defines a unique tensor (independent of the choice of the basis) t ∈ V^{⊗r} ⊗ (V^{∗})^{⊗s} such that (1) holds. Rule for physicists: A tensor is a collection of schemes of coefficients that transform according to the transformation formula for tensors.

1.6. Complexes and exact sequences.

Definition 1.25. A (finite or infinite) sequence of homomorphisms

… ⟶ M_{i−1} ⟶^{f_{i−1}} M_{i} ⟶^{f_{i}} M_{i+1} ⟶ …

is called a complex, if f_{i}f_{i−1} = 0 for all i ∈ I (or equivalently Im(f_{i−1}) ⊆ Ke(f_{i})). A complex is called exact or an exact sequence if Im(f_{i−1}) = Ke(f_{i}) for all i ∈ I.

Lemma 1.26. A complex

… ⟶ M_{i−1} ⟶^{f_{i−1}} M_{i} ⟶^{f_{i}} M_{i+1} ⟶ …

is exact if and only if the sequences

0 ⟶ Im(f_{i−1}) ⟶ M_{i} ⟶ Im(f_{i}) ⟶ 0

are exact for all i ∈ I, if and only if the sequences

0 ⟶ Ke(f_{i−1}) ⟶ M_{i−1} ⟶ Ke(f_{i}) ⟶ 0

are exact for all i ∈ I.

Proof. The sequences

0 ⟶ Ke(f_{i}) ⟶ M_{i} ⟶ Im(f_{i}) ⟶ 0

are obviously exact since Ke(f_{i}) ⟶ M_{i} is a monomorphism, M_{i} ⟶ Im(f_{i}) is an epimorphism and Ke(f_{i}) is the kernel of M_{i} ⟶ Im(f_{i}). The sequence

0 ⟶ Im(f_{i−1}) ⟶ M_{i} ⟶ Im(f_{i}) ⟶ 0

is exact if and only if Im(f_{i−1}) = Ke(f_{i}). The sequence

0 ⟶ Ke(f_{i−1}) ⟶ M_{i−1} ⟶ Ke(f_{i}) ⟶ 0

is exact if and only if M_{i−1} ⟶ Ke(f_{i}) is surjective, if and only if Im(f_{i−1}) = Ke(f_{i}). □

Problem 1.9. (1) In the tensor product ℂ ⊗_{ℂ} ℂ we have 1 ⊗ i − i ⊗ 1 = 0.

In the tensor product ℂ ⊗_{ℝ} ℂ we have 1 ⊗ i − i ⊗ 1 ≠ 0.

(2) For each R-module M we have R ⊗_{R} M ≅ M.

(3) Given the ℚ-vector space V = ℚ^{n}.

(a) Determine dim_{ℝ}(ℝ ⊗_{ℚ} V).

(b) Describe explicitely an isomorphism ℝ ⊗_{ℚ} V ≅ ℝ^{n}.

(4) Let V be a ℚ-vector space and W be an ℝ-vector space.

(a) Hom_{ℝ}(.ℝ_{ℚ}, .W) ≅ W in ℚ-Mod.

(b) Hom_{ℚ}(.V, .W) ≅ Hom_{ℝ}(.ℝ ⊗_{ℚ} V, .W).

(c) Let dim_{ℚ}V < ∞ and dim_{ℝ}W < ∞. How can one explain that in 4b we have infinite matrices on the left hand side and finite matrices on the right hand side?

(d) Hom_{ℚ}(.V, Hom_{ℝ}(.ℝ, .W) ≅ Hom_{ℝ}(.ℝ ⊗_{ℚ} V, .W).

(5) ℤ/(18) ⊗_{ℤ} ℤ/(30) ≠ 0.

(6) m : ℤ/(18) ⊗_{ℤ} ℤ/(30) ∋ \overline{x} ⊗ \overline{y} ↦ \overline{xy} ∈ ℤ/(6) is a homomorphism and m is bijective.

(7) For ℚ-vector spaces V and W we have V ⊗_{ℤ} W ≅ V ⊗_{ℚ} W.

(8) For each finite Abelian group M we have ℚ ⊗_{ℤ} M = 0.

(9) ℤ/(m) ⊗_{ℤ} ℤ/(n) ≅ ℤ/(ggT(m, n)).

(10) ℚ ⊗_{ℤ} ℤ/(n) = 0.

(11) Hom_{ℤ}(ℚ, ℤ/(n)) = 0.

(12) Determine explicitely isomorphisms for

\begin{matrix}ℤ ⊗_{ℤ} ℚ ≅ ℚ, \\ 3ℤ ⊗_{ℤ} ℚ ≅ ℚ. \end{matrix}

Show that the following diagram commutes

[figure: ℚ; ℚ; 3·; 3ℤ ⊗_{ℤ} ℚ; ℤ ⊗_{ℤ} ℚ; ≅; ≅]

(13) The homomorphism 2ℤ⊗_{Z} ℤ/(2) ⟶ ℤ⊗_{ℤ}ℤ/(2) is the zero homomorphism, but both modules are different from zero.

2. Algebras and Coalgebras

2.1. Algebras. Let 𝕂 be a commutative ring. We consider all 𝕂-modules as 𝕂-𝕂-bimodules as in Remark 1.21. Tensor products of 𝕂-modules will be simply written as M ⊗ N := M ⊗_{K} N.

Definition 2.1. A 𝕂-algebra is a 𝕂-module A together with a multiplication ∇ : A ⊗ A ⟶ A (𝕂-module homomorphism) that is associative:

[figure: A ⊗ A; A; ∇; A ⊗ A ⊗ A; A ⊗ A; id ⊗∇; ∇⊗id; ∇]

and a unit η : 𝕂 ⟶ A (𝕂-module homomorphism):

[figure: 𝕂 ⊗ A ≅ A ≅ A ⊗ 𝕂; A ⊗ A; id ⊗η; η⊗id; ∇; A ⊗ A; A.; ∇; id]

A 𝕂-algebra A is commutative if the following diagram commutes

[figure: A ⊗ A; A ⊗ A; τ; A.; ∇; ∇]

Let A and B be 𝕂-algebras. A homomorphism of algebras f : A ⟶ B is a 𝕂-module homomorphism such that the following diagrams commute:

[figure: A; B; f; A ⊗ A; B ⊗ B; f⊗f; ∇_{A}; ∇_{B}]

and

[figure: 𝕂; η_{A}; η_{B}; A; B.; f]

Remark 2.2. Every 𝕂-algebra A is a ring with the multiplication

A × A ⟶^{⊗} A ⊗ A ⟶^{∇} A.

The unit element is η(1), where 1 is the unit element of 𝕂. Obviously the composition of two homomorphisms of algebras is again a homomorphism of algebras. Furthermore the identity map is a homomorphism of algebras.

Problem 2.1. (1) Show that End_{𝕂}(V) is a 𝕂-algebra.

(2) Show that (A, ∇ : A ⊗ A ⟶ A, η : 𝕂 ⟶ A) is a 𝕂-algebra if and only if A with the multiplication A × A ⟶^{⊗} A ⊗ A ⟶^{∇} A and the unit η(1) is a ring and η : 𝕂 ⟶ Cent(A) is a ring homomorphism into the center of A, where Cent(A) := {a ∈ A|∀b ∈ A : ab = ba}.

(3) Let V be a 𝕂-module. Show that D(V) := 𝕂×V with the multiplication (r_{1}, v_{1})(r_{2}, v_{2}) := (r_{1}r_{2}, r_{1}v_{2} + r_{2}v_{1}) is a commutative 𝕂-algebra.

Lemma 2.3. Let A and B be algebras. Then A ⊗ B is an algebra with the multiplication (a_{1} ⊗ b_{1})(a_{2} ⊗ b_{2}) := a_{1}a_{2} ⊗ b_{1}b_{2}.

Proof. Certainly the algebra properties can easily be checked by a simple calculation with elements. For later applications we prefer a diagrammatic proof. Let ∇_{A} : A ⊗ A ⟶ A and ∇_{B} : B ⊗ B ⟶ B denote the multiplications of the two algebras. Then the new multiplication is ∇_{A⊗B} := (∇_{A} ⊗ ∇_{B})(1_{A} ⊗ τ ⊗ 1_{B}) : A ⊗ B ⊗ A ⊗ B ⟶ A ⊗ B where τ : B ⊗ A ⟶ A ⊗ B is the symmetry map from Theorem 1.22. Now the following diagram commutes

[figure: A ⊗ B ⊗ A ⊗ B ⊗ A ⊗ B; A ⊗ A ⊗ B ⊗ B ⊗ A ⊗ B; 1⊗τ⊗1^{3}; A ⊗ B ⊗ A ⊗ B; ∇⊗∇⊗1^{2}; A ⊗ B ⊗ A ⊗ A ⊗ B ⊗ B; A ⊗ A ⊗ A ⊗ B ⊗ B ⊗ B; 1⊗τ_{B,A⊗A}⊗1^{2}; A ⊗ A ⊗ B ⊗ B; ∇⊗1⊗∇⊗1; 1^{3}⊗τ⊗1; 1^{2}⊗τ_{B⊗B,A}⊗1; 1⊗τ⊗1; A ⊗ A ⊗ B ⊗ A ⊗ B ⊗ B; 1⊗τ⊗1^{3}; 1^{3}⊗τ⊗1; 1^{2}⊗τ⊗1^{2}; A ⊗ B ⊗ A ⊗ B; A ⊗ A ⊗ B ⊗ B; 1⊗τ⊗1; A ⊗ B; ∇⊗∇; 1^{2}⊗∇⊗∇; 1⊗∇⊗1⊗∇; ∇⊗∇]

In the left upper rectangle of the diagram the quadrangle commutes by the properties of the tensor product and the two triangles commute by inner properties of τ. The right upper and left lower rectangles commute since τ is a natural transformation (use Exercise 1.7 (5)) and the right lower rectangle commutes by the associativity of the algebras A and B. Furthermore we use the homomorphism η = η_{A⊗B} : 𝕂 ⟶ 𝕂 ⊗ 𝕂 ⟶ A ⊗ B in the following commutative diagram

[figure: 𝕂 ⊗ A ⊗ B ≅ A ⊗ B ≅ A ⊗ B ⊗ 𝕂; A ⊗ B ⊗ 𝕂 ⊗ 𝕂; A ⊗ B ⊗ A ⊗ B; 1^{2}⊗η⊗η; 1; A ⊗ 𝕂 ⊗ B ⊗ 𝕂; A ⊗ A ⊗ B ⊗ B; 1⊗η⊗1⊗η; 1⊗τ⊗1; 1⊗τ⊗1; ∇⊗∇; 𝕂 ⊗ 𝕂 ⊗ A ⊗ B; 𝕂 ⊗ A ⊗ 𝕂 ⊗ B; 1⊗τ⊗1; A ⊗ B ⊗ A ⊗ B; A ⊗ A ⊗ B ⊗ B; 1⊗τ⊗1; A ⊗ B.; ∇⊗∇; η⊗η⊗1^{2}; η⊗1⊗η⊗1]
□

2.2. Tensor algebras.

Definition 2.4. Let 𝕂 be a commutative ring. Let V be a 𝕂-module. A 𝕂-algebra T (V) together with a homomorphism of 𝕂-modules ι : V ⟶ T (V) is called a tensor algebra over V if for each 𝕂-algebra A and for each homomorphism of 𝕂-modules f : V ⟶ A there exists a unique homomorphism of 𝕂-algebras g : T (V) ⟶ A such that the diagram

[figure: V; T (V); ι; f; A; g]

commutes. Note: If you want to define a homomorphism g : T (V) ⟶ A with a tensor algebra as domain you should define it by giving a homomorphism of 𝕂-modules defined on V.

Lemma 2.5. A tensor algebra (T (V), ι) defined by V is unique up to a unique isomorphism.

Proof. Let (T (V), ι) and (T′(V), ι′) be tensor algebras over V. Then

[figure: V; ι; ι′; ι; ι′; T (V); T′(V); h; k; T (V); T′(V); h]

implies k = h^{−1}. □

Proposition 2.6. (Rules of computation in a tensor algebra) Let (T (V), ι) be the tensor algebra over V. Then we have

(1) ι : V ⟶ T (V) is injective (so we may identify the elements ι(v) and v for all v ∈ V),

(2) T (V) = \{∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\}, where v_{i_{j}} ∈ V,

(3) if f : V ⟶ A is a homomorphism of 𝕂-modules, A is a 𝕂-algebra, and g : T (V) ⟶ A is the induced homomorphism of 𝕂-algebras, then

g(∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}) = ∑_{n,\overline{i}} f(v_{i_{1}}) · . . . · f(v_{i_{n}}).

Proof. (1) Use the embedding homomorphism j : V ⟶ D(V), where D(V) is defined as in 2.1 (3) to construct g : T (V) ⟶ D(V) such that g ∘ ι = j. Since j is injective so is ι.

(2) Let B := \{∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\}. Obviously B is the subalgebra of T (V) generated by the elements of V. Let j : B ⟶ T (V) be the embedding homomorphism. Then ι : V ⟶ T (V) factors through a 𝕂-module homomorphism ι′ : V ⟶ B. The following diagram

[figure: V; B; ι′; T (V); j; B; T (V); j; ι′; jp; p]

induces a unique p with p ∘ j ∘ ι′ = p ∘ ι = ι′ since ι′ is a homomorphism of 𝕂-modules. Because of jp ∘ ι = j ∘ ι′ = ι = id_{T (V)} ∘ι we get jp = id_{T (V)}, hence the embedding j is surjective and thus j is the identity.

(3) is precisely the definition of the induced homomorphism. □

Proposition 2.7. Given a 𝕂-module V. Then there exists a tensor algebra (T (V), ι).

Proof. Define T^{n}(V) := V ⊗ … ⊗ V = V^{⊗n} to be the n-fold tensor product of V. Define T^{0}(V) := 𝕂 and T^{1}(V) := V. We define

T (V) := ⨁_{i≥0} T^{i}(V) = 𝕂 ⊕ V ⊕ (V ⊗ V) ⊕ (V ⊗ V ⊗ V) ⊕ ….

The components T^{n}(V) of T (V) are called homogeneous components. The canonical isomorphisms T^{m}(V) ⊗ T^{n}(V) ≅ T^{m+n}(V) taken as multiplication

\begin{matrix}∇ : T^{m}(V) ⊗ T^{n}(V) ⟶ T^{m+n}(V) \\ ∇ : T (V) ⊗ T (V) ⟶ T (V) \end{matrix}

and the embedding η : 𝕂 = T^{0}(V) ⟶ T (V) induce the structure of a 𝕂-algebra on T (V). Furthermore we have the embedding ι : V ⟶ T^{1}(V) ⊆ T (V). We have to show that (T (V), ι) is a tensor algebra. Let f : V ⟶ A be a homomorphism of 𝕂-modules. Each element in T (V) is a sum of decomposable tensors v_{1} ⊗ … ⊗ v_{n}. Define g : T (V) ⟶ A by g(v_{1} ⊗ … ⊗ v_{n}) := f(v_{1}) … f(v_{n}) (and (g : T^{0}(V) ⟶ A) = (η : 𝕂 ⟶ A)). By induction one sees that g is a homomorphism of algebras. Since (g : T^{1}(V) ⟶ A) = (f : V ⟶ A) we get g ∘ ι = f. If h : T (V) ⟶ A is a homomorphism of algebras with h ∘ ι = f we get h(v_{1} ⊗ … ⊗ v_{n}) = h(v_{1}) … h(v_{n}) = f(v_{1}) … f(v_{n}) hence h = g. □

Problem 2.2. (1) Let X be a set and V := 𝕂X be the free 𝕂-module over X. Show that X ⟶ V ⟶ T (V) defines a free algebra over X, i.e. for every 𝕂-algebra A and every map f : X ⟶ A there is a unique homomorphism of 𝕂-algebras g : T (V) ⟶ A such that the diagram

[figure: X; T (V); f; A; g]

commutes. We write 𝕂⟨X⟩ := T (𝕂X) and call it the polynomial ring over 𝕂 in the non-commuting variables X.

(2) Let T (V) and ι : V ⟶ T (V) be a tensor algebra. Regard V as a subset of T (V) by ι. Show that there is a unique homomorphism of algebras Δ : T (V) ⟶ T (V) ⊗ T (V) with Δ(v) = v ⊗ 1 + 1 ⊗ v for all v ∈ V.

(3) Show that (Δ ⊗ 1)Δ = (1 ⊗ Δ)Δ : T (V) ⟶ T (V) ⊗ T (V) ⊗ T (V).

(4) Show that there is a unique homomorphism of algebras ε : T (V) ⟶ 𝕂 with ε(v) = 0 for all v ∈ V.

(5) Show that (ε ⊗ 1)Δ = (1 ⊗ ε)Δ = id_{T (V)}.

(6) Show that there is a unique homomorphism of algebras S : T (V) ⟶ T (V)^{op} with S(v) = −v. (T (V)^{op} is the opposite algebra of T (V) with multiplication s ∗ t := ts for all s, t ∈ T (V) = T (V)^{op} and where st denotes the product in T (V).)

(7) Show that the diagrams

[figure: T (V); 𝕂; ε; T (V); η; T (V) ⊗ T (V); T (V) ⊗ T (V); 1⊗S; S⊗1; Δ; ∇]

commute.

2.3. Symmetric algebras.

Definition 2.8. Let 𝕂 be a commutative ring. Let V be a 𝕂-module. A 𝕂-algebra S(V) together with a homomorphism of 𝕂-modules ι : V ⟶ S(V), such that ι(v)·ι(v′) = ι(v′)·ι(v) for all v, v′ ∈ V, is called a symmetric algebra over V if for each 𝕂-algebra A and for each homomorphism of 𝕂-modules f : V ⟶ A, such that f(v)·f(v′) = f(v′)·f(v) for all v, v′ ∈ V, there exists a unique homomorphism of 𝕂-algebras g : S(V) ⟶ A such that the diagram

[figure: V; S(V); ι; f; A; g]

commutes. Note: If you want to define a homomorphism g : S(V) ⟶ A with a symmetric algebra as domain you should define it by giving a homomorphism of 𝕂-modules f : V ⟶ A satisfying f(v) · f(v′) = f(v′) · f(v) for all v, v′ ∈ V.

Lemma 2.9. A symmetric algebra (S(V), ι) defined by V is unique up to a unique isomorphism.

Proof. Let (S(V), ι) and (S′(V), ι′) be symmetric algebras over V. Then

[figure: V; ι; ι′; ι; ι′; S(V); S′(V); h; k; S(V); S′(V); h]

implies k = h^{−1}. □

Proposition 2.10. (Rules of computation in a symmetric algebra) Let (S(V), ι) be the symmetric algebra over V. Then we have

(1) ι : V ⟶ S(V) is injective (we will identify the elements ι(v) and v for all v ∈ V),

(2) S(V) = \{∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\},

(3) if f : V ⟶ A is a homomorphism of 𝕂-modules satisfying f(v) · f(v′) = f(v′) · f(v) for all v, v′ ∈ V, A is a 𝕂-algebra, and g : S(V) ⟶ A is the induced homomorphism 𝕂-algebras, then

g(∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}) = ∑_{n,\overline{i}} f(v_{i_{1}}) · . . . · f(v_{i_{n}}).

Proof. (1) Use the embedding homomorphism j : V ⟶ D(V), where D(V) is the commutative algebra defined in 2.1 (3) to construct g : S(V) ⟶ D(V) such that g ∘ ι = j. Since j is injective so is ι.

(2) Let B := \{∑_{n,\overline{i}} v_{i_{1}} · . . . · v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\}. Obviously B is the subalgebra of S(V) generated by the elements of V. Let j : B ⟶ S(V) be the embedding homomorphism. Then ι : V ⟶ S(V) factors through a 𝕂-module homomorphism ι′ : V ⟶ B. The following diagram

[figure: V; B; ι′; S(V); j; B; S(V); j; ι′; jp; p]

induces a unique p with p ∘ j ∘ ι′ = p ∘ ι = ι′ since ι′ is a homomorphism of 𝕂-modules satisfying ι′(v) · ι′(v′) = ι′(v′) · ι′(v) for all v, v′ ∈ V. Because of jp ∘ ι = j ∘ ι′ = ι = id_{S(V)} ∘ι we get jp = id_{S(V)}, hence the embedding j is surjective and thus the identity.

(3) is precisely the definition of the induced homomorphism. □

Proposition 2.11. Let V be a 𝕂-module. The symmetric algebra (S(V), ι) is commutative and satisfies the following universal property: for each commutative 𝕂-algebra A and for each homomorphism of 𝕂-modules f : V ⟶ A there exists a unique homomorphism of 𝕂-algebras g : S(V) ⟶ A such that the diagram

[figure: V; S(V); ι; f; A; g]

commutes.

Proof. Commutativity follows from the commutativity of the generators: vv′ = v′v which carries over to the elements of the form ∑_{n,\overline{i}} v_{i_{1}} ·. . .·v_{i_{n}}. The universal property follows since the defining condition f(v)·f(v′) = f(v′)·f(v) for all v, v′ ∈ V is automatically satisfied. □

Proposition 2.12. Given a 𝕂-module V. Then there exists a symmetric algebra (S(V), ι).

Proof. Define S(V) := T (V)/I where I = ⟨vv′−v′v|v, v′ ∈ V ⟩ is the two-sided ideal generated by the elements vv′ − v′v. Let ι be the canonical map V ⟶ T (V) ⟶ S(V). Then the universal property is easily verified by the homomorphism theorem for algebras. □

Problem 2.3. (1) Let X be a set and V := 𝕂X be the free 𝕂-module over X. Show that X ⟶ V ⟶ S(V) defines a free commutative algebra over X, i.e. for every commutative 𝕂-algebra A and every map f : X ⟶ A there is a unique homomorphism of 𝕂-algebras g : S(V) ⟶ A such that the diagram

[figure: X; S(V); f; A; g]

commutes. The algebra 𝕂[X] := S(𝕂X) is called the polynomial ring over 𝕂 in the (commuting) variables X.

(2) Let S(V) and ι : V ⟶ S(V) be a symmetric algebra. Show that there is a unique homomorphism of algebras Δ : S(V) ⟶ S(V) ⊗ S(V) with Δ(v) = v ⊗ 1 + 1 ⊗ v for all v ∈ V.

(3) Show that (Δ ⊗ 1)Δ = (1 ⊗ Δ)Δ : S(V) ⟶ S(V) ⊗ S(V) ⊗ S(V).

(4) Show that there is a unique homomorphism of algebras ε : S(V) ⟶ 𝕂 with ε(v) = 0 for all v ∈ V.

(5) Show that (ε ⊗ 1)Δ = (1 ⊗ ε)Δ = id_{S(V)}.

(6) Show that there is a unique homomorphism of algebras S : S(V) ⟶ S(V) with S(v) = −v.

(7) Show that the diagrams

[figure: S(V); 𝕂; ε; S(V); η; S(V) ⊗ S(V); S(V) ⊗ S(V); 1⊗S; S⊗1; Δ; ∇]

commute.

2.4. Exterior algebras.

Definition 2.13. Let 𝕂 be a commutative ring. Let V be a 𝕂-module. A 𝕂-algebra E(V) together with a homomorphism of 𝕂-modules ι : V ⟶ E(V), such that ι(v)^{2} = 0 for all v ∈ V, is called an exterior algebra or Grassmann algebra over V if for each 𝕂-algebra A and for each homomorphism of 𝕂-modules f : V ⟶ A, such that f(v)^{2} = 0 for all v ∈ V, there exists a unique homomorphism of 𝕂-algebras g : E(V) ⟶ A such that the diagram

[figure: V; E(V); ι; f; A; g]

commutes. The multiplication in E(V) is usually denoted by u ∧ v. Note: If you want to define a homomorphism g : E(V) ⟶ A with an exterior algebra as domain you should define it by giving a homomorphism of 𝕂-modules defined on V satisfying f(v)^{2} = 0 for all v, v′ ∈ V.

Problem 2.4. (1) Let f : V ⟶ A be a 𝕂-module homomorphism satisfying f(v)^{2} = 0 for all v ∈ V. Then f(v)f(v′) = −f(v′)f(v) for all v, v′ ∈ V.

(2) Let 2 be invertible in 𝕂 (e.g. 𝕂 a field of characteristic ≠ 2). Let f : V ⟶ A be a 𝕂module homomorphism satisfying f(v)f(v′) = −f(v′)f(v) for all v, v′ ∈ V. Then f(v)^{2} = 0 for all v ∈ V.

Lemma 2.14. An exterior algebra (E(V), ι) defined by V is unique up to a unique isomorphism.

Proof. Let (E(V), ι) and (E′(V), ι′) be exterior algebras over V. Then

[figure: V; ι; ι′; ι; ι′; E(V); E′(V); h; k; E(V); E′(V); h]

implies k = h^{−1}. □

Proposition 2.15. (Rules of computation in an exterior algebra) Let (E(V), ι) be the exterior algebra over V. Then we have

(1) ι : V ⟶ E(V) is injective (we will identify the elements ι(v) and v for all v ∈ V), (2) E(V) = \{∑_{n,\overline{i}} v_{i_{1}} ∧ … ∧ v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\},

(3) if f : V ⟶ A is a homomorphism of 𝕂-modules satisfying f(v) · f(v′) = −f(v′) · f(v) for all v, v′ ∈ V, A is a 𝕂-algebra, and g : E(V) ⟶ A is the induced homomorphism 𝕂-algebras, then

g(∑_{n,\overline{i}} v_{i_{1}} ∧ … ∧ v_{i_{n}}) = ∑_{n,\overline{i}} f(v_{i_{1}}) · . . . · f(v_{i_{n}}).

Proof. (1) Use the embedding homomorphism j : V ⟶ D(V), where D(V) is the algebra defined in 2.1 (3) to construct g : E(V) ⟶ D(V) such that g ∘ ι = j. Since j is injective so is ι.

(2) Let B := \{∑_{n,\overline{i}} v_{i_{1}} ∧ … ∧ v_{i_{n}}|\overline{i} = (i_{1}, …, i_{n}) multiindex of length n\}. Obviously B is the subalgebra of E(V) generated by the elements of V. Let j : B ⟶ E(V) be the embedding homomorphism. Then ι : V ⟶ E(V) factors through a 𝕂-module homomorphism ι′ : V ⟶ B. The following diagram

[figure: V; B; ι′; E(V); j; B; E(V); j; ι′; jp; p]

induces a unique p with p ∘ j ∘ ι′ = p ∘ ι = ι′ since ι′ is a homomorphism of 𝕂-modules satisfying ι′(v)·ι′(v′) = −ι′(v′)·ι′(v) for all v, v′ ∈ V. Because of jp∘ι = j ∘ι′ = ι = id_{E(V)} ∘ι we get jp = id_{E(V)}, hence the embedding j is surjective and thus j is the identity.

(3) is precisely the definition of the induced homomorphism. □

Proposition 2.16. Given a 𝕂-module V. Then there exists an exterior algebra (E(V), ι).

Proof. Define E(V) := T (V)/I where I = ⟨v^{2}|v ∈ V ⟩ is the two-sided ideal generated by the elements v^{2}. Let ι be the canonical map V ⟶ T (V) ⟶ E(V). Then the universal property is easily verified by the homomorphism theorem for algebras. □

Problem 2.5. (1) Let V be a finite dimensional vector space of dimension n. Show that E(V) is finite dimensional of dimension 2^{n}. (Hint: The homogeneous components E^{i}(V) have dimension ( \begin{matrix}n \\ i \end{matrix} ).

(2) Show that the symmetric group S_{n} operates (from the left) on T^{n}(V) by σ(v_{1}⊗…⊗v_{n}) = v_{σ^{−1}(1)} ⊗ … ⊗ v_{σ^{−1}(n)} with σ ∈ S_{n} and v_{i} ∈ V.

(3) A tensor a ∈ T^{n}(V) is called a symmetric tensor if σ(a) = a for all σ ∈ S_{n}. Let Ŝ^{n}(V) be the subspace of symmetric tensors in T^{n}(V).

a) Show that S : T^{n}(V) ∋ a ↦ ∑_{σ∈S_{n}} σ(a) ∈ T^{n}(V) is a linear map (symmetrization).

b) Show that S has its image in Ŝ^{n}(V).

c) Show that Im(S) = Ŝ^{n}(V) if n! is invertible in 𝕂.

d) Show that Ŝ^{n}(V) ↪ T^{n}(V) →^{ν} S^{n}(V) is an isomorphism if n! is invertible in 𝕂 and ν : T^{n}(V) ⟶ S^{n}(V) is the restriction of ν : T (V) ⟶ S(V), where S(V) is the symmetric algebra.

(4) A tensor a ∈ T^{n}(V) is called an antisymmetric tensor if σ(a) = ε(σ)a for all σ ∈ S_{n} where ε(σ) is the sign of the permutation σ. Let Ê^{n}(V) be the subspace of antisymmetric tensors in T^{n}(V).

a) Show that E : T^{n}(V) ∋ a ↦ ∑_{σ∈S_{n}} ε(σ)σ(a) ∈ T^{n}(V) is a 𝕂-module homomorphism (antisymmetrization). b) Show that E has its image in Ê^{n}(V).

c) Show that Im(E) = Ê^{n}(V) if n! is invertible in 𝕂.

d) Show that Ê^{n}(V) ↪ T^{n}(V) →^{ν} E^{n}(V) is an isomorphism if n! is invertible in 𝕂 and ν : T^{n}(V) ⟶ E^{n}(V) is the restriction of ν : T (V) ⟶ E(V), where E(V) is the exterior algebra.

2.5. Left A-modules.

Definition 2.17. Let A be a 𝕂-algebra. A left A-module is a 𝕂-module M together with a homomorphism μ_{M} : A ⊗ M ⟶ M, such that the diagrams

[figure: A ⊗ M; M; μ; A ⊗ A ⊗ M; A ⊗ M; id ⊗μ; ∇⊗id; μ]

and

[figure: M ≅ 𝕂 ⊗ M; A ⊗ M; η⊗id; M; μ; id]

commute. Let _{A}M and _{A}N be left A-modules and let f : M ⟶ N be a 𝕂-linear map. The map f is called a homomorphism of left A-modules if the diagram

[figure: A ⊗ N; N; μ_{N}; A ⊗ M; M; μ_{M}; 1⊗f; f]

commutes.

Problem 2.6. Show that an Abelian group M is a left module over the ring A if and only if M is a 𝕂-module and a left A-module in the sense of Definition 2.17.

2.6. Coalgebras.

Definition 2.18. A 𝕂-coalgebra is a 𝕂-module C together with a comultiplication or diagonal Δ : C ⟶ C ⊗ C (𝕂-module homomorphism) that is coassociative: and a counit or augmentation ϵ : C ⟶ 𝕂 (𝕂-module homomorphism):

[figure: C ⊗ C; C ⊗ C ⊗ C; id ⊗Δ; C; C ⊗ C; Δ; Δ; Δ⊗id]

[figure: C; C ⊗ C; Δ; Δ; id ⊗ϵ; C ⊗ C; 𝕂 ⊗ C ≅ C ≅ C ⊗ 𝕂.; ϵ⊗id; id]

A 𝕂-coalgebra C is cocommutative if the following diagram commutes

[figure: C ⊗ C; C ⊗ C; τ; C; Δ; Δ]

Let C and D be 𝕂-coalgebras. A homomorphism of coalgebras f : C ⟶ D is a 𝕂-module homomorphism such that the following diagrams commute:

[figure: C ⊗ C; D ⊗ D; f⊗f; C; D; f; Δ_{C}; Δ_{D}]

and

[figure: 𝕂; ϵ_{C}; ϵ_{D}; C; D.; f]

Remark 2.19. Obviously the composition of two homomorphisms of coalgebras is again a homomorphism of coalgebras. Furthermore the identity map is a homomorphism of coalgebras.

Problem 2.7. (1) Show that V ⊗ V^{∗} is a coalgebra for every finite dimensional vector space V over a field 𝕂 if the comultiplication is defined by Δ(v ⊗v^{∗}) := ∑_{i=1}^{n} v ⊗v_{i}^{∗} ⊗v_{i} ⊗v^{∗} where {v_{i}} and {v_{i}^{∗}} are dual bases of V resp. V^{∗}.

(2) Show that the free 𝕂-modules 𝕂X with the basis X and the comultiplication Δ(x) = x⊗x is a coalgebra. What is the counit? Is the counit unique?

(3) Show that 𝕂 ⊕ V with Δ(1) = 1 ⊗ 1, Δ(v) = v ⊗ 1 + 1 ⊗ v defines a coalgebra.

(4) Let C and D be coalgebras. Then C ⊗ D is a coalgebra with the comultiplication Δ_{C⊗D} := (1_{C} ⊗ τ ⊗ 1_{D})(Δ_{C} ⊗ Δ_{D}) : C ⊗ D ⟶ C ⊗ D ⊗ C ⊗ D and counit ε = ε_{C⊗D} : C ⊗ D ⟶ 𝕂 ⊗ K ⟶ 𝕂. (The proof is analogous to the proof of Lemma 2.3.)

To describe the comultiplication of a 𝕂-coalgebra in terms of elements we introduce a notation first introduced by Sweedler similar to the notation ∇(a ⊗ b) = ab used for algebras. Instead of Δ(c) = ∑ c_{i} ⊗ c′_{i} we write

Δ(c) = ∑ c_{(1)} ⊗ c_{(2)}.

Observe that only the complete expression on the right hand side makes sense, not the components c_{(1)} or c_{(2)} which are not considered as families of elements of C. This notation alone does not help much in the calculations we have to perform later on. So we introduce a more general notation.

Definition 2.20. (Sweedler Notation) Let M be an arbitrary 𝕂-module and C be a 𝕂coalgebra. Then there is a bijection between all multilinear maps

f : C × … × C ⟶ M

and all linear maps

f′ : C ⊗ … ⊗ C ⟶ M.

These maps are associated to each other by the formula

f(c_{1}, …, c_{n}) = f′(c_{1} ⊗ … ⊗ c_{n}).

For c ∈ C we define

∑ f(c_{(1)}, …, c_{(n)}) := f′(Δ^{n−1}(c)),

where Δ^{n−1} denotes the n − 1-fold application of Δ, for example Δ^{n−1} = (Δ ⊗ 1 ⊗ … ⊗ 1) ∘ … ∘ (Δ ⊗ 1) ∘ Δ. In particular we obtain for the bilinear map ⊗ : C × C ∋ (c, d) ↦ c ⊗ d ∈ C ⊗ C (with associated identity map)

∑ c_{(1)} ⊗ c_{(2)} = Δ(c),

and for the multilinear map ⊗^{2} : C × C × C ⟶ C ⊗ C ⊗ C

∑ c_{(1)} ⊗ c_{(2)} ⊗ c_{(3)} = (Δ ⊗ 1)Δ(c) = (1 ⊗ Δ)Δ(c).

With this notation one verifies easily

∑ c_{(1)} ⊗ … ⊗ Δ(c_{(i)}) ⊗ … ⊗ c_{(n)} = ∑ c_{(1)} ⊗ … ⊗ c_{(n+1)}

and

\begin{aligned}∑ c_{(1)} ⊗ … ⊗ ϵ(c_{(i)}) ⊗ … ⊗ c_{(n)} & = ∑ c_{(1)} ⊗ … ⊗ 1 ⊗ … ⊗ c_{(n−1)} \\ & = ∑ c_{(1)} ⊗ … ⊗ c_{(n−1)} \end{aligned}

This notation and its application to multilinear maps will also be used in more general contexts like comodules.

Proposition 2.21. Let C be a coalgebra and A an algebra. Then the composition f ∗ g := ∇_{A}(f ⊗ g)Δ_{C} defines a multiplication

Hom(C, A) ⊗ Hom(C, A) ∋ f ⊗ g ↦ f ∗ g ∈ Hom(C, A),

such that Hom(C, A) becomes an algebra. The unit element is given by 𝕂 ∋ α ↦ (c ↦ η(αϵ(c))) ∈ Hom(C, A).

Proof. The multiplication of Hom(C, A) obviously is a bilinear map. The multiplication is associative since (f ∗ g) ∗ h = ∇_{A}((∇_{A}(f ⊗ g)Δ_{C}) ⊗ h)Δ_{C} = ∇_{A}(∇_{A} ⊗ 1)((f ⊗ g) ⊗ h)(Δ_{C} ⊗ 1)Δ_{C} = ∇_{A}(1 ⊗ ∇_{A})(f ⊗ (g ⊗ h))(1 ⊗ Δ_{C})Δ_{C} = ∇_{A}(f ⊗ (∇_{A}(g ⊗ h)Δ_{C}))Δ_{C} = f ∗ (g ∗ h). Furthermore it is unitary with unit 1_{Hom(C,A)} = η_{A}ϵ_{C} since η_{A}ϵ_{C} ∗ f = ∇_{A}(η_{A}ϵ_{C} ⊗ f)Δ_{C} = ∇_{A}(η_{A} ⊗ 1_{A})(1_{𝕂} ⊗ f)(ϵ_{C} ⊗ 1_{C})Δ_{C} = f and similarly f ∗ η_{A}ϵ_{C} = f. □

Definition 2.22. The multiplication ∗ : Hom(C, A) ⊗ Hom(C, A) ⟶ Hom(C, A) is called convolution.

Corollary 2.23. Let C be a 𝕂-coalgebra. Then C^{∗} = Hom_{𝕂}(C, 𝕂) is an 𝕂-algebra.

Proof. Use that 𝕂 itself is a 𝕂-algebra. □

Remark 2.24. If we write the evaluation as C^{∗} ⊗ C ∋ a ⊗ c ↦ ⟨a, c⟩ ∈ 𝕂 then an element a ∈ C^{∗} is completely determined by the values of ⟨a, c⟩ for all c ∈ C. So the product of a and b in C^{∗} is uniquely determined by the formula

⟨a ∗ b, c⟩ = ⟨a ⊗ b, Δ(c)⟩ = ∑ a(c_{(1)})b(c_{(2)}).

The unit element of C^{∗} is ϵ ∈ C^{∗}.

Lemma 2.25. Let 𝕂 be a field and A be a finite dimensional 𝕂-algebra. Then A^{∗} = Hom_{𝕂}(A, 𝕂) is a 𝕂-coalgebra.

Proof. Define the comultiplication on A^{∗} by

Δ : A^{∗} ⟶^{∇^{∗}} (A ⊗ A)^{∗} ⟶^{can^{−1}} A^{∗} ⊗ A^{∗}.

The canonical map can : A^{∗} ⊗ A^{∗} ⟶ (A ⊗ A)^{∗} is invertible, since A is finite dimensional. By a diagrammatic proof or by calculation with elements it is easy to show that A^{∗} becomes a 𝕂-coalgebra. □

Remark 2.26. If 𝕂 is an arbitrary commutative ring and A is a 𝕂-algebra, then A^{∗} = Hom_{𝕂}(A, 𝕂) is a 𝕂-coalgebra if A is a finitely generated projective 𝕂-module.

Problem 2.8. Find sufficient conditions for an algebra A resp. a coalgebra C such that Hom(A, C) becomes a coalgebra with co-convolution as comultiplication.

2.7. Comodules.

Definition 2.27. Let C be a 𝕂-coalgebra. A left C-comodule is a 𝕂-module M together with a 𝕂-module homomorphism δ_{M} : M ⟶ C ⊗ M, such that the diagrams

[figure: C ⊗ M; C ⊗ C ⊗ M; id ⊗δ; M; C ⊗ M; δ; δ; Δ⊗id]

and

[figure: M; δ; C ⊗ M; 𝕂 ⊗ M ≅ M.; ϵ⊗id; id]

commute. Let ^{C}M and ^{C}N be C-comodules and let f : M ⟶ N be a 𝕂-module homomorphism. The map f is called a homomorphism of comodules if the diagram

[figure: N; C ⊗ N; δ_{N}; M; C ⊗ M; δ_{M}; f; 1⊗f]

commutes. Let N be an arbitrary 𝕂-module and M be a C-comodule. Then there is a bijection between all multilinear maps

f : C × … × C × M ⟶ N

and all linear maps

f′ : C ⊗ … ⊗ C ⊗ M ⟶ N.

These maps are associated to each other by the formula

f(c_{1}, …, c_{n}, m) = f′(c_{1} ⊗ … ⊗ c_{n} ⊗ m).

For m ∈ M we define

∑ f(m_{(1)}, …, m_{(n)}, m_{(M)}) := f′(δ^{n}(m)),

where δ^{n} denotes the n-fold application of δ, i.e. δ^{n} = (1 ⊗ … ⊗ 1 ⊗ δ) ∘ … ∘ (1 ⊗ δ) ∘ δ. In particular we obtain for the bilinear map ⊗ : C × M ⟶ C ⊗ M

∑ m_{(1)} ⊗ m_{(M)} = δ(m),

and for the multilinear map ⊗^{2} : C × C × M ⟶ C ⊗ C ⊗ M

∑ m_{(1)} ⊗ m_{(2)} ⊗ m_{(M)} = (1 ⊗ δ)δ(c) = (Δ ⊗ 1)δ(m).

Problem 2.9. Show that a finite dimensional vector space V is a comodule over the coalgebra V ⊗V^{∗} as defined in exercise 2.7 (1) with the coaction δ(v) := ∑ v⊗v_{i}^{∗}⊗v_{i} ∈ (V ⊗V^{∗})⊗V where ∑ v_{i}^{∗} ⊗ v_{i} is the dual basis of V in V^{∗} ⊗ V.

Theorem 2.28. (Fundamental Theorem for Comodules) Let 𝕂 be a field. Let M be a left C-comodule and let m ∈ M be given. Then there exists a finite dimensional subcoalgebra C′ ⊆ C and a finite dimensional C′-comodule M′ with m ∈ M′ ⊆ M where M′ ⊆ M is a 𝕂-submodule, such that the diagram

[figure: M; C ⊗ M; δ; M′; C′ ⊗ M′; δ′]

commutes.

Corollary 2.29. (1) Each element c ∈ C of a coalgebra is contained in a finite dimensional subcoalgebra of C.

(2) Each element m ∈ M of a comodule is contained in a finite dimensional subcomodule of M.

Corollary 2.30. (1) Each finite dimensional subspace V of a coalgebra C is contained in a finite dimensional subcoalgebra C′ of C.

(2) Each finite dimensional subspace V of a comodule M is contained in a finite dimensional subcomodule M′ of M.

Corollary 2.31. (1) Each coalgebra is a union of finite dimensional subcoalgebras.

(2) Each comodule is a union of finite dimensional subcomodules.

Proof. (of the Theorem) We can assume that m ≠ 0 for else we can use M′ = 0 and C′ = 0. Under the representations of δ(m) ∈ C ⊗ M as finite sums of decomposable tensors pick one

δ(m) = ∑_{i=1}^{s} c_{i} ⊗ m_{i}

of shortest length s. Then the families (c_{i}|i = 1, …, s) and (m_{i}|i = 1, …, s) are linearly independent. Choose coefficients c_{ij} ∈ C such that

Δ(c_{j}) = ∑_{i=1}^{t} c_{i} ⊗ c_{ij}, ∀j = 1, …, s,

by suitably extending the linearly independent family (c_{i}|i = 1, …, s) to a linearly independent family (c_{i}|i = 1, …, t) and t ≥ s. We first show that we can choose t = s. By coassociativity we have ∑_{i=1}^{s} c_{i} ⊗ δ(m_{i}) = ∑_{j=1}^{s} Δ(c_{j})⊗m_{j} = ∑_{j=1}^{s} ∑_{i=1}^{t} c_{i} ⊗c_{ij} ⊗m_{j}. Since the c_{i} and the m_{j} are linearly independent we can compare coefficients and get

(3)  δ(m_{i}) = ∑_{j=1}^{s} c_{ij} ⊗ m_{j}, ∀i = 1, …, s

and 0 = ∑_{j=1}^{s} c_{ij} ⊗ m_{j} for i > s. The last statement implies

c_{ij} = 0, ∀i > s, j = 1, …, s.

Hence we get t = s and

Δ(c_{j}) = ∑_{i=1}^{s} c_{i} ⊗ c_{ij}, ∀j = 1, …, s.

Define finite dimensional subspaces C′ = ⟨c_{ij}|i, j = 1, …, s⟩ ⊆ C and M′ = ⟨m_{i}|i = 1, …, s⟩ ⊆ M. Then by (3) we get δ : M′ ⟶ C′ ⊗ M′. We show that m ∈ M′ and that the restriction of Δ to C′ gives a 𝕂-module homomorphism Δ : C′ ⟶ C′ ⊗ C′ so that the required properties of the theorem are satisfied. First observe that m = ∑ ε(c_{i})m_{i} ∈ M′ and c_{j} = ∑ ε(c_{i})c_{ij} ∈ C′. Using coassociativity we get

\begin{aligned}∑_{i,j=1}^{s} c_{i} ⊗ Δ(c_{ij}) ⊗ m_{j} & = ∑_{k,j=1}^{s} Δ(c_{k}) ⊗ c_{kj} ⊗ m_{j} \\ & = ∑_{i,j,k=1}^{s} c_{i} ⊗ c_{ik} ⊗ c_{kj} ⊗ m_{j} \end{aligned}

hence

Δ(c_{ij}) = ∑_{k=1}^{s} c_{ik} ⊗ c_{kj}.

□

Remark 2.32. We give a sketch of a second proof of Theorem 2.28 which is somewhat more technical. Since C is a 𝕂-coalgebra, the dual C^{∗} is an algebra. The comodule structure δ : M ⟶ C ⊗ M leads to a module structure by ρ = (ev ⊗1)(1 ⊗ δ) : C^{∗} ⊗ M ⟶ C^{∗} ⊗ C ⊗ M ⟶ M. Consider the submodule N := C^{∗}m. Then N is finite dimensional, since c^{∗}m = ∑_{i=1}^{n}⟨c^{∗}, c_{i}⟩m_{i} for all c^{∗} ∈ C^{∗} where ∑_{i=1}^{n} c_{i}⊗m_{i} = δ(m). Observe that C^{∗}m is a subspace of the space generated by the m_{i}. But it does not depend on the choice of the m_{i}. Furthermore if we take δ(m) = ∑ c_{i} ⊗ m_{i} with a shortest representation then the m_{i} are in C^{∗}m since c^{∗}m = ∑⟨c^{∗}, c_{i}⟩m_{i} = m_{i} for c^{∗} an element of a dual basis of the c_{i}. N is a C-comodule since δ(c^{∗}m) = ∑⟨c^{∗}, c_{i}⟩δ(m_{i}) = ∑⟨c^{∗}, c_{i(1)}⟩c_{i(2)} ⊗ m_{i} ∈ C ⊗ C^{∗}m. Now we construct a subcoalgebra D of C such that N is a D-comodule with the induced coaction. Let D := N ⊗ N^{∗}. By 2.9 N is a comodule over the coalgebra N ⊗ N^{∗}. Construct a 𝕂-module homomorphism ϕ : D ⟶ C by n ⊗ n^{∗} ↦ ∑ n_{(1)}⟨n^{∗}, n_{(N)}⟩. By definition of the dual basis we have n = ∑ n_{i}⟨n_{i}^{∗}, n⟩. Thus we get

(ϕ ⊗ ϕ)Δ_{D}(n ⊗ n^{∗}) = (ϕ ⊗ ϕ)(∑ n ⊗ n_{i}^{∗} ⊗ n_{i} ⊗ n^{∗})
= ∑ n_{(1)}⟨n_{i}^{∗}, n_{(N)}⟩ ⊗ n_{i(1)}⟨n^{∗}, n_{i(N)}⟩
= ∑ n_{(1)} ⊗ n_{i(1)}⟨n^{∗}, n_{i(N)}⟩⟨n_{i}^{∗}, n_{(N)}⟩
= ∑ n_{(1)} ⊗ n_{(2)}⟨n^{∗}, n_{(N)}⟩ = ∑ Δ_{C}(n_{(1)})⟨n^{∗}, n_{(N)}⟩
= Δ_{C}ϕ(n ⊗ n^{∗}).

Furthermore ε_{C}ϕ(n ⊗ n^{∗}) = ε(∑ n_{(1)}⟨n^{∗}, n_{(N)}⟩) = ⟨n^{∗}, ∑ ε(n_{(1)})n_{(N)}⟩ = ⟨n^{∗}, n⟩ = ε(n ⊗ n^{∗}). Hence ϕ : D ⟶ C is a homomorphism of coalgebras, D is finite dimensional and the image C′ := ϕ(D) is a finite dimensional subcoalgebra of C. Clearly N is also a C′-comodule, since it is a D-comodule. Finally we show that the D-comodule structure on N if lifted to the C-comodule structure coincides with the one defined on M. We have

δ_{C}(c^{∗}m) = δ_{C}(∑⟨c^{∗}, m_{(1)}⟩m_{(M)}) = ∑⟨c^{∗}, m_{(1)}⟩m_{(2)} ⊗ m_{(M)}
= ∑⟨c^{∗}, m_{(1)}⟩m_{(2)} ⊗ m_{i}⟨m_{i}^{∗}, m_{(M)}⟩ = ∑⟨c^{∗}, m_{(1)}⟩m_{(2)}⟨m_{i}^{∗}, m_{(M)}⟩ ⊗ m_{i}
= (ϕ ⊗ 1)(∑⟨c^{∗}, m_{(1)}⟩m_{(M)} ⊗ m_{i}^{∗} ⊗ m_{i}) = (ϕ ⊗ 1)(∑ c^{∗}m ⊗ m_{i}^{∗} ⊗ m_{i})
= (ϕ ⊗ 1)δ_{D}(c^{∗}m).

3. Projective Modules and Generators

3.1. Products and coproducts.

Definition 3.1.

(1) Let (M_{i}|i ∈ I) be a family of R-modules. An R-module ∏ M_{i} together with a family of homomorphisms (p_{j} : ∏ M_{i} ⟶ M_{j}|j ∈ I) is called a (direct) product of the M_{i} and the homomorphisms p_{j} : ∏ M_{i} ⟶ M_{j} are called projections, if for each R-module N and for each family of homomorphisms (f_{j} : N ⟶ M_{j}|j ∈ I) there is a unique homomorphism f : N ⟶ ∏ M_{i} such that

[figure: ∏ M_{i}; M_{j}; p_{j}; f_{j}; N; f]

commute for all j ∈ I.

(2) “The dual notion is called coproduct”: Let (M_{i}|i ∈ I) be a family of R-modules. An R-module ∐ M_{i} together with a family of homomorphisms (ι_{j} : M_{j} ⟶ ∐ M_{i}|j ∈ I) is called a coproduct or direct sum of the M_{i} and the homomorphisms ι_{j} : M_{j} ⟶ ∐ M_{i} are called injections, if for each R-module N and for each family of homomorphisms (f_{j} : M_{j} ⟶ N|j ∈ I) there is a unique homomorphism f : ∐ M_{j} ⟶ N such that

[figure: M_{j}; ∐ M_{i}; ι_{j}; f_{j}; N; f]

commute for all j ∈ I.

Remark 3.2. An analogous definition can be given for algebras, coalgebras, comodules, groups, Abelian groups etc.

Note: If you want to define a homomorphism f : N ⟶ ∏ M_{i} with a product as range (codomain) you should define it by giving homomorphisms f_{i} : N ⟶ M_{i}. If you want to define a homomorphism f : ∐ M_{i} ⟶ N with a coproduct as domain you should define it by giving homomorphisms f_{i} : M_{i} ⟶ N.

Lemma 3.3. Products and coproducts are unique up to a unique isomorphism.

Proof. analogous to Proposition1.6. □

Proposition 3.4. (Rules of computation in a product of R-modules) Let (∏ M_{i}, (p_{j})) be a product of the family of R-modules (M_{i})_{i∈I}.

(1) There is a bijection of sets

∏ M_{i} ∋ a ↦ (a_{i}) := (p_{i}(a)) ∈ {(a_{i})|∀i ∈ I : a_{i} ∈ M_{i}}

such that a + b ↦ (a_{i} + b_{i}) and ra ↦ (ra_{i}).

(2) If (f_{i} : N ⟶ M_{i}) is a family of homomorphisms and f : N ⟶ ∏ M_{i} is the induced homomorphism then the family associated to f(n) ∈ ∏ M_{i} is (f_{i}(n)), i.e. (p_{i}(f(n))) = (f_{i}(n)).

Proof. Let a family (a_{i}|i ∈ I) be given. Form φ_{i} : {1} ⟶ M_{i} with φ_{i}(1) = a_{i} for all i ∈ I. Construct g_{i} ∈ Hom_{R}(R, M_{i}) such that the diagrams

[figure: {1}; R; φ_{i}; M_{i}; g_{i}]

commute (R is the free R-module over the set {1}). Then there is a unique g : R ⟶ ∏ M_{i} with

[figure: ∏ M_{i}; M_{j}; p_{j}; g_{j}; R; g]

for all j ∈ I. The homomorphism g is completely and uniquely determined by g(1) =: a and by the commutative diagram

[figure: {1}; R; ∏ M_{i}; M_{j}; φ_{j}; g; p_{j}; g_{j}]

where p_{j}(a) = φ_{j}(1) = a_{j}. So we have found a ∈ ∏ M_{i} with (p_{i}(a)) = (a_{i}). Hence the map given in the proposition is surjective. Given a and b in ∏ M_{i} with (p_{i}(a)) = (p_{i}(b)) then φ_{j}(1) := p_{j}(a) and ψ_{j}(1) := p_{j}(b) define equal maps φ_{j} = ψ_{j}, hence the induced maps g_{j} : R ⟶ M_{j} and h_{j} : R ⟶ M_{j} are equal so that g = h and hence a = g(1) = h(1) = b. Hence the map given in the proposition is bijective. Since a is uniquely determined by the p_{j}(a) = a_{j} we have p_{j}(a + b) = p_{j}(a) + p_{j}(b) = a_{j} + b_{j} and p_{j}(ra) = rp_{j}(a) = ra_{j} The last statement is p_{i}f = f_{i}. □

Remark 3.5. Observe that this construction can always be performed if there is a free object (algebra, coalgebra, comodule, group, Abelian group, etc.) R over the set {1} i.e. if

[figure: {1}; R; T]

has a universal solution.

Proposition 3.6. (Rules of computation in a coproduct of R-modules) Let (∐ M_{i}, (ι_{j})) be a coproduct of the family of R-modules (M_{i})_{i∈I}.

(1) The homomorphisms ι_{j} : M_{j} ⟶ ∐ M_{i} are injective.

(2) For each element a ∈ ∐ M_{i} there are finitely many a_{i} ∈ M_{i} with a = ∑_{i=1}^{n} ι_{i}(a_{i}). The a_{i} ∈ M_{i} are uniquely determined by a.

Proof. (1) To show the injectivity of ι_{i} define f_{i} : M_{i} ⟶ M_{j} by

f_{i} := \{ \begin{matrix}id, & i = j, \\ 0, & else. \end{matrix}

Then the diagram

[figure: M_{i}; ∐ M_{j}; ι_{i}; f_{i}; M_{j}; f]

defines a uniquely determined homomorphism f. For i = j this implies fι_{i} = id_{M_{i}}, hence ι_{i} is injective.

(2) Define M̃:= ∑ ι_{j}(M_{j}) ⊆ ∐ M_{j}. Then the following diagram commutes with both 0 and ν

[figure: M_{i}; ∐ M_{j}; ι_{i}; ∐ M_{j}/M̃; 0; 0; ν]

Hence ν = 0 and ∐ M_{j} = M̃. Let a = ∑ ι_{j}(a_{j}). Define f as in (1). Then we have f(a) = f(∑ ι_{j}(a_{j})) = ∑ fι_{j}(a_{j}) = ∑ f_{j}(a_{j}) = a_{i}, hence the a_{i} are uniquely determined by a. □

Propositions 3.4 and 3.6 give already an indication of how to construct products and coproducts.

Proposition 3.7. Let (M_{i}|i ∈ I) be a family of R-modules. Then there exist a product (∏ M_{i}, (p_{j} : ∏ M_{i} ⟶ M_{j}|j ∈ I)) and a coproduct (∐ M_{i}, (ι_{j} : M_{j} ⟶ ∏ M_{i}|j ∈ I)).

Proof. 1. Define

∏ M_{i} := {a : I ⟶ ∪_{i∈I}M_{i}| ∀j ∈ I : a(j) = a_{j} ∈ M_{j}}

and p_{j} : ∏ M_{i} ⟶ M_{j}, p_{j}(a) := a(j) = a_{j} ∈ M_{j}. It is easy to check that ∏ M_{i} is an R-module with componentwise operations and that the p_{j} are homomorphisms. If (f_{j} : N ⟶ M_{j}) is a family of homomorphisms then there is a unique map f : N ⟶ ∏ M_{i}, f(n) = (f_{i}(n)|i ∈ I) such that p_{j}f = f_{j} for all j ∈ I. The following families are equal: (p_{j}f(n + n′)) = (f_{j}(n + n′)) = (f_{j}(n) + f_{j}(n′)) = (p_{j}f(n) + p_{j}f(n′)) = (p_{j}(f(n) + f(n′))), hence f(n + n′) = f(n) + f(n′). Analogously one shows f(rn) = rf(n). Thus f is a homomorphism and ∏ M_{i} is a product.

2. Define

∐ M_{i} := {a : I ⟶ ∪_{i∈I}M_{i}|∀j ∈ I : a(j) ∈ M_{j}; a with finite support}

(the notion with finite support means that all but a finite number of the a(j)’s are zero) and ι_{j} : M_{j} ⟶ ∐ M_{i}, ι_{j}(a_{j})(i) := δ_{ij}a_{i}. Then ∐ M_{i} ⊆ ∏ M_{i} is a submodule and the ι_{j} are homomorphisms. Given (f_{j} : M_{j} ⟶ N|j ∈ I). Define f(a) = f(∑ ι_{i}a_{i}) = ∑ fι_{i}(a_{i}) = ∑ f_{i}(a_{i}). Then f is an R-module homomorphism and we have fι_{i}(a_{i}) = f_{i}(a_{i}) hence fι_{i} = f_{i}. If gι_{i} = f_{i} for all i ∈ I then g(a) = g(∑ ι_{i}a_{i}) = ∑ gι_{i}a_{i} = ∑ f_{i}(a_{i}) hence f = g. □

Proposition 3.8. Let (M_{i}|i ∈ I) be a family of submodules of M. The following statements are equivalent:

(1) (M, (ι_{i} : M_{i} ⟶ M)) is a coproduct of R-modules.

(2) M = ∑_{i∈I} M_{i} and (∑ m_{i} = 0 ⟹ ∀i ∈ I : m_{i} = 0).

(3) M = ∑_{i∈I} M_{i} and (∑ m_{i} = ∑ m′_{i} ⟹ ∀i ∈ I : m_{i} = m′_{i}).

(4) M = ∑_{i∈I} M_{i} and ∀i ∈ I : M_{i} ∩ ∑_{j≠i,j∈I} M_{j} = 0.

Definition 3.9. Is one of the equivalent conditions of Proposition 3.8 is satisfied then M is called an internal direct sum of the M_{i} and we write M = ⊕_{i∈I}M_{i}.

Proof of Proposition 3.8: (1) ⟹ (2): Use the commutative diagram

[figure: M_{j}; M; ι_{j}; M/ ∑ M_{i}; 0; 0; ν]

to conclude ν = 0 and M = ∑ M_{i}. If ∑ m_{i} = 0 then use the diagram

[figure: M_{j}; M; ι_{j}; δ_{jk}; M_{k}; p_{k}]

to show 0 = p_{k}(0) = p_{k}(∑ m_{j}) = ∑_{j} p_{k}ι_{j}(m_{j}) = ∑_{j} δ_{jk}(m_{j}) = m_{k}.

(2) ⟹ (3): trivial.

(3) ⟹ (4): Let m_{i} = ∑_{j≠i} m_{j}. Then m_{i} = 0 and m_{j} = 0 for all j ≠ i.

(4) ⟹ (2): If ∑ m_{j} = 0 then m_{i} = ∑_{j≠i} −m_{j} = 0 ∈ M_{i} ∩ ∑_{j≠i} M_{j}.

(3) ⟹ (1): Define f for the diagram

[figure: M_{i}; M; ι_{i}; f_{i}; N; f]

by f(∑ m_{i}) := ∑ f_{i}(m_{i}). Then f is a well defined homomorphism and we have fι_{j}(m_{j}) = f(m_{j}) = f_{j}(m_{j}). Furthermore f is uniquely determined since gι_{j} = f_{j} ⟹ g(∑ m_{i}) = ∑ g(m_{i}) = ∑ gι_{i}(m_{i}) = ∑ f_{i}(m_{i}) = f(∑ m_{i}) ⟹ f = g. □

Proposition 3.10. Let (∐ M_{i}, (ι_{j} : M_{j} ⟶ ∐_{i≠j} M_{i})) be a coproduct of R-modules. Then ∐ M_{i} is an internal direct sum of the ι_{j}(M_{j}).

Proof. ι_{j} is injective ⟹ M_{j} ≅ ι_{j}(M_{j}) ⟹

[figure: M_{j} ≅ ι_{j}(M_{j}); ∐ M_{i}; N]

defines a coproduct. By 3.8 we have an internal direct sum. □

Definition 3.11. A submodule M ⊆ N is called a direct summand of N if there is a submodule M′ ⊆ N such that N = M ⊕ M′ is an internal direct sum.

Proposition 3.12. For a submodule M ⊆ N the following are equivalent:

(1) M is a direct summand of N.

(2) There is p ∈ Hom_{R}(.N, .M) with

[figure: (M; N; ι; M) = id_{M} .; p]

(3) There is f ∈ Hom_{R}(.N, .N) with f^{2} = f and f(N) = M.

Proof. (1) ⟹ (2): Let M_{1} := M and M_{2} ⊆ N with N = M_{1} ⊕ M_{2}. We define p = p_{1} : N ⟶ M_{1} by

[figure: M_{i}; N; ι_{i}; δ_{ij}; M_{j}; p_{j}]

where δ_{ij} = 0 for i ≠ j and δ_{ij} = id_{M_{i}} for i = j. Then p_{1}ι_{1} = δ_{11} = id_{M}.

(2) ⟹ (3): For f := ιp : N ⟶ N we have f^{2} = ιpιp = ιp = f since pι = id. Furthermore f(N) = ιp(N) = M since p is surjective.

(3) ⟹ (1): Let M′ = Ke(f). We first show N = M + M′. Take n ∈ N. Then we have n = f(n) + (n − f(n)) with f(n) ∈ M. Since f(n − f(n)) = f(n) − f^{2}(n) = 0 we get n − f(n) ∈ Ke(f) = M′ so that N = M + M′. Now let n ∈ M ∩ M′. Then f(n) = 0 and n = f(n′) for n′ ∈ N hence n = f(n′) = f^{2}(n′) = f(n) = 0. □

Problem 3.1. Discuss the definition and the properties of products of groups.

Problem 3.2. Show that the tensor product of two commutative 𝕂-algebras is a coproduct.

Problem 3.3. Show that the disjoint union of two sets is a coproduct.

3.2. Projective modules.

Definition 3.13. An R-module P is called projective if for each epimorphism f : M ⟶ N and for each homomorphism g : P ⟶ N there exists a homomorphism h : P ⟶ M such that the diagram

[figure: M; N; f; h; P; g]

commutes.

Example 3.14. All vector spaces are projective. ℤ/nℤ (n > 1) is not a projective ℤ-module.

Lemma 3.15. Let P = ⊕_{i∈I}P_{i}. P is projective iff all P_{i}, i ∈ I are projective.

Proof. Let P be projective. We show that P_{i} is projective. Let f : M ⟶ N be an epimorphism and g : P_{i} ⟶ N be a homomorphism. Consider the diagram

[figure: P_{i}; P; ι_{i}; P_{i}; p_{i}; M; N; f; hι_{i}; h; g]

where p_{i} and ι_{i} are projections and injections of the direct sum, in particular p_{i}ι_{i} = id_{P_{i}}. Since f is an epimorphism there is an h : P ⟶ M with fh = gp_{i} hence g = gp_{i}ι_{i} = fhι_{i}. Thus P_{i} is projective. Assume that all P_{i} are projective. Let f : M ⟶ N be an epimorphism and g : P ⟶ N be a homomorphism. Consider the diagram

[figure: P_{i}; P; ι_{i}; M; N; f; h_{i}; g; h]

Since f is surjective there are h_{i} : P_{i} ⟶ M, i ∈ I with fh_{i} = gι_{i}. Since P is the coproduct of the P_{i} there is a (unique) h : P ⟶ M with hι_{i} = h_{i} for all i ∈ I. Thus fhι_{i} = fh_{i} = gι_{i} for all i ∈ I hence fh = g. So P is projective. □

Proposition 3.16. Let P be an R-module. Then the following are equivalent

(1) P is projective.

(2) Each epimorphism f : M ⟶ P splits, i.e. for each R-module M and each epimorphism f : M ⟶ P there is a homomorphism g : P ⟶ M such that fg = id_{P}.

(3) P is isomorphic to a direct summand of a free R-module RX.

Proof. (1) ⟹ (2): The diagram

[figure: M; P; f; g; P; id_{P}]

implies the existence of g with fg = id_{P}.

(2) ⟹ (3): Let ι : P ⟶ RP be the free module over (the set) P with ι a map. Then there is a homomorphism f : RP ⟶ P such that

[figure: P; RP; ι; id_{P}; P; f]

commutes. Obviously f is surjective. By (2) there is a homomorphism g : P ⟶ RP with fg = id_{P}. By 3.12 P is a direct summand of RP (up to an isomorphism).

(3) ⟹ (1): Let f : M ⟶ N be surjective. Let ι : X ⟶ RX be a free module and let g : RX ⟶ N be a homomorphism. In the following diagram let k = gι : X ⟶ N. Since f is surjective there is a map h : X ⟶ M with fh = k. Hence there is a homomorphism l : RX ⟶ M with lι = h. This implies flι = fh = k = gι and thus fl = g since RX is free. So RX is projective. The transition to a direct summand follows from 3.15.

[figure: X; RX; ι; M; N; f; h; g; k; l]
□

3.3. Dual basis.

Remark 3.17. Let P_{R} be a right R-module. Then E := End_{R}(P.) = Hom_{R}(P., P.) is a ring and P is an E-R-bimodule because of f(pr) = (fp)r. Let P^{∗} := Hom_{R}(P., R.) be the dual of P. Then P^{∗} = _{R} Hom_{R}(_{E}P.,_{R}R.)_{E} is an R-E-bimodule. The following maps are bimodule homomorphisms

ev : _{R}P^{∗} ⊗_{E} P_{R} ∋ f ⊗ p ↦ f(p) ∈ _{R}R_{R},

the evaluation homomorphism, and

db : _{E}P ⊗_{R} P_{E}^{∗} ⟶ _{E}E_{E} = _{E} End_{R}(P.)_{E}

with db(p⊗f)(q) = pf(q), the dual basis homomorphism. We check the bilinearity: ev(fe, p) = (fe)(p) = f(e(p)) = ev(f, ep) and db(pr, f)(q) = (pr)f(q) = p(rf(q)) = db(p, rf)(q). We also check that db is a bimodule homomorphism: db(ep ⊗ f)(q) = e(p)f(q) = e(pf(q)) = e db(p ⊗ f)(q) and db(p ⊗ fe)(q) = pfe(q) = db(p ⊗ f)e(q).

Lemma 3.18. The following diagrams commute

[figure: R ⊗_{R} P^{∗}; P^{∗}; μ; P^{∗} ⊗_{E} P ⊗_{R} P^{∗}; P^{∗} ⊗_{E} E; 1 ⊗ db; ev ⊗1; μ; E ⊗_{E} P; P; μ; P ⊗_{R} P^{∗} ⊗_{E} P; P ⊗_{R} R; 1 ⊗ ev; db ⊗1; μ]

Proof. The proof follows from the associative law: μ(1 ⊗ db)(f ⊗ p ⊗ g)(q) = μ(f ⊗ pg)(q) = f(pg)(q) = f(pg(q)) = f(p)g(q) = μ(f(p)⊗g)(q) = μ(ev ⊗1)(f ⊗p⊗g)(q) and μ(db ⊗1)(p⊗ f ⊗ q) = μ(pf ⊗ q) = pf(q) = μ(p ⊗ f(q)) = μ(1 ⊗ ev)(p ⊗ f ⊗ q). □

Proposition 3.19. (dual basis Lemma) Let P_{R} be a right R-module. Then the following are equivalent:

(1) P is finitely generated and projective,

(2) (dual basis) There are f_{1}, …, f_{n} ∈ Hom_{R}(P., R.) = P^{∗} and p_{1}, …, p_{n} ∈ P so that

p = ∑ p_{i}f_{i}(p)

for all p ∈ P

(3) The dual basis homomorphism

db : P ⊗_{R} P^{∗} ⟶ Hom_{R}(P., P.)

is an isomorphism.

Proof. (1) ⟹ (2): Let P be generated by {p_{1}, …, p_{n}}. Let RX be a free right R-module over the set X = {x_{1}, …, x_{n}}. Let π_{i} : RX ⟶ R be the projections induced by

[figure: X; RX; ι; σ_{i}; R; π_{i}]

where σ_{i}(x_{j}) = δ_{ij}. By Proposition 1.10 we have z = ∑ x_{i}π_{i}(z) for all z ∈ RX. Let g : RX ⟶ P be the R-module homomorphism with g(x_{i}) = p_{i}. Since the p_{i} generate P as a module, the homomorphism g is surjective. P is projective hence there is a homomorphism h : P ⟶ RX with gh = id_{P} by 3.16. Define f_{i} := π_{i}h. Then ∑ p_{i}π_{i}h(p) = ∑ g(x_{i})π_{i}h(p) = g(∑ x_{i}π_{i}(h(p))) = gh(p) = p.

(2) ⟹ (3): The homomorphism ψ : Hom_{R}(P., P.) ⟶ P ⊗_{R}P^{∗} defined by ψ(e) = ∑ e(p_{i})⊗f_{i} is the inverse map of db. In fact we have db ∘ψ(e)(p) = ∑ e(p_{i})f_{i}(p) = e(∑ p_{i}f_{i}(p)) = e(p), hence db ∘ψ = id. Furthermore we have ψ ∘ db(p ⊗ f) = ψ(pf) = ∑ pf(p_{i}) ⊗ f_{i} = p ⊗ ∑ f(p_{i})f_{i} = p ⊗ f since ∑ f(p_{i})f_{i}(q) = f(∑ p_{i}f_{i}(q)) = f(q), hence we have also ψ ∘ db = id.

(3) ⟹ (2): ∑ p_{i} ⊗ f_{i} = db^{−1}(id_{P}) is a dual basis, because ∑ p_{i}f_{i}(p) = db(∑ p_{i} ⊗ f_{i})(p) = id_{P} (p) = p.

(2) ⟹ (1): The p_{i} generate P since ∑ p_{i}f_{i}(p) = p for all p ∈ P. Thus P is finitely generated. Furthermore the homomorphism g : RX ⟶ P with g(x_{i}) = p_{i} is surjective. Let h : P ⟶ RX be defined by h(p) = ∑ x_{i}f_{i}(p). Then gh(p) = p, hence P is a direct summand of RX, and consequently P is projective. □

Remark 3.20. Observe that analogous statements hold for left R- modules. The problem that in that situation two rings R and End_{R}(.P) operate from the left on P is best handled by considering P as a right End_{R}(.P)^{op}- module where End_{R}(.P)^{op} has the opposite multiplication ∗ given by f ∗g := g ∘f. We leave it to the reader to verify the details. The evaluation and dual basis homomorphisms are in this case ev : _{R}P ⊗_{E^{op}} P_{R}^{∗} ∋ p ⊗ f ↦ f(p) ∈ _{R}R_{R}, and db : P^{∗} ⊗_{R} P ⟶ Hom_{R}(.P, .P).

Proposition 3.21. Let R be a commutative ring and P be an R-module. Then the following are equivalent

(1) _{R}P is finitely generated and projective,

(2) there exists an R-module P′, and homomorphisms db′ : R ⟶ P ⊗_{R}P′ and ev : P′⊗_{R}P ⟶ R such that

[figure: (P; P ⊗_{R} P′ ⊗_{R} P; db′ ⊗ id; P) = id_{P} ,; id ⊗ ev; (P′; P′ ⊗_{R} P ⊗_{R} P′; id ⊗ db′; P′) = id_{P′} .; ev ⊗ id]

Proof. “⟸”: ev ∈ Hom_{R}(P′ ⊗_{R} P, R) ≅ Hom_{R}(P′, Hom_{R}(P, R)) induces a homomorphism ϵ : P′ ⟶ P^{∗} by ϵ(f)(p) = ev(f ⊗ p) = fp for f ∈ P′. Let db′(1) = ∑ p_{i} ⊗ f_{i}. Then p = id_{P} (p) = (id ⊗_{R} ev)(db′ ⊗_{R} id)(p) = (id ⊗_{R} ev)(∑ p_{i} ⊗ f_{i} ⊗ p) = ∑ p_{i}f_{i}p. By 3.19 P is finitely generated and projective. “⟹”: Define P′ := P^{∗} and (ev : P′⊗_{R}P ⟶ R) = (ev : P^{∗}⊗_{R}P ⟶ R). Let db′(1) = ∑ p_{i}⊗ f_{i} be the dual basis for P. Then we have (id ⊗_{R} ev)(db′ ⊗_{R} id)(p) = (id ⊗_{R} ev)(∑ p_{i}⊗f_{i}⊗p) = ∑ p_{i}f_{i}(p) = p. Furthermore we have ∑ f(p_{i})f_{i}(p) = f(∑ p_{i}f_{i}(p)) = f(p), hence ∑ f(p_{i})f_{i} = f. This implies (ev ⊗_{R} id)(id ⊗_{R} db′)(f) = (ev ⊗_{R} id)(∑ f ⊗ p_{i} ⊗ f_{i}) = ∑ f(p_{i})f_{i} = f. □

Example 3.22. of a projective module, that is not free:

Let S^{2} = 2-sphere = {(x_{1}, x_{2}, x_{3}) ∈ ℝ^{3}|x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1}. Let R be the ring of all continuous real-valued functions on S^{2}. Let F = {f : S^{2} ⟶ ℝ^{3}|f continuous} = {(f_{1}, f_{2}, f_{3})|f_{i} ∈ R} = R^{3} be the free R-module on three generators. F is a set of vector valued functions, the vectors starting in the point of S^{2} where their counterimage is. These are vector fields over S^{2}. Let P = {tangential vector fields} and Q = {normal vector fields}. Then F = P ⊕ Q as R-modules. So P and Q are projective. Furthermore Q ≅ R. Hence F ≅ P ⊕ R. Suppose P were free. Evaluating all elements of P in a given point p ∈ S^{2} we get the tangent plane at p which is ℝ^{2}. If P is free then it has a basis e_{1}, e_{2} (see later remarks on the rank of free modules over a commutative ring). For p ∈ S^{2} we have e_{1}(p), e_{2}(p) generates the tangent plane, hence is a basis for the tangent plane. So e_{1}(p) ≠ 0 for all p ∈ S^{2}. By the “egg theorem” this is impossible.

3.4. Generators.

Definition 3.23. A right R-module G_{R} is called a generator if for each homomorphism f : M ⟶ N with f ≠ 0 there exists a homomorphism g : G ⟶ M such that fg ≠ 0.

Proposition 3.24. Let G_{R} be an R-module. The following are equivalent (1) G is a generator,

(2) for each R-module M_{R} there is a set I and an epimorphism h : ∐_{I} G ⟶ M,

(3) R is isomorphic to a direct summand of ∐_{I} G (for an appropriate set I),

(4) there are f_{1}, …, f_{n} ∈ G^{∗} = Hom_{R}(G., R.) and q_{1}, …, q_{n} ∈ G with ∑ f_{i}(q_{i}) = 1.

Proof. (1) ⟹ (2): Define I := Hom_{R}(G., M.). Then the diagram

[figure: G = G_{f}; ∐_{I} G_{f}; ι_{f}; f; M; h]

defines a unique homomorphism h with hι_{f} = f for all f ∈ I. Let N = Im(h). Consider ν : M ⟶ M/N. If N ≠ M then ν ≠ 0. Since G is a generator there exists an f such that νf ≠ 0. This implies νh ≠ 0 a contradiction to N = Im(h). Hence N = M so that h is an epimorphism.

(2) ⟹ (3): Let ∐ G ⟶ R be an epimorphism. Since R is a free module hence projective, 3.16 implies that R is a direct summand of ∐ G up to isomorphism.

(3) ⟹ (4): Since R is (isomorphic to) a direct summand of ∐_{I} G there is p : ∐_{I} G ⟶ R with pι = id_{R}. Let p((g_{i})) = 1 and f_{i} = pι_{i} : G ⟶ R. Then 1 = p((g_{i})) = p(∑ ι_{i}(g_{i})) = ∑ pι_{i}(g_{i}) = ∑ f_{i}(q_{i}).

(4) ⟹ (1): Assume (g : M ⟶ N) ≠ 0. Then there is an m ∈ M with g(m) ≠ 0. Define f : R ⟶ M by f(1) = m, f(r) = rm. Let f_{i}, q_{i} be given with ∑ f_{i}(q_{i}) = 1. Then we have 0 ≠ g(m) = gf(1) = ∑ gff_{i}(q_{i}), so we have the existence of a homomorphism ff_{i} : G ⟶ M with gff_{i} ≠ 0. □

4. Categories and Functors

4.1. Categories. In the preceding sections we saw that certain constructions like products can be performed for different kinds of mathematical structures, e.g. modules, rings, Abelian groups, groups, etc. In order to indicate the kind of structure that one uses the notion of a category has been invented.

Definition 4.1. Let C consist of (1) a class Ob C whose elements A, B, C, … ∈ Ob C are called objects,

(2) a family (Mor_{C}(A, B)|A, B ∈ Ob C) of mutually disjoint sets whose elements f, g, … ∈ Mor_{C}(A, B) are called morphisms, and

(3) a family (Mor_{C}(A, B) × Mor_{C}(B, C) ∋ (f, g) ↦ gf ∈ Mor_{C}(A, C)|A, B, C ∈ Ob C) of

maps called compositions. C is called a category if the following axioms hold for C

(1) Associative Law:

∀A, B, C, D ∈ Ob C, f ∈ Mor_{C}(A, B), g ∈ Mor_{C}(B, C), h ∈ Mor_{C}(C, D) :

h(gf) = (hg)f;

(2) Identity Law:

∀A ∈ Ob C ∃1_{A} ∈ Mor_{C}(A, A) ∀B, C ∈ Ob C, ∀f ∈ Mor_{C}(A, B), ∀g ∈ Mor_{C}(C, A) :

1_{A}g = g  and  f1_{A} = f.

Examples 4.2. (1) The category of sets Set.

(2) The categories of R-modules R-Mod, 𝕂-vector spaces 𝕂-Vec or 𝕂-Mod, groups Gr, Abelian groups Ab, monoids Mon, commutative monoids cMon, rings Ri, fields Field, topological spaces Top.

(3) The left A-modules in the sense of Definition 2.17 and their homomorphisms form the category A-Mod of A-modules.

(4) The 𝕂-algebras in the sense of Definition 2.1 and their homomorphisms form the category 𝕂-Alg of 𝕂-algebras.

(5) The category of commutative 𝕂-algebras will be denoted by 𝕂-cAlg.

(6) The 𝕂-coalgebras in the sense of Definition 2.18 and their homomorphisms form a category 𝕂-Coalg of 𝕂-coalgebras.

(7) The category of cocommutative 𝕂-coalgebras will be denoted by 𝕂-cCoalg.

For arbitrary categories we adopt many of the customary notations.

Notation 4.3. f ∈ Mor_{C}(A, B) will be written as f : A ⟶ B or A ⟶^{f} B. A is called the domain, B the range of f. The composition of two morphisms f : A ⟶ B and g : B ⟶ C is written as gf : A ⟶ C or as g ∘ f : A ⟶ C.

Definition and Remark 4.4. A morphism f : A ⟶ B is called an isomorphism if there exists a morphism g : B ⟶ A in C such that fg = 1_{B} and gf = 1_{A}. The morphism g is uniquely determined by f since g′ = g′fg = g. We write f^{−1} := g. An object A is said to be isomorphic to an object B if there exists an isomorphism f : A ⟶ B. If f is an isomorphism then so is f^{−1}. If f : A ⟶ B and g : B ⟶ C are isomorphisms in C then so is gf : A ⟶ C. We have: (f^{−1})^{−1} = f and (gf)^{−1} = f^{−1}g^{−1}. The relation of being isomorphic between objects is an equivalence relation.

Example 4.5. In the categories Set, R-Mod, k-Vec, Gr, Ab, Mon, cMon, Ri, Field the isomorphisms are exactly those morphisms which are bijective as set maps.

In Top the set M = {a, b} with 𝔗_{1} = {∅, {a}, {b}, {a, b}} and with 𝔗_{2} = {∅, M} defines two different topological spaces. The map f = id : (M, 𝔗_{1}) ⟶ (M, 𝔗_{2}) is bijective and continuous. The inverse map, however, is not continuous, hence f is no isomorphism (homeomorphism).

Many well known concepts can be defined for arbitrary categories. We are going to introduce some of them. Here are two examples.

Definition 4.6. (1) A morphism f : A ⟶ B is called a monomorphism if ∀C ∈ Ob C, ∀g, h ∈ Mor_{C}(C, A) :

fg = fh ⟹ g = h (f is left cancellable).

(2) A morphism f : A ⟶ B is called an epimorphism if ∀C ∈ Ob C, ∀g, h ∈ Mor_{C}(B, C) :

gf = hf ⟹ g = h (f is right cancellable).

Definition 4.7. Given A, B ∈ C. An object A×B in C together with morphisms p_{A} : A×B ⟶ A and p_{B} : A × B ⟶ B is called a (categorical) product of A and B if for every (test) object T ∈ C and every pair of morphisms f : T ⟶ A and g : T ⟶ B there exists a unique morphism (f, g) : T ⟶ A × B such that the diagram

[figure: T; f; g; A; A × B; p_{A}; B; p_{B}; (f,g)]

commutes. An object E ∈ C is called a final object if for every (test) object T ∈ C there exists a unique morphism e : T ⟶ E (i.e. Mor_{C}(T, E) consists of exactly one element). A category C which has a product for any two objects A and B and which has a final object is called a category with finite products.

Remark 4.8. If the product (A × B, p_{A}, p_{B}) of two objects A and B in C exists then it is unique up to isomorphism. If the final object E in C exists then it is unique up to isomorphism.

Problem 4.1. Let C be a category with finite products. Give a definition of a product of a family A_{1}, …, A_{n} (n ≥ 0). Show that products of such families exist in C.

Definition and Remark 4.9. Let C be a category. Then C^{op} with the following data Ob C^{op} := Ob C, Mor_{C^{op}}(A, B) := Mor_{C}(B, A), and f ∘_{op} g := g ∘ f defines a new category, the dual category of C.

Remark 4.10. Any notion expressed in categorical terms (with objects, morphisms, and their composition) has a dual notion, i.e. the given notion in the dual category. Monomorphisms f in the dual category C^{op} are epimorphisms in the original category C and conversely. A final object I in the dual category C^{op} is an initial object in the original category C.

Definition 4.11. The coproduct of two objects in the category C is defined to be a product of the objects in the dual category C^{op}.

Remark 4.12. Equivalent to the preceding definition is the following definition. Given A, B ∈ C. An object A ⨿ B in C together with morphisms j_{A} : A ⟶ A ⨿ B and j_{B} : B ⟶ A ⨿ B is a (categorical) coproduct of A and B if for every (test) object T ∈ C and every pair of morphisms f : A ⟶ T and g : B ⟶ T there exists a unique morphism [f, g] : A ⨿ B ⟶ T such that the diagram

[figure: T; f; g; A; A ⨿ B; j_{A}; B; j_{B}; [f,g]]

commutes. The category C is said to have finite coproducts if C^{op} is a category with finite products. In particular coproducts are unique up to isomorphism.

4.2. Functors.

Definition 4.13. Let C and D be categories. Let F consist of (1) a map Ob C ∋ A ↦ F(A) ∈ Ob D,

(2) a family of maps

(F_{A,B} : Mor_{C}(A, B) ∋ f ↦ F_{A,B}(f) ∈ Mor_{D}(F(A), F(B))|A, B ∈ C)

[ or (F_{A,B} : Mor_{C}(A, B) ∋ f ↦ F_{A,B}(f) ∈ Mor_{D}(F(B), F(A))|A, B ∈ C)]

F is called a covariant [contravariant] functor if (1) F_{A,A}(1_{A}) = 1_{F(A)} for all A ∈ Ob C,

(2) F_{A,C}(gf) = F_{B,C}(g)F_{A,B}(f) for all A, B, C ∈ Ob C.

[ F_{A,C}(gf) = F_{A,B}(f)F_{B,C}(g) for all A, B, C ∈ Ob C ]. Notation: We write

A ∈ C instead of  A ∈ Ob C
f ∈ C instead of f ∈ Mor_{C}(A, B)
F(f) instead of  F_{A,B}(f).

Examples 4.14. The following define functors (1) Id : Set ⟶ Set;

(2) Forget : R-Mod ⟶ Set;

(3) Forget : Ri ⟶ Ab;

(4) Forget : Ab ⟶ Gr;

(5) P : Set ⟶ Set, P(M) := power set of M. P(f)(X) := f^{−1}(X) for f : M ⟶ N, X ⊆ N is a contravariant functor;

(6) Q : Set ⟶ Set, Q(M) := power set of M. Q(f)(X) := f(X) for f : M ⟶ N, X ⊆ M is a covariant functor;

(7) - ⊗_{R} N : Mod-R ⟶ Ab;

(8) M ⊗_{R} - : R-Mod ⟶ Ab;

(9) - ⊗_{R} - : Mod-R × R-Mod ⟶ Ab;

(10) the embedding functor ι : 𝕂-Mod ⟶ 𝕂-Mod-𝕂.

(11) the tensor product over 𝕂 in 𝕂-Mod-𝕂 can be restricted to 𝕂-Mod so that the following diagram of functors commutes:

[figure: 𝕂-Mod; 𝕂-Mod-𝕂.; ι; 𝕂-Mod × 𝕂-Mod; 𝕂-Mod-𝕂 × 𝕂-Mod-𝕂; ι × ι; ⊗_{𝕂}; ⊗_{𝕂}]

Proof of (9). (f × g) ∘ (f′ × g′) = ff′ × gg′ implies (f ⊗_{R} g) ∘ (f′ ⊗_{R} g′) = ff′ ⊗_{R} gg′. Furthermore 1_{M} × 1_{N} = 1_{M×N} implies 1_{M} ⊗_{R} 1_{N} = 1_{M⊗_{R}N}. □

Lemma 4.15. (1) Let X ∈ C. Then

Ob C ∋ A ↦ Mor_{C}(X, A) ∈ Ob Set
Mor_{C}(A, B) ∋ f ↦ Mor_{C}(X, f) ∈ Mor_{Set}(Mor_{C}(X, A), Mor_{C}(X, B)),

with Mor_{C}(X, f) : Mor_{C}(X, A) ∋ g ↦ fg ∈ Mor_{C}(X, B) or Mor_{C}(X, f)(g) = fg is a covariant functor Mor_{C}(X, -).

(2) Let X ∈ C. Then

Ob C ∋ A ↦ Mor_{C}(A, X) ∈ Ob Set

Mor_{C}(A, B) ∋ f ↦ Mor_{C}(f, X) ∈ Mor_{Set}(Mor_{C}(B, X), Mor_{C}(A, X)) with Mor_{C}(f, X) : Mor_{C}(B, X) ∋ g ↦ gf ∈ Mor_{C}(A, X) or Mor_{C}(f, X)(g) = gf is a contravariant functor Mor_{C}(-, X).

Proof. (1) Mor_{C}(X, 1_{A})(g) = 1_{A}g = g = id(g), Mor_{C}(X, f) Mor_{C}(X, g)(h) = fgh = Mor_{C}(X, fg)(h).

(2) analogously. □

Remark 4.16. The preceding lemma shows that Mor_{C}(-, -) is a functor in both arguments. A functor in two arguments is called a bifunctor. We can regard the bifunctor Mor_{C}(-, -) as a covariant functor

Mor_{C}(-, -) : C^{op} × C ⟶ Set .

The use of the dual category removes the fact that the bifunctor Mor_{C}(-, -) is contravariant in the first variable. Obviously the composition of two functors is again a functor and this composition is associative. Furthermore for each category C there is an identity functor Id_{C}. Functors of the form Mor_{C}(X, -) resp. Mor_{C}(-, X) are called representable functors (covariant resp. contravariant) and X is called the representing object (see also section 5).

4.3. Natural Transformations.

Definition 4.17. Let F : C ⟶ D and G : C ⟶ D be two functors. A natural transformation or a functorial morphism φ : F ⟶ G is a family of morphisms (φ(A) : F(A) ⟶ G(A)|A ∈ C) such that the diagram

[figure: F(B); G(B); φ(B); F(A); G(A); φ(A); F(f); G(f)]

commutes for all f : A ⟶ B in C, i.e. G(f)φ(A) = φ(B)F(f).

Lemma 4.18. Given covariant functors F = Id_{Set} : Set ⟶ Set and

G = Mor_{Set}(Mor_{Set}(−, A), A) : Set ⟶ Set

for a set A. Then φ : F ⟶ G with

φ(B) : B ∋ b ↦ (Mor_{Set}(B, A) ∋ f ↦ f(b) ∈ A) ∈ G(B)

is a natural transformation.

Proof. Given g : B ⟶ C. Then the following diagram commutes

[figure: B; Mor_{Set}(Mor_{Set}(B, A), A); φ(B); C; Mor_{Set}(Mor_{Set}(C, A), A); φ(C); g; Mor_{Set}(Mor_{Set}(g, A), A)]

since

φ(C)F(g)(b)(f) = φ(C)g(b)(f) = fg(b) = φ(B)(b)(fg)
= [φ(B)(b) Mor_{Set}(g, A)](f) = [Mor_{Set}(Mor_{Set}(g, A), A)φ(B)(b)](f).
□

Lemma 4.19. Let f : A ⟶ B be a morphism in C. Then Mor_{C}(f, -) : Mor_{C}(B, -) ⟶ Mor_{C}(A, -) given by Mor_{C}(f, C) : Mor_{C}(B, C) ∋ g ↦ gf ∈ Mor_{C}(A, C) is a natural transformation of covariant functors. Let f : A ⟶ B be a morphism in C. Then Mor_{C}(-, f) : Mor_{C}(-, A) ⟶ Mor_{C}(-, B) given by Mor_{C}(C, f) : Mor_{C}(C, A) ∋ g ↦ fg ∈ Mor_{C}(C, B) is a natural transformation of contravariant functors.

Proof. Let h : C ⟶ C′ be a morphism in C. Then the diagrams

[figure: Mor_{C}(B, C′); Mor_{C}(A, C′); Mor_{C}(f,C′); Mor_{C}(B, C); Mor_{C}(A, C); Mor_{C}(f,C); Mor_{C}(B,h); Mor_{C}(A,h)]

and

[figure: Mor_{C}(C, A); Mor_{C}(C, B); Mor_{C}(C,f); Mor_{C}(C′, A); Mor_{C}(C′, B); Mor_{C}(C′,f); Mor_{C}(h,A); Mor_{C}(h,B)]

commute. □

Remark 4.20. The composition of two natural transformations is again a natural transformation. The identity id_{F}(A) := 1_{F(A)} is also a natural transformation.

Definition 4.21. A natural transformation φ : F ⟶ G is called a natural isomorphism if there exists a natural transformation ψ : G ⟶ F such that φ ∘ ψ = id_{G} and ψ ∘ φ = id_{F}. The natural transformation ψ is uniquely determined by φ. We write φ^{−1} := ψ. A functor F is said to be isomorphic to a functor G if there exists a natural isomorphism φ : F ⟶ G.

Remark 4.22. The isomorphisms given in Theorem 1.22 for_{R}M_{S},_{S}N_{T}, and_{T} P_{U} are natural isomorphisms:

(1) Associativity Law: α : (M ⊗_{S} N) ⊗_{T} P ≅ M ⊗_{S} (N ⊗_{T} P) with α((m ⊗ n) ⊗ p) := m ⊗ (n ⊗ p);

(2) Law of the Left Unit: λ : R ⊗_{R} M ≅ M with λ(r ⊗ m) := rm;

(3) Law of the Right Unit: ρ : M ⊗_{S} S ≅ M with ρ(m ⊗ r) := mr;

(4) Symmetry Law: τ : M ⊗N ≅ N ⊗M for 𝕂-modules M and N with τ(m⊗n) := n⊗m; (5) Inner Hom-Functors:

ϕ : Hom_{S-T} (.P ⊗_{R} M., .N.) ≅ Hom_{S-R}(.P., . Hom_{T} (M., N.).)

with ϕ(f)(p)(m) := f(p ⊗ m) and

ψ : Hom_{S-T} (.P ⊗_{R} M., .N.) ≅ Hom_{R-T} (.M., . Hom_{S}(.P, .N).)

with ψ(f)(m)(p) := f(p ⊗ m) for bimodules _{R}M_{T}, _{S}N_{T}, and _{S}P_{R}.

Problem 4.2. (1) Let F, G : C ⟶ D be functors. Show that a natural transformation φ : F ⟶ G is a natural isomorphism if and only if φ(A) is an isomorphism for all objects A ∈ C.

(2) Let (A×B, p_{A}, p_{B}) be the product of A and B in C. Then there is a natural isomorphism

Mor(-, A × B) ≅ Mor_{C}(-, A) × Mor_{C}(-, B).

(3) Let C be a category with finite products. For each object A in C show that there exists a morphism Δ_{A} : A ⟶ A × A satisfying p_{1}Δ_{A} = 1_{A} = p_{2}Δ_{A}. Show that this defines a natural transformation. What are the functors?

(4) Let C be a category with finite products. Show that there is a bifunctor - × - : C × C ⟶ C such that (- × -)(A, B) is the object of a product of A and B. We denote elements in the image of this functor by A × B := (- × -)(A, B) and similarly f × g.

(5) With the notation of the preceding problem show that there is a natural transformation α(A, B, C) : (A × B) × C ≅ A × (B × C). Show that the diagram (coherence or constraints)

[figure: ((A × B) × C) × D; (A × (B × C)) × D; α(A,B,C)×1; A × ((B × C) × D); α(A,B×C,D); α(A×B,C,D); 1×α(B,C,D); (A × B) × (C × D); A × (B × (C × D)); α(A,B,C×D)]

commutes.

(6) With the notation of the preceding problem show that there are a natural transformations λ(A) : E × A ⟶ A and ρ(A) : A × E ⟶ A such that the diagram (coherence or constraints)

[figure: (A × E) × B; A × (E × B); α(A,E,B); A × B; ρ(A)×1; 1×λ(B)]

Definition 4.23. Let C and D be categories. A covariant functor F : C ⟶ D is called an equivalence of categories if there exists a covariant functor G : D ⟶ C and natural isomorphisms φ : GF ≅ Id_{C} and ψ : FG ≅ Id_{D}. A contravariant functor F : C ⟶ D is called a duality of categories if there exists a contravariant functor G : D ⟶ C and natural isomorphisms φ : GF ≅ Id_{C} and ψ : FG ≅ Id_{D}. A category C is said to be equivalent to a category D if there exists an equivalence F : C ⟶ D. A category C is said to be dual to a category D if there exists a duality F : C ⟶ D.

Problem 4.3. (1) Show that the dual category C^{op} is dual to the category C.

(2) Let D be a category dual to the category C. Show that D is equivalent to the dual category C^{op}.

(3) Let F : C ⟶ D be an equivalence with respect to G : D ⟶ C, φ : GF ≅ Id_{C}, and ψ : FG ≅ Id_{D}. Show that G : D ⟶ C is an equivalence. Show that G is uniquely determined by F up to a natural isomorphism.

5. Representable and Adjoint Functors, the Yoneda Lemma

5.1. Representable functors.

Definition 5.1. Let F : C ⟶ Set be a covariant functor. A pair (A, x) with A ∈ C, x ∈ F(A) is called a representing (generic, universal) object for F and F is called a representable functor, if for each B ∈ C and y ∈ F(B) there exists a unique f ∈ Mor_{C}(A, B) such that F(f)(x) = y:

[figure: A; B; f; F(A); F(B); F(f); ∋ x; ∋ y]

Proposition 5.2. Let (A, x) and (B, y) be representing objects for F. Then there exists a unique isomorphism h : A ⟶ B such that F(h)(x) = y.

[figure: A; B; h; A; k; B; h; 1_{A}; 1_{B}; F(A); F(B); F(h); F(A); F(k); F(B); F(h); 1_{F(A)}; 1_{F(B)}; x; y; x; y]

Examples 5.3. (1) Let R be a ring. Let X ∈ Set be a set. F : R-Mod ⟶ Set, F(M) := Map(X, M) is a covariant functor. A representing object for F is given by the free R-module (RX, x : X ⟶ RX) with the property, that for all (M, y : X ⟶ M) there exists a unique f ∈ Hom_{R}(RX, M) such that F(f)(x) = Map(X, f)(x) = fx = y

[figure: X; RX; x; y; M.; f]

(2) Given modules M_{R} and_{R}N. Define F : Ab ⟶ Set by F(A) := Bil_{R}(M, N; A). Then F is a covariant functor. A representing object for F is given by the tensor product (M ⊗_{R} N, ⊗ : M × N ⟶ M ⊗_{R} N) with the property that for all (A, f : M × N ⟶ A) there exists a unique g ∈ Hom(M ⊗_{R} N, A) such that F(g)(⊗) = Bil_{R}(M, N; g)(⊗) = g⊗ = f

[figure: M × N; M ⊗_{R} N; ⊗; f; A.; g]

(3) Given a 𝕂-module V. Define F : 𝕂-Alg ⟶ Set by F(A) := Hom(V, A). Then F is a covariant functor. A representing object for F is given by the tensor algebra (T (V), ι : V ⟶ T (V)) with the property that for all (A, f : V ⟶ A) there exists a unique g ∈ Mor_{Alg}(T (V), A) such that F(g)(ι) = Hom(V, g)(ι) = gι = f

[figure: V; T (V); ι; f; A.; g]

(4) Given a 𝕂-module V. Define F : 𝕂-cAlg ⟶ Set by F(A) := Hom(V, A). Then F is a covariant functor. A representing object for F is given by the symmetric algebra (S(V), ι : V ⟶ S(V)) with the property that for all (A, f : V ⟶ A) there exists a unique g ∈ Mor_{cAlg}(S(V), A) such that F(g)(ι) = Hom(V, g)(ι) = gι = f

[figure: V; S(V); ι; f; A.; g]

(5) Given a 𝕂-module V. Define F : 𝕂-Alg ⟶ Set by

F(A) := {f ∈ Hom(V, A)|∀v, v′ ∈ V : f(v)f(v′) = f(v′)f(v)}.

Then F is a covariant functor. A representing object for F is given by the symmetric algebra (S(V), ι : V ⟶ S(V)) with the property that for all (A, f : V ⟶ A) such that f(v)f(v′) = f(v′)f(v) for all v, v′ ∈ V there exists a unique g ∈ Mor_{Alg}(S(V), A) such that F(g)(ι) = Hom(V, g)(ι) = gι = f

[figure: V; S(V); ι; f; A.; g]

(6) Given a 𝕂-module V. Define F : 𝕂-Alg ⟶ Set by

F(A) := {f ∈ Hom(V, A)|∀v ∈ V : f(v)^{2} = 0}.

Then F is a covariant functor. A representing object for F is given by the exterior algebra (E(V), ι : V ⟶ E(V)) with the property that for all (A, f : V ⟶ A) such that f(v)^{2} = 0 for all v ∈ V there exists a unique g ∈ Mor_{Alg}(E(V), A) such that F(g)(ι) = Hom(V, g)(ι) = gι = f

[figure: V; E(V); ι; f; A.; g]

(7) Let 𝕂 be a commutative ring. Let X ∈ Set be a set. F : 𝕂-cAlg ⟶ Set, F(A) := Map(X, A) is a covariant functor. A representing object for F is given by the polynomial ring (𝕂[X], ι : X ⟶ 𝕂[X]) with the property, that for all (A, f : X ⟶ A) there exists a unique g ∈ Mor_{cAlg}(𝕂[X], A) such that F(g)(ι) = Map(X, g)(x) = gι = f

[figure: X; 𝕂[X]; ι; f; A.; g]

(8) Let 𝕂 be a commutative ring. Let X ∈ Set be a set. F : 𝕂-Alg ⟶ Set, F(A) := Map(X, A) is a covariant functor. A representing object for F is given by the noncommutative polynomial ring (𝕂⟨X⟩, ι : X ⟶ 𝕂⟨X⟩) with the property, that for all (A, f : X ⟶ A) there exists a unique g ∈ Mor_{Alg}(𝕂⟨X⟩, A) such that F(g)(ι) = Map(X, g)(x) = gι = f

[figure: X; 𝕂⟨X⟩; ι; f; A.; g]

Problem 5.1. (1) Given V ∈ 𝕂-Mod. For A ∈ 𝕂-Alg define

F (A) := {f : V ⟶ A|f 𝕂-linear, ∀v, w ∈ V : f(v) · f(w) = 0}.

Show that this defines a functor F : 𝕂-Alg ⟶ Set.

(2) Show that F has the algebra D(V) as constructed in Exercise 2.1 (3) as a representing object.

Proposition 5.4. F has a representing object (A, a) if and only if there is a natural isomorphism φ : F ≅ Mor_{C}(A, −) (with a = φ(A)^{−1}(1_{A})).

Proof. ⟹ : The map

φ(B) : F(B) ∋ y ↦ f ∈ Mor_{C}(A, B) with F(f)(a) = y

is bijective with the inverse map

ψ(B) : Mor_{C}(A, B) ∋ f ↦ F(f)(a) ∈ F(B).

In fact we have y ↦ f ↦ F(f)(a) = y and f ↦ y := F(f)(a) ↦ g such that F(g)(a) = y but then F(g)(a) = y = F(f)(a). By uniqueness we get f = g. Hence all φ(B) are bijective with inverse map ψ(B). It is sufficient to show that ψ is a natural transformation. Given g : B ⟶ C. Then the following diagram commutes

[figure: Mor_{C}(A, C); F(C); ψ(C); Mor_{C}(A, B); F(B); ψ(B); Mor_{C}(A,g); F(g)]

since ψ(C) Mor_{C}(A, g)(f) = ψ(C)(gf) = F(gf)(a) = F(g)F(f)(a) = F(g)ψ(B)(f). ⇐: Let A be given. Let a := φ(A)^{−1}(1_{A}). For y ∈ F(B) we get y = φ(B)^{−1}(f) = φ(B)^{−1}(f1_{A}) = φ(B)^{−1} Mor_{C}(A, f)(1_{A}) = F(f)φ(A)^{−1}(1_{A}) = F(f)(a) for a uniquely determined f ∈ Mor_{C}(A, B). □

Proposition 5.5. Let D be a category. Given a representable functor F_{X} : C ⟶ Set for each X ∈ D. Given a natural transformation F_{g} : F_{Y} ⟶ F_{X} for each g : X ⟶ Y (contravariant!) such that F depends functorially on X, i.e. F_{1_{X}} = 1_{F_{X}}, F_{hg} = F_{g}F_{h}. Then the representing objects (A_{X}, a_{X}) for F_{X} depend functorially on X, i.e. for each g : X ⟶ Y there is a unique morphism A_{g} : A_{X} ⟶ A_{Y} (with F_{X}(A_{g})(a_{X}) = F_{g}(A_{Y})(a_{Y})) and the following identities hold A_{1_{X}} = 1_{A_{X}}, A_{hg} = A_{h}A_{g}. So we get a covariant functor D ∋ X ⟶ A_{X} ∈ C.

Proof. Choose a representing object (A_{X}, a_{X}) for F_{X} for each X ∈ D (by the axiom of choice). Then there is a unique morphism A_{g} : A_{X} ⟶ A_{Y} with

F_{X}(A_{g})(a_{X}) = F_{g}(A_{Y})(a_{Y}) ∈ F_{X}(A_{Y}),

for each g : X ⟶ Y because F_{g}(A_{Y}) : F_{Y} (A_{Y}) ⟶ F_{X}(A_{Y}) is given. We have F_{X}(A_{1})(a_{X}) = F_{1}(A_{X})(a_{X}) = a_{X} = F_{X}(1)(a_{X}) hence A_{1} = 1, and F_{X}(A_{hg})(a_{X}) = F_{hg}(A_{Z})(a_{Z}) = F_{g}(A_{Z})F_{h}(A_{Z})(a_{Z}) = F_{g}(A_{Z})F_{Y} (A_{h})(a_{Y}) = F_{X}(A_{h})F_{g}(A_{Y})(a_{Y}) = F_{X}(A_{h})F_{X}(A_{g})(a_{X}) = F_{X}(A_{h}A_{g})(a_{X}) hence A_{h}A_{g} = A_{hg} for g : X ⟶ Y and h : Y ⟶ Z in D. □

Corollary 5.6. (1) Map(X, M) ≅ Hom_{R}(RX, M) is a natural transformation in M (and in X!). In particular Set ∋ X ↦ RX ∈ R-Mod is a functor.

(2) Bil_{R}(M, N; A) ≅ Hom(M ⊗_{R} N, A) is a natural transformation in A (and in (M, N) ∈ Mod-R × R-Mod). In particular Mod-R × R-Mod ∋ M, N ↦ M ⊗_{r} N ∈ Ab is a functor.

(3) R-Mod-S × S-Mod-T ∋ (M, N) ↦ M ⊗_{S} N ∈ R-Mod-T is a functor.

5.2. The Yoneda Lemma.

Theorem 5.7. (Yoneda Lemma) Let C be a category. Given a covariant functor F : C ⟶ Set and an object A ∈ C. Then the map

π : Nat(Mor_{C}(A, -), F) ∋ ϕ ↦ ϕ(A)(1_{A}) ∈ F(A)

is bijective with the inverse map

π^{−1} : F(A) ∋ a ↦ h^{a} ∈ Nat(Mor_{C}(A, -), F),

where h^{a}(B)(f) = F(f)(a).

Proof. For ϕ ∈ Nat(Mor_{C}(A, -), F) we have a map ϕ(A) : Mor_{C}(A, A) ⟶ F(A), hence π with π(ϕ) := ϕ(A)(1_{A}) is a well defined map. For π^{−1} we have to check that h^{a} is a natural transformation. Given f : B ⟶ C in C. Then the diagram

[figure: F(B); F(C); F(f); Mor_{C}(A, B); Mor_{C}(A, C); Mor(A,f); h^{a}(B); h^{a}(C)]

is commutative. In fact if g ∈ Mor_{C}(A, B) then h^{a}(C) Mor_{C}(A, f)(g) = h^{a}(C)(fg) = F(fg)(a) = F(f)F(g)(a) = F(f)h^{a}(B)(g). Thus π^{−1} is well defined. Let π^{−1}(a) = h^{a}. Then ππ^{−1}(a) = h^{a}(A)(1_{A}) = F(1_{A})(a) = a. Let π(ϕ) = ϕ(A)(1_{A}) = a. Then π^{−1}π(ϕ) = h^{a} and h^{a}(B)(f) = F(f)(a) = F(f)(ϕ(A)(1_{A})) = ϕ(B) Mor_{C}(A, f)(1_{A}) = ϕ(B)(f), hence h^{a} = ϕ. □

Corollary 5.8. Given A, B ∈ C. Then the following hold

(1) Mor_{C}(A, B) ∋ f ↦ Mor_{C}(f, -) ∈ Nat(Mor_{C}(B, -), Mor_{C}(A, -)) is a bijective map.

(2) Under the bijective map from (1) the isomorphisms in Mor_{C}(A, B) correspond to natural isomorphisms in Nat(Mor_{C}(B, -), Mor_{C}(A, -)).

(3) For contravariant functors F : C ⟶ Set we have Nat(Mor_{C}(-, A), F) ≅ F(A).

(4) Mor_{C}(A, B) ∋ f ↦ Mor_{C}(-, f) ∈ Nat(Mor_{C}(-, A), Mor_{C}(-, B)) is a bijective map that defines a one-to-one correspondence between the isomorphisms in Mor_{C}(A, B) and the natural isomorphisms in Nat(Mor_{C}(-, A), Mor_{C}(-, B)).

Proof. (1) follows from the Yoneda Lemma with F = Mor_{C}(A, -).

(2) Observe that h^{f}(C)(g) = Mor_{C}(A, g)(f) = gf = Mor_{C}(f, C)(g) hence h^{f} = Mor_{C}(f, -). Since we have Mor_{C}(f, -) Mor_{C}(g, -) = Mor_{C}(gf, -) and Mor_{C}(f, -) = id_{Mor_{C}(A,-)} if and only if f = 1_{A} we get the one-to-one correspondence between the isomorphisms from (1).

(3) and (4) follow by dualizing. □

Remark 5.9. The map π is a natural transformation in the arguments A and F. More precisely: if f : A ⟶ B and ϕ : F ⟶ G are given then the following diagrams commute

[figure: Nat(Mor_{C}(A, -), G); G(A); π; Nat(Mor_{C}(A, -), F); F(A); π; Nat(Mor(A,-),ϕ); ϕ(A)]

[figure: Nat(Mor_{C}(B, -), F); F(B).; π; Nat(Mor_{C}(A, -), F); F(A); π; Nat(Mor(f,-),F); F(f)]

This can be easily checked. Indeed we have for ψ : Mor_{C}(A, -) ⟶ F

π Nat(Mor_{C}(A, -), ϕ)(ψ) = π(ϕψ) = (ϕψ)(A)(1_{A}) = ϕ(A)ψ(A)(1_{A}) = ϕ(A)π(ψ)

and

π Nat(Mor_{C}(f, -), F)(ψ) = π(ψ Mor_{C}(f, -)) = (ψ Mor_{C}(f, -))(B)(1_{B}) = ψ(B)(f)
= ψ(B) Mor_{C}(A, f)(1_{A}) = F(f)ψ(A)(1_{A}) = F(f)π(ψ).

Remark 5.10. By the previous corollary the representing object A is uniquely determined up to isomorphism by the isomorphism class of the functor Mor_{C}(A, -).

Proposition 5.11. Let G : C × D ⟶ Set be a covariant bifunctor such that the functor G(C, -) : D ⟶ Set is representable for all C ∈ C. Then there exists a contravariant functor F : C ⟶ D such that G ≅ Mor_{D}(F-, -) holds. Furthermore F is uniquely determined by G up to isomorphism.

Proof. For each C ∈ C choose an object F(C) ∈ D and an isomorphism ξ_{C} : G(C, -) ≅ Mor_{D}(F(C), -). Given f : C ⟶ C′ in C then let F(f) : F(C′) ⟶ F(C) be the uniquely determined morphism (by the Yoneda Lemma) in D such that the diagram

[figure: G(C′, -); Mor_{D}(F(C′), -); ξ_{C′}; G(C, -); Mor_{D}(F(C), -); ξ_{C}; G(f,-); Mor(F(f),-)]

commutes. Because of the uniqueness of F(f) and because of the functoriality of G it is easy to see that F(fg) = F(g)F(f) and F(1_{C}) = 1_{F(C)} hold and that F is a contravariant functor. If F′ : C ⟶ D is given with G ≅ Mor_{D}(F′-, -) then ϕ : Mor_{D}(F-, -) ≅ Mor_{D}(F′-, -). Hence by the Yoneda Lemma ψ(C) : F(C) ≅ F′(C) is an isomorphism for all C ∈ C. With these isomorphisms induced by ϕ the diagram

[figure: Mor_{D}(F′(C′), -); Mor_{D}(F(C′), -); Mor(ψ(C′),-); Mor_{D}(F′(C), -); Mor_{D}(F(C), -); Mor(ψ(C),-); Mor(F′(f),-); Mor(F(f),-)]

commutes. Hence the diagram

[figure: F(C); F′(C); ψ(C); F(C′); F′(C′); ψ(C′); F′(f); F(f)]

commutes. Thus ψ : F ⟶ F′ is a natural isomorphism. □

5.3. Adjoint functors.

Definition 5.12. Let C and D be categories and F : C ⟶ D and G : D ⟶ C be covariant functors. F is called left adjoint to G and G right adjoint to F if there is a natural isomorphism of bifunctors ϕ : Mor_{D}(F-, -) ⟶ Mor_{C}(-, G-) from C^{op} × D to Set.

Lemma 5.13. If F : C ⟶ D is left adjoint to G : D ⟶ C then F is uniquely determined by G up to isomorphism. Similarly G is uniquely determined by F up to isomorphism.

Proof. We only prove the first claim. Assume that also F′ is left adjoint to G with ϕ′ : Mor_{D}(F′-, -) ⟶ Mor_{C}(-, G-). Then we have a natural isomorphism ϕ′^{−1}ϕ : Mor_{D}(F-, -) ⟶ Mor_{D}(F′-, -). By Proposition 5.11 we get F ≅ F′. □

Lemma 5.14. A functor G : D ⟶ C has a left adjoint functor iff all functors Mor_{C}(C, G-) are representable.

Proof. follows from 5.11. □

Lemma 5.15. Let F : C ⟶ D and G : D ⟶ C be covariant functors. Then

Nat(Id_{C}, GF) ∋ Φ ↦ G-Φ- ∈ Nat(Mor_{D}(F-, -), Mor_{C}(-, G-))

is a bijective map with inverse map

Nat(Mor_{D}(F-, -), Mor_{C}(-, G-)) ∋ ϕ ↦ ϕ(-, F-)(1_{F-}) ∈ Nat(Id_{C}, GF).

Furthermore

Nat(FG, Id_{C}) ∋ Ψ ↦ Ψ-F- ∈ Nat(Mor_{C}(-, G-), Mor_{D}(F-, -))

is a bijective map with inverse map

Nat(Mor_{C}(-, G-), Mor_{D}(F-, -)) ∋ ψ ↦ ψ(G-, -)(1_{G-}) ∈ Nat(FG, Id_{C}).

Proof. The natural transformation G-Φ- is defined as follows. Given C ∈ C, D ∈ D and f ∈ Mor_{D}(F(C), D) then let (G-Φ-)(C, D)(f) := G(f)Φ(C) : C ⟶ GF(C) ⟶ G(D). It is easy to check the properties of a natural transformation. Given Φ then one obtains by applying the two maps G(1_{F(C)})Φ(C) = GF(1_{C})Φ(C) = Φ(C). Given ϕ one obtains

\begin{matrix}G(f)(ϕ(C, F(C))(1_{F(C)}) = Mor_{C}(C, G(f))ϕ(C, F(C))(1_{F(C)}) \\ = ϕ(C, D) Mor_{D}(F(C), f)(1_{F(C)}) = ϕ(C, D)(f). \end{matrix}

So the two maps are inverses of each other. The second part of the lemma is proved similarly. □

Proposition 5.16. Let

ϕ : Mor_{D}(F-, -) ⟶ Mor_{C}(-, G-) and ψ : Mor_{C}(-, G-) ⟶ Mor_{D}(F-, -)

be natural transformations with associated natural transformations (by Lemma 5.15) Φ : Id_{C} ⟶ GF resp. Ψ : FG ⟶ Id_{D}.

(1) Then we have ϕψ = id_{Mor(-,G-)} if and only if (G ⟶^{ΦG} GFG ⟶^{GΨ} G) = id_{G}.

(2) Furthermore we have ψϕ = id_{Mor(F-,-)} if and only if (F ⟶^{FΦ} FGF ⟶^{ΨF} F) = id_{F}.

Proof. We get

GΨ(D)ΦG(D) = GΨ(D)ϕ(G(D), FG(D))(1_{FG(D)})
= Mor_{C}(G(D), GΨ(D))ϕ(G(D), FG(D))(1_{FG(D)})
= ϕ(G(D), D) Mor_{D}(FG(D), Ψ(D))(1_{FG(D)})
= ϕ(G(D), D)(Ψ(D))
= ϕ(G(D), D)ψ(G(D), D)(1_{G(D)})
= ϕψ(G(D), D)(1_{G(D)}).

Similarly we get

\begin{matrix}ϕψ(C, D)(f) = ϕ(C, D)ψ(C, D)(f) = G(Ψ(D)F(f))Φ(C) \\ = GΨ(D)GF(f)Φ(C) = GΨ(D)ΦG(D)f. □ \end{matrix}

Corollary 5.17. Let F : C ⟶ D and G : D ⟶ C be functors. F is left adjoint to G if and only if there are natural transformations Φ : Id_{C} ⟶ GF and Ψ : FG ⟶ Id_{D} such that (GΨ)(ΦG) = id_{G} and (ΨF)(FΦ) = id_{F}.

Definition 5.18. The natural transformations Φ : Id_{C} ⟶ GF and Ψ : FG ⟶ Id_{D} given in 5.17 are called unit and counit resp. for the adjoint functors F and G.

Problem 5.2. (1) Let _{R}M_{S} be a bimodule. Show that the functor M ⊗_{S} - : _{S}M ⟶ _{R}M is left adjoint to Hom_{R}(M, -) : _{R}M ⟶ _{S}M. Determine the associated unit and counit.

(2) Show that there is a natural isomorphism Map(A × B, C) ≅ Map(B, Map(A, C)). Determine the associated unit and counit.

(3) Show that there is a natural isomorphism 𝕂-Alg(𝕂G, A) ≅ Gr(G, U(A)) where U(A) is the group of units of the algebra A and 𝕂G is the group ring (see Section 12 ). Determine the associated unit and counit.

(4) Use Section 12 to show that there is a natural isomorphism

𝕂-Alg(U(𝔤), A) ≅ Lie-Alg(𝔤, A^{L}).

Determine the corresponding left adjoint functor and the associated unit and counit.

5.4. Universal problems.

Definition 5.19. Let G : D ⟶ C be a covariant functor. G generates a (co-)universal problem a follows: Given C ∈ C. Find an object F(C) ∈ D and a morphism ι : C ⟶ G(F(C)) in C such that for each object D ∈ D and for each morphism f : C ⟶ G(D) in C there is a unique morphism g : F(C) ⟶ D in D such that the diagram

[figure: C; G(F(C)); ι; f; G(D); G(g)]

commutes. A pair (F(C), ι) that satisfies the above conditions is called a universal solution of the (co-)universal problem defined by G and C. Let F : C ⟶ D be a covariant functor. F generates a universal problem a follows: Given D ∈ D. Find an object G(D) ∈ C and a morphism ν : F(G(D)) ⟶ D in D such that for each object C ∈ C and for each morphism f : F(C) ⟶ D in D there is a unique morphism g : C ⟶ G(D) in C such that the diagram

[figure: FG(D); D; ν; f; F(C); F(g)]

commutes. A pair (G(D), ν) that satisfies the above conditions is called a universal solution of the universal problem defined by F and D.

Proposition 5.20. Let F : C ⟶ D be left adjoint to G : D ⟶ C. Then F(C) and the unit ι = Φ(C) : C ⟶ GF(C) form a universal solution for the (co-)universal problem defined by G and C. Furthermore G(D) and the counit ν = Ψ(D) : FG(D) ⟶ D form a universal solution for the universal problem defined by F and D.

Proof. By Theorem 5.16 the morphisms ϕ : Mor_{D}(F-, -) ⟶ Mor_{C}(-, G-) and ψ : Mor_{C}(-, G-) ⟶ Mor_{D}(F-, -) are inverses of each other. Using unit and counit they are defined as ϕ(C, D)(g) = G(g)Φ(C) resp. ψ(C, D)(f) = Ψ(D)F(f). Hence for each f : C ⟶ G(D) there is a unique g : F(C) ⟶ D such that f = ϕ(C, D)(g) = G(g)Φ(C) = G(g)ι. The second statement follows analogously. □

Remark 5.21. If G : D ⟶ C and C ∈ C are given then the universal solution (F(C), ι : C ⟶ G(D)) can be considered as the best (co-)approximation of the object C in C by an object D in D with the help of a functor G. The object D ∈ D turns out to be F(C). If F : C ⟶ D and D ∈ D are given then the universal solution (G(D), ν : FG(D) ⟶ D) can be considered as the best approximation of the object D in D by an object C in C with the help of a functor F. The object C ∈ C turns out to be G(D).

Proposition 5.22. Given G : D ⟶ C. Assume that for each C ∈ C the universal problem defined by G and C has a universal solution. Then there is a left adjoint functor F : C ⟶ D to G. Given F : C ⟶ D. Assume that for each D ∈ D the universal problem defined by F and D has a universal solution. Then there is a right adjoint functor G : D ⟶ C to F.

Proof. Assume that the (co-)universal problem defined by G and C is solved by ι : C ⟶ GF(C). Then the map Mor_{C}(C, G(D)) ∋ f ↦ g ∈ Mor_{D}(F(C), D) with G(g)ι = f is bijective. The inverse map is given by g ↦ G(g)ι. This is a natural transformation since the diagram commutes for each h ∈ Mor_{D}(D, D′). In fact we have

[figure: Mor_{D}(F(C), D′); Mor_{C}(C, G(D′)); G(-)ι; Mor_{D}(F(C), D); Mor_{C}(C, G(D)); G(-)ι; Mor_{D}(F(C),h); Mor_{C}(C,G(h))]

Mor_{C}(C, G(h))(G(g)ι) = G(h)G(g)ι = G(hg)ι = G(Mor_{C}(F(C), h)(g))ι.

Hence for all C ∈ C the functor Mor_{C}(C, G(-)) : D ⟶ Set induced by the bifunctor Mor_{C}(-, G(-)) : C^{op} × D ⟶ Set is representable. By Theorem 5.11 there is a functor F : C ⟶ D such that Mor_{C}(-, G(-)) ≅ Mor_{D}(F(-), -). The second statement follows analogously. □

Remark 5.23. One can characterize the properties that G : D ⟶ C (resp. F : C ⟶ D) must have in order to possess a left (right) adjoint functor. One of the essential properties for this is that G preserves limits (and thus preserves direct products and difference kernels).

Proposition 5.24. The construction of tensor algebras T (V) defines a functor T : 𝕂-Mod ⟶ 𝕂-Alg that is left adjoint to the underlying functor U : 𝕂-Alg ⟶ 𝕂-Mod.

Proof. Follows from the universal property and 5.22. □

Proposition 5.25. The construction of symmetric algebras S(V) defines a functor S : 𝕂-Mod ⟶ 𝕂-cAlg that is left adjoint to the underlying functor U : 𝕂-cAlg ⟶ 𝕂-Mod.

Proof. Follows from the universal property and 5.22. □

6. Limits and Colimits, Products and Equalizers

6.1. Limits of diagrams. Limit constructions are a very important tool in category theory. We will introduce the basic facts on limits and colimits in this section.

Definition 6.1. A diagram scheme D is a small category (i. e. the class of objects is a set). Let C be an arbitrary category. A diagram in C over the diagram scheme D is a covariant functor F : D ⟶ C.

Example 6.2. (for diagram schemes)

(1) The empty category D.

(2) The category with precisely one object D and precisely one morphism 1_{D}.

(3) The category with two objects D_{1}, D_{2} and one morphism f : D_{1} ⟶ D_{2} (apart from the two identities).

(4) The category with two objects D_{1}, D_{2} and two morphisms f, g : D_{1} ⟶ D_{2} between them.

(5) The category with a family of objects (D_{i}|i ∈ I) and the associated identities.

(6) The category with four objects D_{1}, …, D_{4} and morphisms f, g, h, k such that the diagram

[figure: D_{3}; D_{4}; h; D_{1}; D_{2}; f; g; k]

commutes, i. e. kf = hg.

Definition 6.3. Let D be a diagram scheme and C a category. Each object C ∈ C defines a constant diagram K_{C} : D ⟶ C with K_{C}(D) := C for all D ∈ D and K(f) := 1_{C} for all morphisms in D. Each morphism f : C ⟶ C′ in C defines a constant natural transformation K_{f} : K_{C} ⟶ K_{C′} with K_{f}(D) = f. This defines a constant functor K : C ⟶ Funct(D, C) from the category C into the category of diagrams Funct(D, C). Let F : D ⟶ C be a diagram. An object C together with a natural transformation π : K_{C} ⟶ F is called a limit or a projective limit of the diagram F with the projection π if for each object C′ ∈ C and for each natural transformation φ : K_{C′} ⟶ F there is a unique morphism f : C′ ⟶ C such that

[figure: K_{C}; F; π; φ; K_{C′}; K_{f}]

commutes, this means in particular that the diagrams

[figure: C; F(D_{i}); π_{i}; π_{j}; F(D_{j}); F(g)]

commute for all morphisms g : D_{i} ⟶ D_{j} in D (π is a natural transformation) and the diagrams

[figure: C; F(D_{i}).; π_{i}; φ_{i}; C′; f]

commute for all objects D_{i} in D. A category C has limits for diagrams over a diagram scheme D if for each diagram F : D ⟶ C over D there is a limit in C. A category C is called complete if each diagram in C has a limit.

Example 6.4. (1) Let D be a diagram scheme consisting of two objects D_{1}, D_{2} and the identities. A diagram F : D ⟶ C is defined by giving two objects C_{1} and C_{2} in C. An object C_{1} × C_{2} together with two morphisms π_{1} : C_{1} × C_{2} ⟶ C_{1} and π_{2} : C_{1} × C_{2} ⟶ C_{2} is called a product of the two objects if C_{1} × C_{2}, π : K_{C_{1}×C_{2}} ⟶ F is a limit, i. e. if for each object C′ in C and for any two morphisms φ_{1} : C′ ⟶ C_{1} and φ_{2} : C′ ⟶ C_{2} there is a unique morphism f : C′ ⟶ C_{1} × C_{2} such that

[figure: φ_{1}; φ_{2}; C′; f; C_{1}; C_{1} × C_{2}; π_{1}; C_{2}; π_{2}]

commutes. The two morphisms π_{1} : C_{1} × C_{2} ⟶ C_{1} and π_{2} : C_{1} × C_{2} ⟶ C_{2} are called the projections from the product to the two factors.

(2) Let D a diagram scheme consisting of a finite (non empty) set of objects D_{1}, …, D_{n} and the associated identities. A limit of a diagram F : D ⟶ C is called a finite product of the objects C_{1} := F(D_{1}), …, C_{n} := F(D_{n}) and is denoted by C_{1} × … × C_{n} = ∏_{i=1}^{n} C_{i}.

(3) A limit over a discrete diagram (i. e. D has only the identities as morphisms) is called product of the C_{i} := F(D_{i}), i ∈ I and is denoted by ∏_{I} C_{i}.

(4) Let D be the empty diagram scheme and F : D ⟶ C the (only possible) empty diagram. The limit C, π : K_{C} ⟶ F of F is called the final object. It has the property that for each object C′ in C (the uniquely determined natural transformation φ : K_{C′} ⟶ F does not have to be mentioned) there is a unique morphism f : C′ ⟶ C. In Set the one-point set is a final object. In Ab, Gr, Vec the zero group 0 is a final object.

(5) Let D be the diagram scheme from 6.2 (4) with two objects D_{1}, D_{2} and two morphisms (different from the two identities) a, b : D_{1} ⟶ D_{2}. A diagram over D consists of two objects C_{1} and C_{2} and two morphisms g, h : C_{1} ⟶ C_{2}. The limit of such a diagram is called equalizer of the two morphisms and is given by an object Ke(g, h) and a morphism π_{1} : Ke(g, h) ⟶ C_{1}. The second morphism to C_{2} arises from the composition π_{2} = gπ_{1} = hπ_{1}. The equalizer has the following universal property. For each object C′ and each morphism φ_{1} : C′ ⟶ C_{1} with gφ_{1} = hφ_{1}(= φ_{2}) there is a unique morphism f : C′ ⟶ Ke(g, h) with π_{1}f = φ_{1} (and thus π_{2}f = φ_{2}), i. e. the diagram

[figure: C′; f; φ_{1}; Ke(g, h); C_{1}; π_{1}; g; C_{2}; h]

commutes.

Problem 6.1. (1) Let F : D ⟶ Set be a discrete diagram. Show that the Cartesian product over F coincides with the categorical product.

(2) Let D be a pair of morphisms as in 6.4 (5) and let F : D ⟶ Set be a diagram. Show that the set {x ∈ F(D_{1})|F(f)(x) = F(g)(x)} with the inclusion map into F(D_{1}) is an equalizer of F : D ⟶ Set.

(3) Let F : D ⟶ Set be a diagram. Show that the set

{(x_{D}|D ∈ Ob D, x_{D} ∈ F(D))|∀(f : D ⟶ D′) ∈ D : F(f)(x_{D}) = x_{D′}}

with the projections into the single components of the families is the limit of F.

(4) Given a homomorphism f : M ⟶ N in R-Mod. Show that (K, ι : K ⟶ M) is a kernel of f iff it is the equalizer of the pair of homomorphisms f, 0 : M ⟶ N iff the sequence

0 ⟶ K \underrightarrow{ι} M \underrightarrow{f} N

is exact.

6.2. Colimits of diagrams.

Definition 6.5. Let F : D ⟶ C be a diagram. An object C and a natural transformation ι : F ⟶ K_{C} is called colimit or inductive limit of the diagram F with the injection ι if for each object C′ ∈ C and for each natural transformation φ : F ⟶ K_{C′} there is a unique morphism f : C ⟶ C′ such that

[figure: F; K_{C}; ι; φ; K_{C′}; K_{f}]

commutes, i. e. the diagram

[figure: F(D_{j}); C; ι_{j}; ι_{i}; F(D_{i}); F(g)]

commutes for all morphisms g : D_{i} ⟶ D_{j} in D (ι is a natural transformation) and the diagram

[figure: F(D_{i}); C; ι_{i}; φ_{i}; C′.; f]

commutes for all objects D_{i} in D. The special colimits that can be formed over the diagrams as in Example 6.4 are called coproduct, initial object, resp. coequalizer.

Example 6.6. In 𝕂-Vec the object 0 is an initial object. In 𝕂-Alg the object 𝕂 is an initial object. In 𝕂-Alg the object {a ∈ A|f(a) = g(a)} is the equalizer of the two algebra homomorphisms f : A ⟶ B and g : A ⟶ B. In 𝕂-Alg the Cartesian (set of pairs) and the categorical products coincide.

Remark 6.7. A colimit of a diagram C is a limit of the corresponding (dual) diagram in the dual category C^{op}. Thus theorems about limits in arbitrary categories automatically also produce (dual) theorems about colimits. However, observe that theorems about limits in a particular category (for example the category of vector spaces) translate only into theorems about colimits in the dual category, which most often is not too useful.

Proposition 6.8. Limits and colimits of diagrams are unique up to isomorphism.

Proof. Let F : D ⟶ C be a diagram and let C, π and C̃, π̃ be limits of F. Then there are unique morphisms f : C̃ ⟶ C and g : C ⟶ C̃ with πK_{f} = π̃ and π̃K_{g} = π. This implies πK_{1_{C}} = π id_{K_{C}} = π = π̃K_{g} = πK_{f}K_{g} = πK_{fg} and analogously π̃K_{1C̃} = π̃K_{gf}. Because of the uniqueness this implies 1_{C} = fg and 1C̃ = gf. □

Remark 6.9. Now that we have the uniqueness of the limit resp. colimit (up to isomorphism) we can introduce a unified notation. The limit of a diagram F : D ⟶ C will be denoted by \underleftarrow{lim}(F), the colimit by \underrightarrow{lim}(F).

Problem 6.2. Given a homomorphism f : M ⟶ N in R-Mod. Show that (Q, ν : N ⟶ Q) is a cokernel of f iff it is the coequalizer of the pair of homomorphisms f, 0 : M ⟶ N iff the sequence

M \underrightarrow{f} N \underrightarrow{ν} Q ⟶ 0

is exact.

6.3. Completeness.

Theorem 6.10. If C has arbitrary products and equalizers then C has arbitrary limits. In this case we say that C is complete.

Proof. Let D be a diagram scheme and F : D ⟶ C a diagram. First we form the products ∏_{D∈Ob D} F(D) and ∏_{f∈Mor D} F(Codom(f)) where Codom(f) is the codomain (range) of the morphism f : D′ ⟶ D′′ in D so in this case Codom(f) = D′′. We define for each morphism f : D′ ⟶ D′′ two morphisms as follows

p_{f} := π_{F(D′′)} : ∏_{D∈Ob D} F(D) ⟶ F(D′′) = F(Codom(f))

and

q_{f} := F(f)π_{F(D′)} : ∏_{D∈Ob D} F(D) ⟶ F(D′) ⟶ F(D′′) = F(Codom(f)).

These two families of morphisms induce two morphisms into the corresponding product

p, q : ∏_{D∈Ob D} F(D) ⟶ ∏_{f∈Mor D} F(Codom(f))

with π_{f}q = q_{f} and π_{f}p = p_{f}. Now we show that the equalizer of these two morphisms

[figure: Ke(p, q); ∏_{D∈Ob D} F(D); ψ; ∏_{f∈Mor D} F(Codom(f)); p; q]

is the limit of the diagram F : D ⟶ C. We have pψ = qψ. The morphism ρ(D) := π_{F(D)}ψ : Ke(p, q) ⟶ ∏_{D∈Ob D} F(D) ⟶ F(D) defines a family of morphisms for D ∈ Ob D. If f : D′ ⟶ D′′ is in D then the diagram

[figure: Ke(p, q); F(D′); ρ(D′); ρ(D′′); F(D′′); F(f)]

is commutative because of F(f)ρ(D′) = F(f)π_{F(D′)}ψ = q_{f}ψ = π_{f}qψ = π_{f}pψ = p_{f}ψ = π_{F(D′′)}ψ = ρ(D′′). Thus we have obtained a natural transformation ρ : K_{Ke(p,q)} ⟶ F. Now let an object C′ and a natural transformation φ : K_{C′} ⟶ F be given. Then this defines a unique morphism g : C′ ⟶ ∏_{D∈Ob D} F(D) with π_{F(D)}g = φ(D) for all D ∈ D. Since φ is a natural transformation we have φ(D′′) = F(f)φ(D′) for each morphism f : D′ ⟶ D′′. Thus we obtain π_{f}pg = p_{f}g = π_{F(D′′)}g = φ(D′′) = F(f)φ(D′) = F(f)π_{F(D′)}g = q_{f}g = π_{f}qg for all morphisms f : D′ ⟶ D′′ hence pg = qg. Thus g can be uniquely factorized through the equalizer ψ : Ke(p, q) ⟶ ∏_{D∈Ob D} F(D) in the form g = ψh with h : C′ ⟶ Ke(p, q). Then we have ρ(D)h = π_{F(D)}ψh = π_{F(D)}g = φ(D) for all D ∈ D hence ρK_{h} = φ. Finally let another morphism h′ : C′ ⟶ Ke(p, q) with ρK_{h′} = φ be given. Then we have π_{F(D)}ψh′ = ρ(D)h′ = φ(D) = ρ(D)h = π_{F(D)}ψh hence ψh′ = ψh = g. Because of the uniqueness of the factorization of g through ψ we get h = h′. Thus (Ke(p, q), ρ) is the limit of F. □

Remark 6.11. The proof of the preceding Theorem gives an explicit construction of the limit of F as an equalizer

[figure: Ke(p, q); ∏_{D∈Ob D} F(D); ψ; ∏_{f∈Mor D} F(Codom(f)); p; q]

Hence the limit can be represented as a subobject of a suitable product. Dually the colimit can be represented as a quotient object of a suitable coproduct.

6.4. Adjoint functors and limits. Another fact is very important for us, the fact that certain functors preserve limits resp. colimits. We say that a functor G : C ⟶ C′ preserves limits over the diagram scheme D if \underleftarrow{lim}(GF) ≅ G(\underleftarrow{lim}(F)) for each diagram F : D ⟶ C.

Proposition 6.12. Covariant representable functors preserve limits. Contravariant representable functors map colimits into limits.

Proof. We only prove the first assertion. The second assertion is dual to the first one. For a diagram F : D ⟶ Set the set

{(x_{D}|D ∈ Ob D, x_{D} ∈ F(D))|∀(f : D ⟶ D′) ∈ D : F(f)(x_{D}) = x_{D′}}

is a limit of F by Problem 6.1 (3). Now let a diagram F : D ⟶ C be given and let \underleftarrow{lim}(F) be the limit. Furthermore let Mor_{C}(C′, -) : C ⟶ Set be a representable functor. By the definition of the limit of F there is a unique morphism f : C′ ⟶ \underleftarrow{lim}(F) with πK_{f} = φ for each natural transformation φ : K_{C′} ⟶ F. This defines an isomorphism Nat(K_{C′}, F) ≅ Mor_{C}(C′, \underleftarrow{lim}(F)). Hence we have

\underleftarrow{lim}(Mor_{C}(C′, F)) ≅
{(φ(D) : C′ ⟶ F(D)|D ∈ D)|∀(f : D ⟶ D′) ∈ D : F(f)φ(D) = φ(D′)}
= Nat(K_{C′}, F) ≅ Mor_{C}(C′, \underleftarrow{lim}(F)). □

Corollary 6.13. Let F : C ⟶ C′ be left adjoint to G : C′ ⟶ C. Then F preserves colimits and G preserves limits.

Proof. For a diagram H : D ⟶ C we have

Mor_{C}(-, \underleftarrow{lim}(GH)) ≅ \underleftarrow{lim} Mor_{C}(-, GH) ≅ \underleftarrow{lim} Mor_{C′}(F-, H) ≅

Mor_{C′}(F-, \underleftarrow{lim}(H)) ≅ Mor_{C}(-, G(\underleftarrow{lim}(H))),

hence \underleftarrow{lim}(GH) ≅ G(\underleftarrow{lim}(H)) as representing objects. The proof for the left adjoint functor is analogous. □

7. The Morita Theorems

Throughout this section let 𝕂 be a commutative ring.

Definition 7.1. A category C is called a 𝕂-category, if Mor_{C}(M, N) is a 𝕂-module and Mor_{C}(f, g) is a homomorphism of 𝕂-modules for all M, N, f, g ∈ C. A functor F : C ⟶ D between 𝕂-categories C and D is called a 𝕂-functor, if F : Mor_{C}(M, N) ⟶ Mor_{D}(F(M), F(N)) for all M, N ∈ C is a homomorphism of 𝕂-modules. If 𝕂 = ℤ, then 𝕂-categories are called (pre-)additive categories and 𝕂-functors are called additive functors.

Remark 7.2. In this section 7 we always write homomorphisms at the opposite side from where ring elements act on the modules: f : _{R}M ⟶ _{R}N with (rm)f = r(mf).

Let A and B be 𝕂-algebras. Let _{A}M be a left A-module. Then it is also a 𝕂-module by κm := (κ·1_{A})·m. Analogously a right B-module is also a 𝕂-module. We redefine the notion of a bimodule as follows:

Definition 7.3. A 𝕂-bimodule _{A}M_{B} is an A-B-bimodule satisfying (κ · 1_{A}) · m = κm = mκ = m · (1_{B} · κ) i.e. the induced right and left structures of a 𝕂-module coincide.

Definition 7.4. A Morita context consists of a 6-tuple (A, B,_{A}P_{B},_{B}Q_{A}, f, g) with 𝕂-algebras A, B, 𝕂-bimodules _{A}P_{B}, _{B}Q_{A} and homomorphisms of 𝕂-bimodules

f : _{A}P ⊗_{B} Q_{A} ⟶ _{A}A_{A}, g : _{B}Q ⊗_{A} P_{B} ⟶ _{B}B_{B},

such that:

(1) qf(p ⊗ q′) = g(q ⊗ p)q′ oder q(pq′) = (qp)q′,

(2) f(p ⊗ q)p′ = pg(q ⊗ p′) oder (pq)p′ = p(qp′), where we will use the following notation pq := f(p ⊗ q) and qp := g(q ⊗ p).

Remark 7.5. With this convention all products are associative e.g. (pb)q = p(bq), (qa)p = q(ap).

Lemma 7.6. Let A be a 𝕂-algebra and _{A}P be an A-module. Then (A, B, P, Q, f = ev, g = db) is a Morita context with

\begin{matrix}B := Hom_{A}(.P, .P) & _{B}Q_{A} := _{B} Hom_{A}(.P_{B}, .A_{A})_{A} \\ f(p ⊗ q) := (p)q & (p′)[g(q ⊗ p)] := (p′)qp. \end{matrix}

Proof. as in 3.18. □

Definition 7.7. A 𝕂-equivalence of 𝕂-categories C and D consists of a pair of 𝕂-functors F : C ⟶ D, G : D ⟶ C such that Id_{D} ≅ FG and Id_{C} ≅ GF.

Theorem 7.8. (Morita I) Let (A, B, P, Q, f, g) be a Morita context. Let f and g be epimorphisms. Then the following statements hold

(1) P is a finitely generated projective generator in A-Mod and in Mod-B. Q is a finitely generated projective generator in Mod-A and in B-Mod.

(2) f and g are isomorphisms.

(3) Q ≅ Hom_{A}(.P, .A) ≅ Hom_{B}(P., B.)

P ≅ Hom_{B}(.Q, .B) ≅ Hom_{A}(Q., A.)

as bimodules.

(4) A ≅ Hom_{B}(.Q, .Q) ≅ Hom_{B}(P., P.)

B ≅ Hom_{A}(.P, .P) ≅ Hom_{A}(Q., Q.) as 𝕂-algebras and as bimodules.

(5) P ⊗_{B} - : B-Mod ⟶ A-Mod and Q ⊗_{A} - : A-Mod ⟶ B-Mod are mutually inverse 𝕂-equivalences. Symmetrically - ⊗_{A} P : Mod-A ⟶ Mod-B and - ⊗_{B} Q : Mod-B ⟶ Mod-A are mutually inverse 𝕂-equivalences. Furthermore the following functors are naturally isomorphic:

P ⊗_{B} - ≅ Hom_{B}(.Q, .-),
Q ⊗_{A} - ≅ Hom_{A}(.P, .-),
- ⊗_{A} P ≅ Hom_{A}(Q., -.),
- ⊗_{B} Q ≅ Hom_{B}(P., -.).

(6) We have the following isomorphisms of lattices (ordered sets):

V(_{A}P) ≅ V(_{B}B),  V(P_{B}) ≅ V(A_{A}),
V(_{B}Q) ≅ V(_{A}A),  V(Q_{A}) ≅ V(B_{B}),
V(_{B}Q_{A}) ≅ V(_{A}A_{A}) ≅ V(_{B}B_{B}) ≅ V(_{A}P_{B}).

(7) The following centers are isomorphic Cent(A) ≅ Cent(B).

Proof. (1) The isomorphisms from Theorem 1.22 (5) map g ∈ Hom_{B-B}(.Q ⊗_{A} P., .B.) to homomorphisms of bimodules g_{1} : P ⟶ Hom_{B}(.Q, .B) and g_{2} : Q ⟶ Hom_{B}(P., B.). Furthermore f induces homomorphisms of bimodules f_{1} : P ⟶ Hom_{A}(Q., A.) and f_{2} : Q ⟶ Hom_{A}(.P, .A). If g is an epimorphism then there is an element ∑ q_{i} ⊗ p_{i} ∈ Q ⊗_{A} P with g(∑ q_{i} ⊗ p_{i}) = 1_{B} = id_{P}. Hence p = ∑ pq_{i}p_{i} = ∑(p)[f_{2}(q_{i})]p_{i} for each p ∈ P. By the dual basis Lemma 3.19 _{A}P is finitely generated and projective. If f is an epimorphism then there is an element ∑ x_{i} ⊗ y_{i} ∈ P ⊗_{B} Q with f(∑ x_{i} ⊗ y_{i}) = 1_{A} = ∑(x_{i})[f_{2}(y_{i})]. By 3.24 _{A}P is a generator. The claims for P_{B}, _{B}Q, and Q_{A} follow by symmetry.

(2) If f(∑ a_{i} ⊗ b_{i}) = 0 then ∑_{i} a_{i} ⊗ b_{i} = ∑_{i,j} a_{i} ⊗ b_{i}f(x_{j} ⊗ y_{j}) = ∑ a_{i} ⊗ g(b_{i} ⊗ x_{j})y_{j} = ∑ a_{i}g(b_{i} ⊗ x_{j}) ⊗ y_{j} = ∑ f(a_{i} ⊗ b_{i})x_{j} ⊗ y_{j} = 0. Hence f is injective. By symmetry we get that g is an isomorphism.

(3) The homomorphism f_{2} : Q ⟶ Hom_{A}(.P, .A) defined as in (1) satisfies (p)[f_{2}(q)] = f(p ⊗ q) = pq. Let φ ∈ Hom_{A}(.P, .A). Then (p)φ = (p ∑ q_{i}p_{i})φ = ∑(pq_{i})(p_{i})φ hence φ = ∑ q_{i}(p_{i})φ = ∑ f_{2}(q_{i}(p_{i})φ). Thus f_{2} is an epimorphism. Let (p)[f_{2}(q)] = pq = 0 for all p ∈ P. Then we get q = 1_{B}q = ∑ q_{i}p_{i}q = 0. Hence f_{2} is an isomorphism.

(4) The structure of a B-module on P induces B ⟶ Hom_{A}(.P, .P). Let pb = 0 for all p ∈ P. Then b = 1_{B} · b = ∑ q_{i}p_{i}b = 0. If φ ∈ Hom_{A}(.P, .P) then we have (p)φ = (p1_{B})φ = (∑ p(q_{i}p_{i}))φ = ∑(pq_{i})(p_{i})φ = ∑ p(q_{i}(p_{i})φ) and thus φ = ∑ q_{i}(p_{i})φ. This shows that we have an isomorphism B ⟶ Hom_{A}(.P, .P) of 𝕂-algebras and bimodules.

(5) _{A}P ⊗_{B} Q ⊗_{A} X ≅ _{A}A ⊗_{A} X ≅ _{A}X is natural in X and _{B}Q ⊗_{A} P ⊗_{B} Y ≅ _{B}B ⊗_{B} Y ≅ BY is natural in Y thus we get the claim. Furthermore _{B}Q ⊗_{A} U ≅ _{B} Hom_{A}(.P, .A) ⊗_{A} U ≅ _{B} Hom_{A}(.P, .A ⊗_{A} U) ≅ _{B} Hom_{A}(.P, .U) is natural in U since the homomorphism φ : Hom_{A}(.P, .A) ⊗_{A} U ⟶ Hom_{A}(.P, .A ⊗_{A} U) with (p)[φ(f ⊗ u)] := ((p)f) ⊗ u is an isomorphism. More generally we show:

Lemma 7.9. If _{A}P is finitely generated projective and _{A}V_{B} and _{B}U are (bi-)modules then the natural transformation (in U and V)

φ : Hom_{A}(.P, .V) ⊗_{B} U ⟶ Hom_{A}(.P, .V ⊗_{B} U)

is an isomorphism.

Proof. Let ∑ f_{i} ⊗ p_{i} ∈ Hom_{A}(.P, .A) ⊗_{A} P be a dual basis for P. Then

φ^{−1} : Hom_{A}(.P, .V ⊗_{B} U) ⟶ Hom_{A}(.P, .V) ⊗_{B} U

defined by φ^{−1}(g) = ∑_{i,j}()f_{i}v_{ij} ⊗ u_{ij} with (p_{i})g =: ∑_{j} v_{ij} ⊗ u_{ij} is inverse to φ defined by (p)[φ(f ⊗u)] = (p)f ⊗u. Since φ is a homomorphism (p)[φ(fb⊗u)] = (p)fb⊗u = (p)f ⊗bu = (p)[φ(f ⊗ bu)] it suffices to show that φ^{−1} is a map. Now we have (p_{i})φ(f ⊗ u) = (p_{i})f ⊗ u hence φ^{−1}φ(f ⊗ u) = ∑()f_{i}(p_{i})f ⊗ u = ∑(()f_{i}p_{i})f ⊗ u = f ⊗ u. Furthermore we have φφ^{−1}(g) = φ(∑()f_{i}(p_{i})g) = ∑()f_{i}(p_{i})g = (∑()f_{i}p_{i})g = g. □

Proof of 7.8: (continued)

(6) Under the equivalence of categories _{A}P is mapped to Hom_{A}(.P, .P) ≅ _{B}B. This implies V(_{A}P) ≅ V(_{B}B). In fact, a submodule of _{A}P is an isomorphism class of monomorphisms AU ⟶ _{A}P, two such isomorphisms being called isomorphic, if there is a (necessarily unique) isomorphism U ≅ U′, such that

[figure: U; P; U′]

commutes. Obviously such subobjects are being preserved under an equivalence of categories. For subobjects of _{A}P_{B} we have furthermore that

[figure: U; P; U; P; ·b; ·b  and; Hom_{A}(.P, .U); B; Hom_{A}(.P, .U); B; ·b; ·b]

commute. Hence _{A}U_{B} ∈ V(_{A}P_{B}) iff Hom_{A}(.P, .U) ∈ V(_{B}B_{B}).

(7) The proof of this part will consist of two steps. We use the algebra End_{funkt}(Id_{A- Mod}) of natural endomorphisms of Id_{A- Mod} with the addition of morphisms and the composition of morphisms as the operations of the algebra. Obviously this defines an algebra. In a first step we show that the center of A is isomorphic to End_{funkt}(Id_{A- Mod}). In a second step we show that End_{funkt}(Id_{A- Mod}) ≅ End_{funkt}(Id_{B- Mod}). This last step is almost trivial since all terms defined by categorical means are preserved by an equivalence. Then we have Cent(A) ≅ End_{funkt}(Id_{A- Mod}) ≅ End_{funkt}(Id_{B- Mod}) ≅ Cent(B). Let z ∈ Z(A). For M = Id_{A- Mod}(M) we have zam = azm hence z = z· ∈ End_{A}(M). Thus z· defines an endomorphism of Id_{A- Mod}(M), for

[figure: N; N; z·; M; M; z·; f; f]

commutes. So we have defined a homomorphism Cent(A) ⟶ End_{funkt}(Id_{A- Mod}). Let φ ∈ End_{funkt}(Id_{A- Mod}). Then the diagram

[figure: M; M; φ(M); A; A; φ(A); f_{m}; f_{m}]

commutes, where (a)f_{m} = am. Each f ∈ Hom_{A}(.A, .M) is of this form. For M = A we have a(1)[φ(A)] = (a)[φ(A)] = (1)[f_{a}φ(A)] = (1)[φ(A)f_{a}] = (1)[φ(A)]a hence (1)[φ(A)] ∈ Z(A). For an arbitrary M ∈ A-Mod we then have (m)[φ(M)] = (1)[f_{m}φ(M)] = (1)[φ(A)f_{m}] = (1)[φ(A)]m i.e. φ(M) is of the form z· with z = (1)[φ(A)]. The maps defined in this way obviously are inverses of each other: z ↦ z· ↦ z · 1 = z and φ ↦ (1)[φ(A)] ↦ (1)[φ(A)]·. In order to show that End_{funkt}(Id_{A- Mod}) and End_{funkt}(Id_{B- Mod}) are isomorphic, let φ ∈ End_{funkt}(Id_{A- Mod}) =: E(A). We define φ′ ∈ E(B) by

[figure: ST (_{B}M); ST (_{B}M); SφT (M); _{B}M; _{B}M; φ′(M); β(M); β(M)]

where S : A-Mod ⟶ B-Mod, T : B-Mod ⟶ A-Mod are the mutually inverse equivalences from (5), and α : Id_{A- Mod} ⟶ T S and β : Id_{B- Mod} ⟶ ST resp. are the associated isomorphisms. Analogously we associate with each ψ ∈ E(B) an element ψ′ ∈ E(A) by

[figure: T S(_{A}N); T S(_{A}N); T ψS(N); _{A}N; _{A}N; ψ′(N); α(N); α(N)].

The compositions of ψ ↦ ψ′ and φ ↦ φ′ in each direction define isomorphisms, hence each single map is an isomorphism. One of the two compositions is contained in the following diagram.

[figure: N; N; φ′′(N); T S(N); T S(N); T φ′S(N); T (ST)S(N); T (ST)S(N); T SφT S(N); T S(N); T S(N); T Sφ(N); N; N; φ(N); α(N); α(N); T βS(N); T βS(N); T Sα(N); T Sα(N); α(N); α(N)]

Thus the map φ ↦ φ′′ is an inner automorphism of E(A), hence it is bijective. □

Theorem 7.10. (Morita II) Let S : A-Mod ⟶ B-Mod and T : B-Mod ⟶ A-Mod be mutually inverse 𝕂-equivalences. Let _{A}P_{B} := T (B) and _{B}Q_{A} := S(A). Then there are isomorphisms f : _{A}P ⊗_{B} Q_{A} ⟶ _{A}A_{A} and g : _{B}Q ⊗_{A} P_{B} ⟶ _{B}B_{B}, such that (A, B, P, Q, f, g) is a Morita context. Furthermore the following hold S ≅ Q ⊗_{A} - and T ≅ P ⊗_{B} -.

Theorem 7.11. (Morita III) Let P ∈ A-Mod be a finitely generated projective generator (= progenerator). Then the Morita context (A, Hom_{A}(.P, .P), P, Q, f = ev, g = db) is strict, i.e. f and g are epimorphisms.

Proof. Since _{A}P is finitely generated projective, g = db is an isomorphism (3.19). Since _{A}P is a generator, f = ev is an epimorphism (3.24). □

Proof of 7.10: 1. Given S, T. Then S : Hom_{A}(.M, .N) ∋ f ↦ S(f) ∈ Hom_{B}(.SM, .SN) is an isomorphism. Let α : T S ≅ Id_{A- Mod}. Then

[figure: Hom_{A}(.M, .N); Hom_{B}(.SM, .SN); S; T; Hom_{A}(.T SM, .T SN); Hom_{A}(.M, .N); Hom(α^{−1},α)]

is the identity, since Hom(α^{−1}, α)T S(f) = α ∘ T Sf ∘ α^{−1} = f. This holds since

[figure: M; N; f; T SM; T SN; T Df; α; α]

commutes. So S is a monomorphism and Hom(α^{−1}, α) ∘ T is an epimorphism. Since Hom(α^{−1}, α) is an isomorphism, T is an epimorphism where T : Hom_{B}(.SM, .SN) ⟶ Hom_{A}(.T SM, .T SN). By symmetry T is a monomorphism. Hence T is an isomorphism in the above map. Thus S is an isomorphism.

2. Hom_{B}(.SM, .N) ⟶^{T} Hom_{A}(.T SM, .T N) ⟶^{Hom(α^{−1},id)} Hom_{A}(.M, .T N) is a natural isomorphism. It is clear that this is an isomorphism. Since T is a functor, the first map is a natural transformation. The second map is a natural transformation, since α is a natural transformation. In particular, S is left adjoint to T.

3. S(⊕_{i∈I}M_{i}) ≅ ⊕_{i∈I}S(M_{i}), since S is a left adjoint functor and thus preserves direct coproducts.

4. If f ∈ B-Mod is an epimorphism, then T f ∈ A-Mod is an epimorphism, too. In fact, let f : M ⟶ N be an epimorphism. Let g, h ∈ A-Mod be given with g ∘ T f = h ∘ T f. Then we have a commutative diagram

[figure: ST M; ST N; ST f; Sg; SM; Sh; M; N; f; β; β]

with Sg ∘ ST f = Sh ∘ ST f. Since f is an epimorphism this implies Sg = Sh, hence g = h.

5. If P ∈ A-Mod is projective, then SP ∈ B-Mod is projective. In fact given an epimorphism f : M ⟶ N in B-Mod and a homomorphism g : SP ⟶ N. Then T f : T M ⟶ T N is an epimorphism and T g : T SP ⟶ T N is in A-Mod. Since α : T SP ≅ P, there is an h : P ⟶ T M with T f ∘ h = T g ∘ α^{−1} or T f ∘ h ∘ α = T g. We apply S and get ST f ∘ S(h ∘ α) = ST g, where S(h ∘ α) ∈ Hom_{B}(.ST SP, .ST M). Since β : ST M ≅ M, we have an isomorphism Hom(β^{−1}, β) : Hom_{B}(.ST SP, .ST M) ⟶ Hom_{B}(.SP, .M) with inverse Hom(β, β^{−1}). For k : SP ⟶ M with k = β ∘S(h∘α)∘β^{−1} we then have β ∘ST (k) = k ∘β = β ∘ S(h ∘ α) ∘ β^{−1} ∘ β = β ∘ S(h ∘ α), hence ST (k) = S(h ∘ α) and T (k) = h ∘ α. So we get ST f ∘ ST k = ST g = ST (f ∘ k) and thus g = f ∘ k. So SP is projective.

6. SA is finitely generated as a B-module: Since SA is projective, we have SA⊕X ≅ ⨁_{i∈I} B. By (3) applied to T we get A ⊕ T X ≅ T SA ⊕ T X ≅ ⨁_{i∈I} T B. Since A is finitely generated, the image of A in ⨁_{i∈I} T B is already a direct summand in a finite direct subsum ⨁_{i∈E} T B, so A ⊕ Y ≅ ⨁_{i∈E} T B. Hence SA ⊕ SY ≅ ⨁_{i∈E} ST B ≅ ⨁_{i∈E} B and thus SA is finitely generated.

7. If G ∈ A-Mod is a generator then SG ∈ B-Mod is also a generator. In fact let (f : M ⟶ N) ≠ 0 in B-Mod. Then T f ≠ 0, hence there is a g : G ⟶ T M with T f ∘ g ≠ 0. Consequently ST f ∘ Sg ≠ 0 and f ∘ (α ∘ Sg) = α ∘ ST f ∘ Sg ≠ 0.

8. This shows that S(A) is a finitely generated projective generator. (Remark: An equivalence S always maps finitely generated modules to finitely generated modules. We will give the proof further down in Proposition 7.12.)

9. A ≅ Hom_{B}(.SA, .SA) as algebras, since A ≅ Hom_{A}(.A, .A) ⟶^{S} Hom_{B}(.SA, .SA).

10. T B ≅ Hom_{B}(.SA, .B), since Hom_{B}(.SA, .B) ⟶^{T} Hom_{A}(.T SA, .T B) ≅ Hom_{A}(.A, .T B) ≅

T B.

11. (B, A, SA, T B, f, g) defines a strict Morita context by Morita III.

12. The functor S is isomorphic to SA ⊗_{A} −. Infact we have Hom_{B}(.SA ⊗_{A} M, .N) ≅ Hom_{A}(.M, . Hom_{B}(.SA, .N))

≅ Hom_{A}(.M, . Hom_{A}(.A, .T N))
≅ Hom_{A}(.M, .T N)
≅ Hom_{B}(.SM, .N).

The representing object _{B}SM ≅ _{B}SA ⊗_{A} M depends functorially on M by 5.5. □

Proposition 7.12. _{A}M is finitely generated iff in each set of submodules {A_{i}|i ∈ I} with A_{i} ⊆ M and ∑_{i∈I} A_{i} = M there is a finite subset {A_{i}|i ∈ I_{0}} (I_{0} ⊆ I finite) such that ∑_{i∈I_{0}} A_{i} = M.

Proof. Let M = Am_{1} + … + Am_{n}. Each m_{j} is contained in a finite sum of the A_{i}, hence all of the m_{j} and hence the module M itself. Conversely consider {Am|m ∈ M}. Then M = ∑ Am, hence M is a sum of finitely many of the Am and thus is finitely generated. □

Corollary 7.13. Under an equivalence of categories T : A-Mod ⟶ B-Mod finitely generated modules are mapped into finitely generated modules.

Proof. The lattice of submodules V(M) is isomorphic to the lattice of submodules V(T M). □

Problem 7.1. Let A-Mod be equivalent to B-Mod. Show that Mod-A and Mod-B are also equivalent.

Problem 7.2. Show that an equivalence of arbitrary categories preserves monomorphisms.

Problem 7.3. Show that an equivalence of module categories preserves projective modules, but not free modules.

8. Simple and Semisimple rings and Modules

8.1. Simple and Semisimple rings.

Definition 8.1. An ideal _{R}I ⊆ _{R}R is called nilpotent, if there is n ≥ 1 such that I^{n} = 0. A module_{R}M is called Artinian (Emil Artin, 1898-1962), if each non empty set of submodules of M contains a minimal element. A module _{R}M is called Noetherian (Emmy Noether, 1882-1935), if each non empty set of submodules of M contains a maximal element. A ring R is called simple, if _{R}R as a module is Artinian and if R does not have non trivial (≠ 0, R) two sided ideals. A ring R is called semisimple, if _{R}R is Artinian and if R does not have non trivial (≠ 0) nilpotent left ideals.

Lemma 8.2. Each simple ring is semisimple.

Proof. C := ∑(I|_{R}I ⊆ _{R}R nilpotent) is a two sided ideal. In fact take a ∈ I and r ∈ R. Then

(r_{1}ar)(r_{2}ar) … (r_{n}ar) = (r_{1}a)(rr_{2}a) … (rr_{n}a)r ∈ I^{n}R = 0.

Hence we have (Rar)^{n} = 0 ⟹ Rar ⊆ C, so ar ∈ C and C is a two sided ideal. Thus C = 0 or C = R. If C = 0 then there are no non trivial nilpotent ideals. If C = R then there are ideals and elements a_{i} ∈ I_{i} such that 1 = a_{1} + … + a_{n}. The ideal I_{1} + I_{2} is nilpotent since (a_{1} + b_{1})(a_{2} + b_{2}) … (a_{2n} + b_{2n}) consists of monomials either in I_{1}^{n} · R or in I_{2}^{n} · R. But I_{1}^{n} = 0 = I_{2}^{n} ⟹ (I_{1} + I_{2})^{2n} = 0. Hence 1 is nilpotent. Contradiction. □

Definition 8.3. A module _{R}M is called simple iff M ≠ 0 and M has only the modules 0 and M as submodules. An ideal _{R}I is called simple or minimal, if it is simple as a module.

Lemma 8.4. Let R be semisimple. Then each left ideal of R is a direct summand of R.

Proof. Let I be an ideal in R, that is not a direct summand, and let I be minimal with respect to this property. Such an ideal exists, since R Artinian. Case 1: Let I ⊆ R be an ideal that is not minimal (simple), i.e. there is an ideal J ⊆ I with 0 ≠ J ≠ I. Then J is a direct summand of R, i.e. there is a homomorphism f : R ⟶ J with (J ⟶ I ⟶ R ⟶^{f} J) = id_{J}. This implies I = J ⊕ K for K := Ke(I ⟶ R ⟶^{f} J). Since K ≠ I, there is also a g : R ⟶ K with (K ⟶ I ⟶ R ⟶^{g} K) = id_{K}. The map f + g − gf : I ⟶ R ⟶ I satisfies (f + g − gf)(j) = f(j) + g(j) − gf(j) = j + g(j) − g(j) = j for all j ∈ J and (f +g −gf)(k) = f(k)+g(k)−gf(k) = 0+k −0 = k for all k ∈ K, hence (f +g −gf : I ⟶ R ⟶ I) = id_{I}. Thus I is a direct summand of R. Contradiction. Case 2: Let I be a minimal or simple ideal. Since I is not nilpotent and 0 ≠ I^{2} ⊆ I holds, we get I^{2} = I. In particular there exists an a ∈ I with Ia = I, since Ia is also an ideal. Thus ·a : I ⟶ I is an epimorphism and even an isomorphism, for Ke(·a) must be zero as an ideal (see Lemma of Schur 8.5.) So there is an e ∈ I, e ≠ 0 with ea = a. ⟹ (e^{2}−e)a = eea−ea = a − a = 0 ⟹ e^{2} − e = 0 ∈ I ⟹ e^{2} = e ∈ I. From I = Re we get R = Re ⊕ R(1 − e), since R = Re + R(1 − e) and re = s(1 − e) ∈ Re ∩ R(1 − e) ⟹ re = re^{2} = s(1 − e)e = 0. Thus I is a direct summand of R. Contradiction. □

Lemma 8.5. (Schur) Let _{R}M, _{R}N be simple modules. Then the following hold: (1) If M ≅ N̸, then Hom_{R}(.M, .N) = 0.

(2) Hom_{R}(.M, .M) is a skew-field (= division algebra = non commutative field).

Proof. Let f : M ⟶ N be a homomorphism with f ≠ 0. Then Im(f) = N, since N is simple and Ke(f) = 0, since M is simple, hence f is an isomorphism. This implies (1).

Furthermore we have (2), since each endomorphism f : M ⟶ M with f ≠ 0 is invertible under the multiplication of Hom_{R}(.M, .M). Observe that a skew-field is a ring, whose non zero elements form a group under the multiplication. □

Remark 8.6. Let _{R}M be simple. Then End_{R}(.M) = D is a skew-field. Hence the R-module structure of M can be characterized by R ⟶ End_{D}(M.) = M_{n}(D).

Theorem 8.7. (Artin-Wedderburn) The following are equivalent:

(1) R is simple.

(2) R possesses a simple ideal that is an R-progenerator.

(3) R ≅ M_{n}(D) is a full matrix ring over a skew-field D. (n is unique, D is unique up to isomorphism.)

(4) R = I_{1} ⊕ … ⊕ I_{n} with isomorphic simple left ideals I_{1}, …, I_{n}.

Proof. (1) ⟹ (2): Since R is Artinian there is a simple ideal 0 ≠ I ⊆ R. Let J := ∑{I′|I′ ideal in R and I′ ≅ I}. Then J is a two sided ideal, since I′ · r ≠ 0 ⟹ ·r : I′ ⟶ R with Ke(·r) = 0, hence ·r is injective and the image I′ · r is isomorphic to I′ resp. I, hence is in J. Since R is simple we have R = J = ∑ I_{i}. Since 1 ∈ I_{1} + … + I_{n}, there is an epimorphism I_{1} ⊕ … ⊕ I_{n} ⟶ R (exterior direct sum), that splits since R is projective. Hence R is a direct summand of I_{1} ⊕ … ⊕ I_{n} up to isomorphism, and thus I is a generator. Furthermore I is a direct summand of R by 8.4, hence it is finitely generated projective, thus I is an R-progenerator.

(2) ⟹ (3): By the Lemma of Schur End_{R}(.I) =: D is a skew-field. _{R}I_{D} generates an equivalence of categories. Hence R ≅ End_{D}(I.) ≅ M_{n}(D).

(3) ⟹ (4): R ≅ M_{n}(D) ⟹ R ≅ End_{D}(V.) with an n-dimensional D-vector space V. V_{D} is a progenerator. Hence we have V(_{R}R) ≅ V(_{D}V^{∗}). Since V^{∗} ≅ D ⊕ … ⊕ D, we have RR ≅ I_{1} ⊕ … ⊕ I_{n} with I_{1} ≅ … ≅ I_{n} ≅ _{R}V ⊗_{D} D ≅ _{R}V.

(4) ⟹ (2): I_{1} is obviously an R-progenerator.

(2) ⟹ (1): R-Mod ≅ D-Mod with D ≅ End_{R}(I). Hence V(_{R}R) ≅ V(_{D} Hom_{D}(I.,_{D}D.)) is Artinian, and we have V(_{R}R_{R}) ≅ V(_{D}D_{D}) = {0, D}. Thus R is simple. □

Corollary 8.8. Let R be a simple ring and let _{R}M ≠ 0 be finitely generated. Then the following hold

(1) _{R}M is an R-progenerator.

(2) S := End_{R}(.M) is a simple ring.

(3) Cent(R) ≅ Cent(End_{R}(.M)).

(4) R ≅ End_{S}(M.).

Proof. (1) The claim follows from the fact that R-Mod ≅ D-Mod and since each finitely generated D-module is a progenerator.

(2) S-Mod ≅ R-Mod ≅ D-Mod implies that V(_{S}S) ≅ V(_{D}P) is Artinian. Furthermore V(_{S}S_{S}) ≅ V(_{D}D_{D}), hence S is a simple ring. (3)+(4) follow from the Morita theorems. □

8.2. Injective Modules.

Definition and Remark 8.9. An R-module _{R}J is called injective, if for each monomorphism f : M ⟶ N and for each homomorphism g : M ⟶ J there exists a homomorphism h : N ⟶ J with hf = g

[figure: M; N; f; J; g; h].

Vector spaces are injective. _{ℤ}ℤ is not injective. The injective ℤ-modules are exactly the divisible Abelian groups. _{ℤ}ℚ is injective.

Theorem 8.10. (The Baer criterion): The following are equivalent for Q ∈ R-Mod:

(1) Q is injective.

(2) ∀_{R}I ⊆ _{R}R, ∀g : I ⟶ Q ∃h : R ⟶ Q with hι = g

[figure: I; R; ι; Q; g; h].

(3) Each monomorphism f : Q ⟶^{f} M splits, i.e. there is an epimorphism g : M ⟶ Q with gf = 1_{Q}.

Proof. (1) ⟹ (2): follows immediately from the definition.

(1) ⟹ (3): The diagram

[figure: Q; M; f; Q; 1_{Q}; g]

defines the required g.

(3) ⟹ (1): In the diagram

[figure: M; N; f; Q; P; φ; ρ; g; ψ]

assume that f is a monomorphism and P := N ⊕ Q/{(f(m), −g(m))|m ∈ M} with φ resp. ψ are canonical maps to the left resp. the right components: φ(q) := \overline{(0, q)}, ψ(n) := \overline{(n, 0)}. Since ψf(m) = \overline{(f(m), 0)} = \overline{(0, g(m))} = φg(m) we have ψf = φg. Let φ(q) = \overline{(0, q)} = 0. Then there exists an m ∈ M with f(m) = 0 and g(m) = q. Since f is an injective map, we have m = 0 and thus φ injective. By (3) there is a ρ with ρφ = 1_{Q}. Then ρψf = ρφg = g, and thus Q is injective.

(2) ⟹ (1): Given a monomorphism f : M ⟶ N and a homomorphism g : N ⟶ Q. Consider the set S := {(N_{i}, φ_{i})}, where N_{i} ⊆ N is a submodule with Im(f) ⊆ N_{i} and φ_{i} : N_{i} ⟶ Q is a homomorphism such that

[figure: M; N_{i}; f; N; Q; g; φ_{i}]

commutes. We have S ≠ ∅, since (Im(f), gf^{−1}) ∈ S. Furthermore S is ordered by (N_{i}, φ_{i}) ≤ (N_{j}, φ_{j}) if N_{i} ⊆ N_{j} and φ_{j}|_{N_{i}} = φ_{i}. Let {(N_{i}, φ_{i})|i ∈ J} be a chain in S. Then ∪N_{i} ⊆ N is a submodule. ψ : ∪N_{i} ⟶ Q with ψ(n_{i}) = φ_{i}(n_{i}) is a well defined homomorphism and (∪N_{i}, ψ) ∈ S. Furthermore we have (N_{j}, φ_{j}) ≤ (∪N_{i}, ψ) for all j ∈ J. By Zorn’s Lemma there exists a maximal element (N′, φ′) in S. We show that N′ = N, for then the continuation of g to N exists. Let x ∈ N \ N′. Then N′ ⫋ N′ + Rx. Let I := {r ∈ R|rx ∈ N′}. Then I is an ideal and we have a commutative diagram

[figure: I; R; ι; M; N′; f; N′ + Rx; g; Q; φ′; σ; ·x; ρ; τ]

with ρ(r) := r · x. Then we have ρ(I) ⊆ N′. Thus by (2) there is a homomorphism σ : R ⟶ Q with σι = φ′ ∘ (·x). We define τ : N′ + Rx ⟶ Q by τ(n′ + rx) := φ′(n′) + σ(r). This is a well defined map, for if n′ + rx = n′_{1} + r_{1}x then (r − r_{1})x = n′_{1} − n′ ∈ N′ hence r − r_{1} ∈ I. Thus σ(r − r_{1}) = φ′((r − r_{1})x) = φ′(n′_{1} − n′) and φ′(n′) + σ(r) = φ′(n′_{1}) + σ(r_{1}). It is easy to see that τ is also a homomorphism. Since τ|_{N′} = φ′ holds we have (N′ + Rx, τ) ∈ S and (N′, φ′) ≨ (N′ + Rx, τ) a contradiction to the maximality of (N′, φ′). Thus N′ = N. □

Corollary 8.11. If R is a semisimple ring then each R-module is projective and injective.

Proof. Let Q be an R-module. By 8.4 each ideal is a direct summand of R. The following diagram together with the Baer criterion shows that Q is injective:

[figure: I; R; Q.]

Let f : N ⟶ P be surjective. Since Ke(f) ⊆ N is a submodule and injective there is a g : N ⟶ Ke(f) with g(n) = n for all n ∈ Ke(f). We define k : P ⟶ N by k(p) = n − g(n) for n ∈ N with f(n) = p. If also f(n′) = p then f(n − n′) = 0 hence n − n′ ∈ Ke(f) and g(n−n′) = n−n′. This implies n−g(n) = n′−g(n′). So k is a well defined map. Furthermore fk(p) = f(n − g(n)) = f(n) − fg(n) = p − 0, hence fk = 1_{P}. In order to show that k is a homomorphism let f(n) = p, f(n′) = p′. Then we get f(rn + r′n′) = rp + r′p′. This implies k(rp + r′p′) = rn + r′n′ − g(rn + r′n′) = r(n − g(n)) + r′(n′ − g(n′)) = rk(p) + r′k(p′). Thus P is projective. □

Lemma 8.12. Let 0 ⟶ M ⟶^{f} N ⟶^{g} P ⟶ 0 be a short exact sequence. M and P are Artinian if and only if N is Artinian. In particular if M and N are Artinian then M ⊕ N is Artinian.

Proof. Let N be Artinian. This implies immediately that M is Artinian. If {L_{i}} is a set of submodules of P then {g^{−1}(L_{i})} is a set of submodules of N. Let g^{−1}(L_{0}) be minimal in this set. Since gg^{−1}(L_{i}) = L_{i} we have that L_{0} is minimal in {L_{i}}. Let M and P be Artinian. Let {L_{i}} be a set of submodules of N. Let L_{0} be chosen such that g(L_{0}) is minimal in the set {g(L_{i})}. Let L be chosen such that f^{−1}(L) is minimal in the set {f^{−1}(L_{j})|L_{j} ∈ {L_{i}} and g(L_{j}) = g(L_{0})}. We show that L is minimal in {L_{i}}. Let L′ ∈ {L_{i}} with L ⊇ L′. Then g(L_{0}) = g(L) ⊇ g(L′), hence g(L′) = g(L_{0}). Furthermore we have f^{−1}(L) ⊇ f^{−1}(L′), hence L = L′. □

8.3. Simple and Semisimple Modules.

Lemma 8.13. Let R_{1}, …, R_{n} be semisimple rings. Then R_{1} ×…×R_{n} is a semisimple ring.

Proof. (Only for the case R_{1} × R_{2}) By Lemma 8.12 R_{1} × R_{2} is Artinian. Let I ⊆ R be nilpotent. From I^{n} = 0 we get for each a ∈ I the equation (Ra)^{n} = 0. From a = (a_{1}, a_{2}) follows 0 = (Ra)^{n} = (R_{1}a_{1}, R_{2}a_{2})^{n}. Hence R_{1}a_{1} = 0 and R_{2}a_{2} = 0, i.e. Ra = 0 and thus I = 0. □

Lemma 8.14. Each proper submodule N of a finitely generated module M is contained in a maximal submodule of M. In particular M possesses a simple quotient module.

Proof. Let N ⫋ M be a proper submodule of M. Let M be the set of submodules U with N ⊆ U ⫋ M. M is ordered by inclusion. Let (U_{i}) be a chain in M and U′ := ∪U_{i}. Then U′ is again a submodule and N ⫅ U′. If U′ = M then all generating elements m_{1}, …, m_{t} are in U′, hence there is a module U_{i} with m_{1}, …, m_{t} ∈ U_{i}. Thus U_{i} = M. This is impossible. So U′ ≠ M and thus in M. Furthermore U′ is an upper bound of (U_{i}). By Zorn’s Lemma there is a maximal submodule of M (in M), that contains N. □

Lemma 8.15. (1) If X ⊆ _{ℤ}ℚ is a set of generating elements of ℚ over ℤ and x ∈ X

then X \ {x} is also a set of generating elements of ℚ.

(2) _{ℤ}ℚ possesses no maximal submodules.

Proof. (1) Let B = ⟨X\{x}⟩. Then ℚ = ℤx+B. There is a y ∈ ℚ with 2y = x. We represent y as y = nx+b with n ∈ ℤ, b ∈ B. This implies x = 2y = 2nx+2b and thus (1−2n)x = 2b ∈ B. Furthermore there is a z ∈ ℚ with (1−2n)z = x, since obviously 1−2n ≠ 0. We represent z as z = mx+b′. This implies x = (1−2n)z = (1−2n)mx+(1−2n)b′ = 2mb+(1−2n)b′ ∈ B. Thus B = ℚ and we can omit x from the set of generating elements.

(2) Let N ⊆ ℚ be a maximal submodule and x ∈ ℚ \ N. Then N ∪ {x} is a set of generating elements of ℚ, hence also N. Contradiction. □

Lemma 8.16. Let _{R}M be a module in which each submodule is a direct summand. Then each submodule 0 ≠ N ⊆ M contains a simple submodule. Furthermore M is a sum of simple submodules.

Proof. Let x ∈ N, x ≠ 0. It suffices to show that Rx has a simple submodule. Since Rx is finitely generated Rx possesses a maximal submodule L. Since L is a direct summand of M, there is f : M ⟶ L with (L ⟶ Rx ⟶ M ⟶^{f} L) = 1_{L}, hence L ⊕ I = Rx, where I = Ke(Rx ⟶ M ⟶ L). If 0 ≠ J ⫋ I then L ⫋ L + J ⫋ Rx in contradiction to L maximal in Rx. Hence I is simple with I ⊆ Rx ⊆ N.

Let N := ∑ I_{j} be the sum of all simple submodules of M. Then M = N ⊕ K. If K ≠ 0 then K contains a simple submodule I and we have I ⊆ N ∩ K. Contradiction. Thus K = 0 and M = ∑ I_{j}. □

Lemma 8.17. Let _{R}M be a sum of simple submodules: M = ∑_{j∈X} I_{j}. Let N ⊆ M be a submodule. Then there is a set Y ⊆ X with M = N ⊕ ⨁_{j∈Y} I_{j} and a set Z ⊆ X with N ≅ ⨁_{j∈Z} I_{j}. In particular each submodule N of M is a direct sum of simple submodules.

Proof. Let S = {Z ⊆ X|N + (∑_{j∈Z} I_{j}) = N ⊕ (⨁_{j∈Z} I_{j})}. The set S is ordered by inclusion and not empty since ∅ ∈ S. Let (Z_{i}) be a chain in S. Then Z′ := ∪Z_{i} ∈ S. In order to show this let n + ∑_{j∈Z′} a_{j} = 0. Then at most finitely many a_{j} ∈ I_{j} are different from 0. Hence there is a Z_{i} in the chain with j ∈ Z_{i} for all a_{j} ≠ 0 in the sum. From N +(∑_{j∈Z_{i}} I_{j}) = N ⊕(⨁_{j∈Z_{i}} I_{j}) we get n = 0 = a_{j} for all j ∈ Z′. By Zorn’s Lemma there is a maximal element Z′′ ∈ S, and we have P := N + (∑_{j∈Z′′} I_{j}) = N ⊕ (⨁_{j∈Z′′} I_{j}). Let I_{k} be simple with k ∈ X \ Z′′. If P + I_{k} = P ⊕ I_{k}, then N + (∑_{j∈Z′′} I_{j}) + I_{k} = N ⊕ (⨁_{j∈Z′′} I_{j}) ⊕ I_{R} in contradiction to the maximality of Z′′. Hence 0 ≠ P ∩ I_{k} ⊆ I_{k}, or I_{k} ⊆ P. This implies P = N + ∑_{j∈X} I_{j} = M. Now we apply the first claim to ⨁_{j∈Y} I_{j} and obtain N ⊕ (⨁_{j∈Y} I_{j}) = M = (⨁_{j∈Y} I_{j}) ⊕ (⨁_{j∈Z} I_{j}). This implies N ≅ M/(⨁_{j∈Y} I_{j}) ≅ ⨁_{j∈Z} I_{j}. □

Theorem 8.18. (Structure Theorem for Semisimple Modules): For _{R}M the following are equivalent

(1) Each submodule of M is a sum of simple submodules.

(2) M is a sum of simple submodules.

(3) M is a direct sum of simple submodules.

(4) Each submodule of M is a direct summand.

Proof. (1) ⟹ (2): trivial.

(2) ⟹ (3): Lemma 8.17.

(3) ⟹ (1): Lemma 8.17.

(2) ⟹ (4): Lemma 8.17.

(4) ⟹ (2): Lemma 8.16. □

Definition 8.19. A module _{R}M is called semisimple, if it satisfies one of the equivalent conditions of Theorem 8.18.

Corollary 8.20. (1) Each submodule of a semisimple module is semisimple. (2) Each quotient (residue class) module of a semisimple module is semisimple.

(3) Each sum of semisimple modules is semisimple.

Proof. (1) trivial.

(2) Let N ⊆ M. Then M ≅ N ⊕ M/N, in particular M/N is isomorphic to a submodule of M.

(3) trivial. □

Remark 8.21. With the notion of a semisimple module we have obtained a particularly suitable generalization of the notion of a vector space. Important theorems of linear algebra have been generalized in Theorem 8.18. The simple modules over a field are exactly the one dimensional vector spaces. Condition (2) of Theorem 8.18 is trivially satisfied since each vector space is the sum of simple (one dimensional) vector spaces, one simply has to form V = ∑_{v∈V \{0}} Kv or V = ∑_{v∈E} Kv for an arbitrary set of generating elements E of V. Thus each vector space V is semisimple. So condition (3) holds. It says that each set of generating elements E contains a basis. (4) is the important statement that each subspace of a vector space has a direct complement. Lemma 8.17 also contains claims about the dimension of vector spaces, subspaces and quotient spaces.

Theorem 8.22. (Wedderburn) The following are equivalent for R:

(1) _{R}R is semisimple (as a ring).

(2) Each R-module is projective.

(3) Each R-module is injective.

(4) Each R-module is semisimple.

(5) _{R}R is semisimple (as an R-module).

(6) R is a direct sum of simple left ideals.

(7) R ≅ R_{1} × … × R_{n} with simple rings R_{i} (i = 1, …, n).

(8) R ≅ B_{1} ⊕ … ⊕ B_{n}, where the B_{i} are minimal two sided ideals and _{R}R is Artinian.

(9) R_{R} is semisimple (as a ring).

Proof. (1) ⟹ (3): Corollary 8.11.

(3) ⟹ (4): Theorem 8.18 (4) and Theorem 8.10 (3).

(4) ⟹ (5): Specialization.

(5) ⟹ (6): Theorem 8.18 (3).

(6) ⟹ (3): Theorem 8.18 (4) and 8.11.

(6) ⟹ (2): Theorem 8.18 (4) and 8.11.

(2) ⟹ (4): Let N ⊆ M be a submodule. Then M/N is projective, so there is f : M/N ⟶ M with (M/N ⟶ M ⟶ M/N) = id or (M ⟶ M/N ⟶ M) = p with p^{2} = p. Hence M = Ke(p) ⊕ Im(p) and Ke(p) = N.

(6) ⟹ (8): Let R = I_{11} ⊕ … ⊕ I_{1i_{1}} ⊕ I_{21} ⊕ … ⊕ I_{2i_{2}} ⊕ … ⊕ I_{n1} ⊕ … ⊕ I_{ni_{n}} be a direct sum of simple ideals, finitely many, since R is finitely generated, and let I_{ij} ≅ I_{ik} for all i, j, k and I_{i1} ≅ I̸_{j1} for i ≠ j. Let B_{k} := ⨁_{j=1}^{i_{k}} I_{kj}. Let I ⊆ R be simple. Let p_{k} : R ⟶ B_{k} be the projection onto B_{k} w.r.t. R = B_{1} ⊕ … ⊕ B_{n}. Then there is at least one k with p_{k}(I) ≠ 0. Then I ≅ p_{k}(I) = J ⊆ B_{k} is a simple ideal. Because of 8.17 we get I ⊕ (⨁_{j=r+1}^{m} I_{kj}) = B_{k} = I_{k1} ⊕ … ⊕ I_{kr} ⊕ (⨁_{j=r+1}^{m} I_{kj}) using a suitable numbering. Hence J ≅ I_{k1} ⊕ … ⊕ I_{kr} and thus r = 1 and I ≅ J ≅ I_{k1}. So there is a unique k with p_{k}(I) ≠ 0. In particular we have I ⊆ B_{k}. If f : _{R}R ⟶ _{R}R with f(I) ≠ 0 is given, then f(I) ≅ I is simple and f(I) ⊆ B_{k} for one k. So f(B_{k}) ⊆ B_{k} holds for all f ∈ Hom_{R}(.R, .R) ≅ R, and B_{k} is a two sided ideal. Observe that B_{i}B_{j} ⊆ B_{i} ∩ B_{j} = 0. For 1 ∈ R = B_{1} ⊕ … ⊕ B_{n} let 1 = e_{1} + … + e_{n} with e_{i} ∈ B_{i}. For b ∈ B_{i} we get e_{i}b = (e_{1} + … + e_{n})(0 + … + b + … + 0) = b = be_{i}. Thus B_{i} can be considered as ring with unit e_{i}. (B_{i} is not a subring of R but a quotient ring of R.) Since B_{i}B_{j} = 0 we have that L ⊆ B_{i} is a (one sided resp. two sided) B_{i}-ideal of B_{i} iff L is an R-ideal. Since B_{i} = I_{1} ⊕ … ⊕ I_{n} is a direct sum of simple R-ideals resp. B_{i}-ideals and since I_{j} ≅ I_{k} holds, B_{i} is a simple ring by Theorem 8.7. In particular B_{i} has no two sided nontrivial ideals, i.e. the two sided ideals B_{i} ⊆ R are minimal. 8.12 implies that R is Artinian.

(8) ⟹ (7): Since B_{i}B_{j} ⊆ B_{i} ∩ B_{j} = 0 the B_{i} are simple rings as above, hence R = R_{1} × … × R_{n} with R_{i} = B_{i}, because addition and multiplication are performed in the B_{i} (componentwise).

(7) ⟹ (1): Lemma 8.12.

(7) ⟹ (9): In order to have condition (7) symmetric in the sides, it suffices to show that a simple ring R is right Artinian. But R ≅ M_{n}(D) ≅ Hom_{D}(V^{∗}., V.^{∗}) is left and right Artinian. □

8.4. Noetherian Modules.

Definition 8.23. A module _{F} M is called Noetherian (Emmy Noether 1882-1935), if each nonempty set of submodules of M has a maximal element.

Theorem 8.24. For _{R}M the following are equivalent:

(1) M is Noetherian.

(2) Each ascending chain M_{i} ⊆ M_{i+1}, i ∈ ℕ of submodules of M becomes stationary, i.e. there is an n ∈ ℕ with M_{n} = M_{n+i} for all i ∈ ℕ.

(3) Each submodule of M is finitely generated.

Proof. (2) ⟹ (1): Let M be a nonempty set of submodules without a maximal element. Using the axiom of choice we choose for each N ∈ M an N′ ∈ M with N ⫋ N′. For N ∈ M we then have an ascending chain M_{1} = N, M_{i+1} = M′_{i} with

M_{1} ⫋ M_{2} ⫋ … ⫋ M_{i} ⫋ M_{i+1} ⫋ …

This is impossible by (2).

(1) ⟹ (3): Let M′ ⊆ M. Then {N|N ⊆ M′, N finitely generated} ≠ ∅ has a maximal element N′. If N′ ≠ M′, then there is an m ∈ M′ \ N′. So N′ + Rm ⊆ M′ is finitely generated and N′ ⫋ N′ + Rm in contradiction to the maximality of N′. Hence N′ = M′, i.e. M′ is finitely generated.

(3) ⟹ (2): Let M_{1} ⊆ M_{2} ⊆ … ⊆ M_{n} ⊆ … ⊆ M be an ascending chain of submodules of M. Let N := ⋃_{i∈ℕ} M_{i}. N is a finitely generated submodule of M, i.e. N = Ra_{1} + … + Ra_{n}. Then there is an M_{r} with a_{1}, …, a_{n} ∈ M_{r}. This implies M_{r} = N = M_{r+i} for all i ∈ ℕ, i.e. the chain becomes stationary. □

Lemma 8.25. Let 0 ⟶ M ⟶^{f} N ⟶^{g} P ⟶ 0 be a short exact sequence. M and P are Noetherian iff N is Noetherian. In particular if M and N are Noetherian then so is M ⊕ N.

Proof. Let N be Noetherian. Then it is clear that M Noetherian. If {L_{i}} is a set of submodules of P then {g^{−1}(L_{i})} is a set of submodules of N. Let g^{−1}(L_{0}) be maximal in this set. With gg^{−1}(L_{i}) = L_{i} we get that L_{0} is maximal in {L_{i}}. Let M and P be Noetherian. Let {L_{i}} be a set of submodules of N. Let L_{0} be chosen such that g(L_{0}) is maximal in the set {g(L_{i})}. Let L be chosen such that f^{−1}(L) is maximal in the set {f^{−1}(L_{j})|L_{j} ∈ {L_{i}} and g(L_{j}) = g(L_{0})}. We show that L is maximal in {L_{i}}. Let L′ ∈ {L_{i}} with L ⊆ L′. Then g(L_{0}) = g(L) ⊆ g(L′) hence g(L′) = g(L_{0}). Furthermore we have f^{−1}(L) ⊆ f^{−1}(L′) hence L = L′. □

Corollary 8.26. _{R}R is Noetherian as a left R-module iff all finitely generated left R-modules are Noetherian.

Proof. ⟸: trivial. ⟹: If M is finitely generated then there is a short exact sequence 0 ⟶ K ⟶ R ⊕ … ⊕ R ⟶ M ⟶ 0. Since R is Noetherian R⊕…⊕R Noetherian, too, so that M is Noetherian. □

Theorem 8.27. (Hilbert Basis Theorem) If R is left Noetherian then R[x] is left Noetherian.

Proof. Let J ⊆ R[x] be an ideal. We have to show that J finitely generated. Let J_{0} := {r ∈ R|∃p(x) ∈ J with highest coefficient r}. (The highest coefficient of the zero polynomial is 0 by definition.) J_{0} ⊆ R is an ideal, hence J_{0} = ⟨r_{1}, …, r_{n}⟩. For the r_{i} choose p_{i}(x) ∈ J with highest coefficients r_{i}. Let m ≥ deg(p_{i}(x)) for i = 1, …, n. Let g ∈ J with deg(g) ≥ m. Then g = sx^{t} + ∑_{i≤t} s_{i}x^{i}. Since s ∈ J_{0} we have s = ∑_{j=1}^{n} λ_{j}r_{j}. This implies g_{1} := g − ∑_{j=1}^{n} λ_{j}p_{j}(x)x^{t−deg(p_{j}(x))} ∈ J and deg(g_{1}) ≤ t − 1. By induction we have g = g_{0} + \overline{g} with g_{0} ∈ ∑_{j=1}^{n} R[x]p_{j}(x) and deg(\overline{g}) < m. This implies \overline{g} ∈ J ∩ (R + Rx + … + Rx^{m−1}) ⊆ R + Rx + … + Rx^{m−1}. Both R-modules are finitely generated hence \overline{g} = ∑_{i=1}^{k} μ_{i}q_{i}(X) with ⟨q_{1}(x), …, q_{k}(x)⟩ = J ∩ (R + Rx + … + Rx^{m−1}). Thus {p_{1}(x), …, p_{n}(x), q_{1}(x), …, q_{k}(x)} form a set of generating elements of J. □

Corollary 8.28. Let R be a commutative Noetherian ring and let S be a commutative Ralgebra. Let S be finitely generated as an R-algebra (i.e. there are s_{1}, …, s_{n} ∈ S such that for all s ∈ S there are representations s = ∑ r_{i_{1},...,i_{n}}s_{1}^{i_{1}} … s_{n}^{i_{n}}). Then S is Noetherian.

Proof. 1. By induction we have R[x_{1}, …, x_{n}] Noetherian.

2. There is an epimorphism R[x_{1}, …, x_{n}] ⟶ S. Thus S is a Noetherian R[x_{1}, …, x_{n}]-module hence it is also a Noetherian S-module. □

Proposition 8.29. Let R be commutative or M be Noetherian. Let M be finitely generated. Let f : N ⟶ M be an epimorphism where N ⊆ M is a submodule. Then f is an isomorphism.

Proof. 1. Let M be Noetherian. We construct an ascending chain K_{0} ⊆ K_{1} ⊆ K_{2} ⊆ … by K_{0} := Ke(f) = f^{−1}(0), K_{i} := f^{−1}(K_{i−1}). We have K_{0} = f^{−1}(0) ⊆ f^{−1}(K_{0}) = K_{1}. If K_{i−2} ⊆ K_{i−1} then we have K_{i−1} = f^{−1}(K_{i−2}) ⊆ f^{−1}(K_{i−1}) = K_{i}. Since M is Noetherian the chain becomes stationary K_{n} = K_{n+1} = . . .. Let x_{0} ∈ K_{0}. We want to show x_{0} = 0. There is x_{1} ∈ K_{1} with f(x_{1}) = x_{0}, since f is an epimorphism. Similarly there are x_{0}, x_{1}, x_{2}, … with f(x_{i}) = x_{i−1} and f^{n+1}(x_{n+1}) = f^{n}(x_{n}) = … = f(x_{1}) = x_{0}. Since the chain becomes stationary we get x_{n+1} ∈ K_{n}, which implies f(x_{n+1}) ∈ K_{n−1} and thus f^{n}(x_{n+1}) ∈ K_{0}. Hence x_{0} = f^{n+1}(x_{n+1}) = 0. This proves that f is a monomorphism.

2. Let R commutative. Let M = Ry_{1} + … + Ry_{n}. Let x_{i} ∈ N_{i} with f(x_{i}) = y_{i}. Let x_{0} ∈ N with f(x_{0}) = 0. Then there are coefficients r_{ij} ∈ R with x_{i} = ∑_{j=1}^{n} r_{ij}y_{j}, i = 0, …, n. We consider R′ := ℤ[r_{ij}] ⊆ R, the subring of R generated by the r_{ij}. Since ℤ is Noetherian and R′ is finitely generated as a ℤ-algebra R′ is Noetherian. Let M′ := ∑_{i=1}^{n} R′y_{i} ⊆ M and N′ = ∑_{i=0}^{n} R′x_{i} ⊆ N. Then N′ ⊆ M′ is an R′-submodule, M′ as an R′-module is finitely generated, hence Noetherian, and the f(x_{i}) = y_{i}, f(x_{0}) = 0 generate a homomorphism of R′-modules f′ : N′ ⟶ M′. Since f′ is surjective f′ is injective and thus x_{0} = 0 so that f is injective. □

Problem 8.1. Where does the commutativity of R enter the second part of the proof of Proposition 8.29?

Corollary 8.30. Let R be commutative or _{R}M be Noetherian. Let M = Ry_{1} + … + Ry_{m}. Let N ⊆ M be a free submodule with the free generating elements x_{1}, …, x_{n}. Then n ≤ m. If n = m then M is free over y_{1}, …, y_{m}.

Proof. Since N is free there is a homomorphism f : N ⟶ M with f(x_{i}) = y_{i} for i = 1, …, min(m, n) and f(x_{i}) = 0 else. If n ≥ m then f is surjective, hence bijective. Thus we have n ≤ m. If n = m then f is bijective and M free with the generating elements y_{1}, …, y_{n}. □

Corollary 8.31. Let R be commutative or Noetherian. Let M be free over x_{1}, …, x_{n} and free over y_{1}, …, y_{m}. Then we have m = n.

Proof. If R is Noetherian then M is also Noetherian. Thus the claim follows from 8.30. □

Definition 8.32. Let R be commutative or Noetherian. The rank of a finitely generated free module _{R}M is the number of free generating elements uniquely determined by 8.31.

Example 8.33. The endomorphism ring of a countably infinite dimensional vector space is neither left nor right Noetherian.

Proof. From ap + bq = 1, pa = 1, qb = 1, pb = 0, qa = 0 we get (as in the exercise 1.4) RR = _{R}Rp ⊕_{R}Rq free and R_{R} = aR_{R} ⊕ bR_{R} free. □

Definition 8.34. An element r ∈ R in a ring R is called a left unit (right unit), if rR = R (Rr = R). r ∈ R is called a unit, if Rr = R = rR.

Lemma 8.35. If r ∈ R is a unit, then there is a unique s ∈ R with sr = 1. Furthermore we have rs = 1 and s is a unit.

Proof. Let sr = s′r = 1 and let rt = 1. Then s = s1 = srt = 1t = t and analogously s′ = t. □

Corollary 8.36. In each left Noetherian ring R each right unit x ∈ R (i.e. Rx = R) is also a left unit and conversely.

Proof. Let Rx = R. Then ·x : R ⟶ R is an epimorphism, hence an isomorphism. So there is an inverse isomorphism g : R ⟶ R with g ∈ Hom_{R}(.R, .R) ≅ R, hence g = ·y. This implies 1 · x · y = 1 and 1 · y · x = 1, i.e. x^{−1} = y and x is a unit. If xR = R then there is a y ∈ R with xy = 1. So y is a right unit hence y is a unit. By 8.36 x is the unique inverse of y, hence x is a unit. □

9. Radical and Socle

Definition 9.1. (1) N ⊆ M is called large (essential) iff

∀U ⊆ M : N ∩ U = 0 ⟹ U = 0.

(2) N ⊆ M is called small (superfluous) iff

∀U ⊆ M : N + U = M ⟹ U = M.

Lemma 9.2. Let N ⊆ M ⊆ P, U ⊆ P be submodules. Then the modular law holds:

N + (U ∩ M) = (N + U) ∩ M.

Proof. ⊆: From n + u ∈ N + U with n ∈ N and u ∈ U ∩ M ⊆ M it follows that n + u ∈ M and hence n + u ∈ (N + U) ∩ M. ⊇: From n + u = m ∈ (N + U) ∩ M it follows that u = m − n ∈ M ∩ U and hence n + u ∈ N + (U ∩ M). □

Lemma 9.3. (1) Let N ⊆ N′ ⊆ M′ ⊆ M be submodules and let N be large in M. Then

N′ is large in M′.

(2) Let N ⊆ N′ ⊆ M′ ⊆ M be submodules and let N′ be small in M′. Then N is small in M.

(3) Let N, N′ ⊆ M be large submodules in M. Then N ∩ N′ is large in M.

(4) Let N, N′ ⊆ M be small submodules in M. Then N + N′ is small in M.

Proof. (1) Let U ⊆ M′ with N′ ∩ U = 0. Then N ∩ U = 0 hence U = 0.

(2) Let U ⊆ M with N +U = M, then N′ +U = M. From N′ +(U ∩M′) = (N′ +U)∩M′ = M ∩ M′ = M′ we get U ∩ M′ = M′ and thus M′ ⊆ U which implies N ⊆ U. Now from N + U = M we get U = M.

(3) Let (N ∩ N′) ∩ U = 0. Then N ∩ (N′ ∩ U) = 0 hence N′ ∩ U = 0 and thus U = 0.

(4) Let (N +N′)+U = M. Then N +(N′+U) = M hence N′+U = M and thus U = M. □

Lemma 9.4. Let N, U ⊆ M be submodules.

(1) If N is maximal w.r.t. the condition N ∩ U = 0 then N + U ⊆ M is a large submodule.

(2) If N is minimal w.r.t. the condition N + U = M then N ∩ U ⊆ M is a small submodule.

(3) There is a submodule N that is maximal w.r.t. N ∩ U = 0.

Proof. (1) Let V ⊆ M with (N + U) ∩ V = 0 be given. We have N ∩ U = 0. Let n+v = u ∈ (N +V)∩U. This implies v = u−n ∈ (N +U)∩V = 0 hence n = u ∈ N ∩U = 0 and (N + V) ∩ U = 0. Thus N + V = N, since N is maximal w.r.t. N ∩ U = 0. This implies V ⊆ N hence V ⊆ (N + U) ∩ V = 0 and V = 0. So we get that N + U ⊆ M is large.

(2) Let V ⊆ M with (N ∩ U) + V = M. We have N + U = M. Let m ∈ M with m = n + u ∈ N + U. Furthermore let n = n′ + v with n′ ∈ N ∩ U and v ∈ V (since n ∈ M). This implies v ∈ V ∩ N and m = (n′ + u) + v ∈ U + (V ∩ N) and thus (N ∩ V) + U = M. Since N is minimal w.r.t. N + U = M we have N = N ∩ V hence N ⊆ V. From this and from (N ∩ U) + V = M we get V = M. Thus N ∩ U ⊆ M is small.

(3) The set V := {V ⊆ M|V ∩ U = 0} is inductively ordered, for let (V_{i})_{i∈I} be a chain in V and let x ∈ (∪V_{i}) ∩ U. Then there is an i ∈ I with x ∈ V_{i} ∩ U hence x = 0. Thus ∪V_{i} in V is an upper bound of the V_{i}. Consequently there is a submodule N of M that is maximal w.r.t. N ∩ U = 0. □

Lemma 9.5. N ⊆ M is large if and only if the following holds

∀m ∈ M \ {0}∃r ∈ R : rm ∈ N \ {0}.

Proof. N ⊆ M large ⟺ [∀U ⊆ M : N ∩ U = 0 ⟹ U = 0] ⟺ [∀U ⊆ M : U ≠ 0 ⟹

N ∩ U ≠ 0] ⟺^{(∗)} [∀Rm ⊆ M : Rm ≠ 0 ⟹ N ∩ Rm ≠ 0] ⟺ [∀m ∈ M \ {0}∃r ∈

R : rm ∈ N \ {0}]. Only one direction (∗) needs an additional argument. If U ≠ 0 and the right hand side of (∗) holds, then there exists an m ∈ U with Rm ≠ 0. Hence we get 0 ≠ N ∩ Rm ⊆ N ∩ U. □

Lemma 9.6. Let Rm ⊆ M be not small. Then there exists a submodule N ⊆ M that is a maximal submodule and that does not contain m.

Proof. The set S := {U ⫋ M|Rm + U = M} is not empty since Rm is not small in M. S is inductively ordered. In fact let (U_{i}|i ∈ I) be a chain in S. Then we have m ∉ U_{i} for all i ∈ I. Hence ∪U_{i} ⫋ M and obviously Rm + (∪U_{i}) = M. Then there is a maximal element N in S. Let N ⫋ N′ ⊆ M. Then Rm + N′ = M. Since N′ ∉ S we get N′ = M hence N is a maximal submodule. Furthermore we have obviously m ∉ N. □

Definition 9.7. (1) Radical(M) = Rad(M) := ∩{U ⫋ M|U maximal submodule}, (2) Socle(M) = Soc(M) := ∑{U ⊆ M|U simple submodule}.

Proposition 9.8. (1) Rad(M) = ∑{V ⊆ M small}.

(2) Soc(M) = ∩{V ⊆ M large}.

Proof. (1) ⊇: Let V ⊆ M small. For all maximal submodules U ⊆ M we have U ⊆ U + V ⫋ M since V is small and U ≠ M. This implies U = U + V and V ⊆ U. Thus V ⊆ ∩U and thus ∑ V ⊆ ∩U. ⊆: If Rm is not small in M then by 9.6 there is a maximal submodule N in M with m ∉ N. So we have m ∉ ∩U = Rad(M) ⊆ N. If also m ∈ Rad(M) holds then Rm is small in M. So we get m ∈ ∑{V ⊆ M small}.

(2) ⊆: Let V be large in M and let U be simple. Then we have V ∩U ≠ 0 so that V ∩U = U and thus U ⊆ V. This implies ∑ U ⊆ ∩V. ⊇: First we show that each submodule of ∩V_{i} is a direct summand of ∩V_{i}. Let N ⊆ ∩V_{i} be given. Let X be maximal in M with N ∩ X = 0 (Lemma 9.4 (3)). Then N + X = V ⊆ M is large by Lemma 9.4 (1). This implies N + (X ∩ (∩V_{i})) = (N + X) ∩ (∩V_{i}) (Lemma 9.2) = V ∩ (∩V_{i}) = ∩V_{i} and N ∩ (X ∩ (∩V_{i})) = 0. So we have N ⊕ (X ∩ (∩V_{i})) = ∩V_{i}. Theorem 8.16 implies that ∩V_{i} is a sum of simple submodules of ∩V_{i}. Thus ∩V_{i} is contained in the sum of all simple submodules of M, i.e. in the socle of M. □

Remark 9.9. A module M is semisimple if and only if it coincides with its socle.

Corollary 9.10. m ∈ Rad(M) iff Rm ⊆ M is small.

Proof. ⟸: by Proposition 9.8. ⟹: was explicitly noted in the proof of Proposition 9.8. □

Corollary 9.11. Each finitely generated submodule of Rad(M) is small in M.

Proof. By 9.10 the modules Rm_{1}, …, Rm_{n} ⊆ M are small, if m_{1}, …, m_{n} ∈ Rad(M). By 9.3 (4) we then get that ∑_{i=1}^{n} Rm_{i} is small in M. □

Proposition 9.12. Let M be finitely generated. Then Rad(M) is small in M.

Proof. Since M is finitely generated each proper submodule of M is contained in a maximal submodule (8.14). Let N ⫋ M and let U be a maximal submodule with N ⊆ U ⫋ M. Then Rad(M) ⊆ U thus Rad(M) + N ⊆ U ⫋ M. So Rad(M) is small in M. □

Proposition 9.13. Let f ∈ Hom_{R}(M, N). Then we have (1) f(Rad(M)) ⊆ Rad(N).

(2) f(Soc(M)) ⊆ Soc(N).

Proof. (1) Let U ⊆ M be small. Let V ⊆ N with f(U) + V = N. This implies f^{−1}(f(U) + V) = f^{−1}(N) = M = U + f^{−1}(V), because f(x) = f(u) + v implies f(x − u) = v, x − u ∈ f^{−1}(V) and thus x ∈ U + f^{−1}(V), so f^{−1}(f(U) + V) ⊆ U + f^{−1}(V). Since U is small we get f^{−1}(V) = M. This implies f(f^{−1}(V)) = f(M) ⊆ V, hence f(U) ⊆ V and V = N. So we have f(U) small in M. This shows f(Rad(M)) = ∑_{U small} f(U) ⊆ ∑_{V small} V = Rad(N).

(2) Let U ⊆ M be simple. Then f(U) ⊆ N is simple or 0. So we have f(∑ U_{i}) ⊆ Soc(N). □

Corollary 9.14. Rad and Soc are covariant subfunctors of Id : R-Mod ⟶ R-Mod.

Corollary 9.15. (1) Let U ⊆ M be small and f ∈ Hom_{R}(M, N). Then f(U) ⊆ N is

small.

(2) Let U ⊆ N be large and f ∈ Hom_{R}(M, N). Then f^{−1}(U) ⊆ M is large.

Proof. (1) was proved in Proposition 9.13 (1).

(2) Let V ⊆ M and f^{−1}(U) ∩ V = 0. Then f(f^{−1}(U) ∩ V) = 0 = ff^{−1}(U) ∩ f(V), because if x ∈ ff^{−1}(U) ∩ f(V) with x = f(v), then f(v) ∈ U by ff^{−1}(U) ⊆ U. This implies v ∈ f^{−1}(U) ∩ V, so x ∈ f(f^{−1}(U) ∩ V) = 0. Now this implies 0 = ff^{−1}(U) ∩ f(V) = U ∩ Im(f) ∩ f(V) = U ∩ f(V) and thus f(V) = 0, because U is large in N. So we have V ⊆ Ke(f) ⊆ f^{−1}(U). From f^{−1}(U) ∩ V = 0 we get V = 0. Thus f^{−1}(U) is large in M. □

Corollary 9.16. (1) Rad(_{R}R)M ⊆ Rad(M).

(2) Soc(_{R}R)M ⊆ Soc(M).

Proof. Let m ∈ M. Then (R ∋ r ↦ rm ∈ M) ∈ Hom_{R}(R, M). This implies Rad(_{R}R)m ⊆ Rad(M), Soc(_{R}R)m ⊆ Soc(M) and that implies the claim. □

Corollary 9.17. Rad(_{R}R) and Soc(_{R}R) are two sided ideals.

Proposition 9.18. Let f ∈ Hom_{R}(M, N) and Ke(f) ⊆ Rad(M). Then we have

f(Rad(M)) = Rad(f(M)).

Proof. ⊆: follows from 9.13. ⊇: Let f(m) ∈ Rad(f(M)). If Rm ⊆ M is small then m ∈ Rad(M) and f(m) ∈ f(Rad(M)). If Rm ⊆ M is not small then by 9.6 there is a maximal submodule U ⫋ M with m ∉ U. We have Rm + U = M and thus f(U) + Rf(m) = f(M). From f(m) ∈ Rad(f(M)) we get that Rf(m) ⊆ f(M) is small. This implies f(U) = f(M) and thus U + Ke(f) = M. From the assumption Ke(f) ⊆ Rad(M) ⊆ U we get U = M, a contradiction. □

Corollary 9.19. Let N ⊆ M be a submodule. Then the following hold (1) (Rad(M) + N)/N ⊆ Rad(M/N).

(2) N ⊆ Rad(M) ⟹ Rad(M)/N = Rad(M/N).

Proof. (1) f : M ⟶ M/N implies f(Rad(M)) ⊆ Rad(M/N) and f(Rad(M)) = (Rad(M) + N)/N.

(2) From N = Ke(f) ⊆ Rad(M) the claim follows. □

Corollary 9.20. Rad(M) is the smallest submodule U ⊆ M with Rad(M/U) = 0.

Proof. We have Rad(M/ Rad(M)) = Rad(M)/ Rad(M) = 0. If Rad(M/U) = 0 then Rad(M) + U/U = 0 and thus Rad(M) + U = U so that Rad(M) ⊆ U. □

Lemma 9.21. If Soc(M) = M then Rad(M) = 0.

Proof. If Soc(M) = M holds then M is semisimple. So no submodule is small and thus Rad(M) = 0. □

Lemma 9.22. Let M be Artinian. Then we have

Rad(M) = 0 ⟺ Soc(M) = M.

Proof. Let M be Artinian and Rad(M) = 0. Let U ⊆ M and N be minimal with N +U = M. By 9.4 (2) we have N ∩ U ⊆ M small so that N ∩ U = 0. Thus U is a direct summand of M, M is semisimple and M = Soc(M). □

Proposition 9.23. The following are equivalent for M:

(1) M is finitely generated and semisimple.

(2) M is Artinian and Rad(M) = 0.

Proof. It suffices to show the following: If M is semisimple, then M is finitely generated iff M is Artinian. Let M be semisimple. Then M = ⊕U_{i} with simple modules U_{i}. M is finitely generated if and only if the direct sum has only finitely many summands (≠ 0). If M is Artinian then the direct sum has only finitely many summands. If the direct sum has only finitely many summands, then each descending chain N_{1} ⊇ N_{2} ⊇ … in M can only have finitely many direct complements by 8.17. Thus such a chain must become stationary, i.e. M is Artinian. □

Proposition 9.24. (Lemma of Nakayama) For _{R}I ⊆ _{R}R the following are equivalent: (1) I ⊆ Rad(_{R}R).

(2) 1 + I contains only right units.

(3) 1 + I contains only units.

(4) 1 + IR contains only units.

(5) IM = M ⟹ M = 0 for all finitely generated modules _{R}M.

(6) IM + U = M ⟹ U = M for all finitely generated modules _{R}M.

(7) IM ⊆ Rad(_{R}M) for all finitely generated modules _{R}M.

Proof. (1) ⟹ (2): Rad(R) ⊆ R is small. Thus I ⊆ R is small. From R(1 + i) + I = R it follows R(1 + i) = R. Thus 1 + i is a right unit.

(2) ⟹ (3): Let k(1 + i) = 1. This implies ki = 1 − k ∈ I and thus k − 1 ∈ I. So k = 1 + (k − 1) is a right unit. Since k is also a left unit, we get (1 + i)k = 1, so that 1 + i is a unit.

(3) ⟹ (4): Given i ∈ I and r ∈ R. Then 1 + ri is a unit with inverse (1 + ri)^{−1}. Since (1 − i(1 + ri)^{−1}r)(1 + ir) = 1 + ir − i(1 + ri)^{−1}(r + rir) = 1 + ir − i(1 + ri)^{−1}(1 + ri)r = 1 + ir − ir = 1 and symmetrically (1 + ir)(1 − i(1 + ri)^{−1}r) = 1 we get that 1 + ir is a unit. If a is a unit and i ∈ I, r ∈ R then a + ir is a unit, since a(1 + a^{−1}ir) = (a + ir) is a product of two units by a^{−1}i ∈ I. If ∑_{k=1}^{n} i_{k}r_{k} ∈ IR then 1 + ∑ i_{k}r_{k} is a unit, since 1 + ∑ i_{k}r_{k} = (((1 + i_{1}r_{1}) + i_{2}r_{2}) … + i_{n}r_{n}) and each of the bracketed terms is a unit.

(4) ⟹ (5): Let M be finitely generated and IM = M. Let t be the minimal length of a system of generators of M = Rm_{1} + … + Rm_{t}. By IM = M each element in M can be represented as a finite sum of the form ∑ i′_{j}m′_{j}; the m′_{j} can be represented as a linear combination of the m_{i}. So there are coefficients i_{k}r_{k} ∈ I with m_{1} = ∑_{k=1}^{t} i_{k}r_{k}m_{k}. This implies (1 − i_{1}r_{1})m_{1} = ∑_{k=2}^{t} i_{k}r_{k}m_{k}. Since also 1 − i_{1}r_{1} is a unit, we get m_{1} = ∑_{k=2}^{t}(1 − i_{1}r_{1})^{−1}i_{k}r_{k}m_{k} ∈ Rm_{2} + … + Rm_{t} a contradiction to the minimality of t. So we have M = 0.

(5) ⟹ (6): IM + U = M ⟹ I(M/U) = (IM + U)/U = M/U ⟹ M/U = 0 ⟹ M = U.

(6) ⟹ (7): IM small in M ⟹ IM ⊆ Rad(M).

(7) ⟹ (1): M = R ⟹ IR ⊆ Rad(_{R}R). □

Corollary 9.25. Rad(_{R}R) = Rad(R_{R}).

Proof. Let I = Rad(_{R}R). Then 1 + I consists of units. Since I is a right ideal, we get I ⊆ Rad(R_{R}). By symmetry we get Rad(_{R}R) = Rad(R_{R}). □

Lemma 9.26. R left Artinian ⟹ R/ Rad(R) semisimple.

Proof. By 8.12 R/ Rad(R) is Artinian. By 9.20 Rad(R/ Rad(R)) = 0 and by 9.23 R/ Rad(R) is semisimple. □

Lemma 9.27. R Artinian ⟹ Rad(R) nilpotent.

Proof. Let I := Rad(R). Since R is Artinian, the chain I ⊇ I^{2} ⊇ I^{3} ⊇ … ⊇ I^{t+1} = … becomes stationary. Assume I^{t} ≠ 0. Since also I^{t}I ≠ 0 there is a minimal module K ⊆ I w.r.t. I^{t}K ≠ 0. So there exists an x ∈ K with I^{t}x ≠ 0, i.e. we have K = Rx. Because of I^{t}K = I^{t+1}K = I^{t}(IK) ≠ 0 and IK ⊆ K we get IK = K. By the Lemma of Nakayama we get K = 0, a contradiction, so I^{t} = 0. □

Theorem 9.28. (Hopkins) Let _{R}R be Artinian. Then _{R}R is Noetherian.

Proof. Let I := Rad(R) and I^{n+1} = 0. Then I^{i}/I^{i+1} is an R/I-module and it is Artinian as an R-module. So I^{i}/I^{i+1} is also Artinian as R/I-module. By 9.26 R/I is semisimple hence I^{i}/I^{i+1} is also semisimple, i.e. I^{i}/I^{i+1} = ⊕_{k∈X}E_{k} with simple R/I-modules E_{k}. Since I^{i}/I^{i+1} is Artinian the direct sum is finite hence I^{i}/I^{i+1} are Noetherian (as R/I-module and as Rmodule). With the exact sequences 0 ⟶ I^{i+1} ⟶ I^{i} ⟶ I^{i}/I^{i+1} ⟶ 0, with I^{n+1} = 0, I^{0} = R and with 8.25 we get by induction that R is Noetherian. □

Corollary 9.29. If _{R}I ⊆ _{R}R is nilpotent then I ⊆ Rad(R).

Proof. Let I^{n} = 0 and i ∈ I. Then (1 + i) · (1 − i + i^{2} − … ± i^{n+1}) = 1 hence (1 + i) is a unit. By the Lemma of Nakayama we get I ⊆ Rad(R). □

Proposition 9.30. _{R}M is finitely generated if and only if (1) Rad(M) ⊆ M is small, and

(2) M/ Rad(M) is finitely generated.

Proof. ⟹: trivial by 9.12 . ⟸: Let \{\overline{x}_{i} = x_{i} + Rad(M)|i = 1, …, n\} be a set of generating elements of M/ Rad(M). Then M = Rx_{1} + … + Rx_{n} + Rad(M) which implies by (1) that M = Rx_{1} + … + Rx_{n}. □

Corollary 9.31. M is Noetherian if and only if for all submodules U ⊆ M the following hold:

(1) Rad(U) ⊆ U is small.

(2) U/ Rad(U) is finitely generated

10. Localization

10.1. Local rings.

Definition 10.1. Let R be a ring. An element r ∈ R is called a non unit, if r is not a unit. The element r is called invertible, if r is a left or a right unit. R is called a local ring, if the sum of any two non invertible elements is a non unit.

Lemma 10.2. Let r be an idempotent (r^{2} = r) in a local ring R. Then r = 0 or r = 1.

Proof. We have (1 − r)^{2} = 1 − 2r + r^{2} = 1 − r. Since 1 = (1 − r) + r is a unit, r or 1 − r is invertible. If r is invertible, e.g. by sr = 1, then we have r = sr^{2} = sr = 1. If 1 − r is invertible e.g. by s(1 − r) = 1, then we have 1 − r = 1, thus r = 0. □

Lemma 10.3. Let R be a ring with the unique idempotents 0 and 1. Then each invertible element in R is a unit.

Proof. Let r be invertible e.g. by sr = 1. Then (rs)^{2} = rsrs = rs, so rs ∈ {0, 1}. If rs = 0, then we have 1 = (sr)^{2} = srsr = 0, a contradiction. So we have rs = 1, i.e. r is a unit. □

Corollary 10.4. In a local ring R all non units are not invertible.

Proposition 10.5. Let R be a local ring. Then the following hold:

(1) All non units are not invertible and form a two sided ideal N.

(2) N is the only maximal (one sided and two sided) and largest ideal of R.

Proof. (1) Let N be the set of the non units of R. Since R is local, so non units are not invertible, N is closed w.r.t. to the addition. Given s ∈ N and r ∈ R. We show that also rs ∈ N holds. In fact if rs ∉ N then rs is a unit, so there is a t ∈ R with trs = 1. Because of 10.3 s is also a unit in contradiction to s ∈ N. Thus N is a two sided ideal.

(2) Obviously we have N ⫋ R. If I ⫋ R and s ∈ I, then Rs ⫋ R, so s is a non unit and thus s ∈ N. So I ⊆ N holds. □

Proposition 10.6. R is local, if and only if R possesses a unique maximal (largest) left ideal.

Proof. ⟹: follows from 10.5. ⟸: Let N be the only maximal ideal of R. Then N = Rad(R) is a two sided ideal. Let r ∈ R \ N. Then N + Rr = R. Since N = Rad(R) is small in R, we have Rr = R, so there is a t with tr = 1. If t is a right unit, then also r is a unit by Lemma 8.35. But if t is not a right unit, then Rt ≠ R, so Rt ⊆ N and thus t ∈ N. Since N is a two sided ideal we have also 1 = tr ∈ N, a contradiction. Thus each r ∈ R \ N is a unit. So each non unit lies in N. If x, y are non units, then it follows from x, y ∈ N that x + y ∈ N hence x + y is a non unit and thus R is local. □

Lemma 10.7. Let R be a local ring with maximal ideal 𝔪 ⫋ R. Let M be a finitely generated module. If M/𝔪M = 0 then M = 0.

Proof. From 𝔪 = Rad(R) and 𝔪M = M it follows that M = 0 by the Lemma of Nakayama.

□

10.2. Localization. In this section let R be always a commutative ring. Recall from Basic Algebra: A set S with ∅ ⫋ S ⊂ R is called multiplicatively closed, if

∀s, s′ ∈ S : ss′ ∈ S  and  0 ∉ S.

On R × S define an equivalence relation by

(r, s) ∼ (r′, s′) :⟺ ∃t ∈ S : tsr′ = ts′r.

R[S^{−1}] = S^{−1}R := R × S/ ∼ is a commutative ring with unit element. The elements are denoted by

\frac{r}{s} := \overline{(r, s)}.

The map

φ : R ∋ r ↦ \frac{sr}{s} ∈ R[S^{−1}]

is a homomorphism of rings. It is independent of the choice of s ∈ S. If R has no zero divisors, then φ is injective.

Proposition 10.8. Let S ⊆ R be a multiplicatively closed set. Let _{R}M be an R-module. Then the relation

(m, s) ∼ (m′, s′) :⟺ ∃t ∈ S : tsm′ = ts′m

on M × S is an equivalence relation. Furthermore

S^{−1}M := M × S/ ∼ with the elements \frac{m}{s} := \overline{(m, s)}

is an S^{−1}R-module with the operations

\frac{m}{s} + \frac{m′}{s′} = \frac{s′m + sm′}{ss′} and \frac{r}{s} \frac{m}{s}′ = \frac{rm}{ss′}.

Proof. as in Basic Algebra for S^{−1}R. □

Problem 10.1. Give a complete proof of Proposition 10.8.

Lemma 10.9. \frac{m}{s} = 0 holds in S^{−1}M if and only if there is a t ∈ S with tm = 0.

Proof. (m, s) ∼ (0, s′) ⟺ ∃t′ ∈ S : t′s′m = 0 ⟺ ∃t′s′ ∈ S : t′s′m = 0. □

Lemma 10.10. (1) φ_{M} : M ∋ m ↦ \frac{sm}{s} ∈ S^{−1}M is a homomorphism of groups indepen-

dent of s ∈ S.

(2) φ_{M} is injective if and only if S contains no zero divisors for M, i.e. sm = 0 ⟹ m = 0.

(3) φ_{M} is bijective if and only if the map M ∋ m ↦ sm ∈ M is bijective for all s ∈ S.

(4) φ_{R} is a homomorphism of rings.

(5) φ_{M} : M ⟶ S^{−1}M is φ_{R}-semilinear, i.e. φ_{M}(rm) = φ_{R}(r)φ_{M}(m).

Proof. (1) t′(tsm − stm) = 0 implies \frac{sm}{s} = \frac{tm}{t}.

(2) φ_{M}(m) = 0 ⟺ \frac{sm}{s} = 0 ⟺ ∃t ∈ S : tm = 0 by 10.9.

(3) φ_{M} surjective ⟺ ∀\frac{m}{s} ∈ S^{−1}M ∃m′ ∈ M : \frac{sm′}{s} = \frac{m}{s} ⟺ ∀m ∈ M, s ∈ S ∃m′ ∈ M : sm′ = m ⟺ ∀s ∈ S : (s· : M ⟶ M) surjective.

(4) + (5) φ_{M}(rm) = \frac{s^{2}rm}{s^{2}} = \frac{sr}{s} \frac{sm}{s} = φ_{R}(r)φ_{M}(m). □

Corollary 10.11. S^{−1} : R-Mod ⟶ S^{−1}R-Mod is an additive functor.

Proof. For f ∈ Hom_{R}(M, N) we form S^{−1}f ∈ Hom_{S^{−1}R}(S^{−1}M, S^{−1}N) by S^{−1}f(\frac{m}{s}) := \frac{f(m)}{s}. In order to show that S^{−1}f is a well defined map assume (m, s) ∼ (m′, s′). Then ts′m = tsm′ for a t ∈ S and thus ts′f(m) = tsf(m′). This implies \frac{f(m)}{s} = \frac{f(m′)}{s′}. With the usual rules for calculations with fractions one proves that S^{−1}f is an S^{−1}Rhomomorphism and that S^{−1} id_{M} = id_{S^{−1}M}, S^{−1}(fg) = S^{−1}(f)S^{−1}(g) and S^{−1}(f + g) = S^{−1}(f) + S^{−1}(g) hold. □

Localization 83

Proposition 10.12. The map

α(M) : S^{−1}R ⊗_{R} M ∋ \frac{r}{s} ⊗ m ↦ \frac{rm}{s} ∈ S^{−1}M

defines a functorial isomorphism

α : S^{−1}R ⊗_{R} M ≅ S^{−1}M

of functors S^{−1}R ⊗_{R} -, S^{−1}- : R-Mod ⟶ S^{−1}R-Mod.

Proof. α(M) is a well defined map, for α̃(M) : S^{−1}R × M ∋ (\frac{r}{s}, m) ↦ \frac{rm}{s} ∈ S^{−1}M is well defined: (\frac{r}{s}, m) = (\frac{r′}{s′} , m) ⟹ ∃t ∈ S : ts′r = tsr′ ⟹ ts′rm = tsr′m ⟹ \frac{rm}{s} = \frac{r′m}{s′}. Furthermore α̃(M) is obviously additive in both arguments. Finally we have α̃(M)(\frac{r}{s}t, m) = \frac{rtm}{s} = α̃(M)(\frac{r}{s}, tm), i.e. α̃(M) is R-bilinear. We define an inverse map β(M) : S^{−1}M ∋ \frac{m}{s} ↦ \frac{t}{st} ⊗m ∈ S^{−1}R⊗_{R}M. The map β(M) is well defined, since \frac{m}{s} = \frac{m′}{s′} ⟹ ∃t′ ∈ S : t′s′m = t′sm′ ⟹ \frac{t}{st} ⊗ m = \frac{ts′t′}{sts′t′} ⊗ m = \frac{t}{sts′t′} ⊗ s′t′m = \frac{t}{sts′t′} ⊗ st′m = \frac{tst′}{sts′t′} ⊗ m′ = \frac{t}{s′t} ⊗ m′. We have βα = id, since β(M)α(M)(\frac{r}{s} ⊗ m) = β(M)(\frac{rm}{s}) = \frac{t}{st} ⊗ rm = \frac{rt}{st} ⊗ m = \frac{r}{s} ⊗ m. Similarly we have αβ = id, since α(M)β(M)(\frac{m}{s}) = α(M)(\frac{t}{st} ⊗ m) = \frac{tm}{st} = \frac{m}{s}. α is an S^{−1}R-homomorphism, since α(M)(\frac{r′}{s′} \frac{r}{s} ⊗ m) = α(M)(\frac{r′r}{s′s} ⊗ m) = \frac{r′rm}{s′s} = \frac{r′}{s′} \frac{rm}{s} = \frac{r′}{s′} α(M)(\frac{r}{s} ⊗ m). α is a functorial homomorphism. In fact the diagram

[figure: S^{−1}R ⊗_{R} N; S^{−1}N; α(N); S^{−1}R ⊗_{R} M; S^{−1}M; α(M); S^{−1}R ⊗_{R} f; S^{−1}f]

commutes since we have S^{−1}f ∘ α(M)(\frac{r}{s} ⊗ m) = S^{−1}f(\frac{rm}{s}) = \frac{f(rm)}{s} = \frac{rf(m)}{s} = α(N)(\frac{r}{s} ⊗ f(m)) = α(N) ∘ S^{−1}R ⊗_{R} f(\frac{r}{s} ⊗ m). □

Definition 10.13. An additive functor T : R-Mod ⟶ S-Mod is called exact, if for each exact sequence

… ⟶ M_{i−1} ⟶^{f_{i−1}} M_{i} ⟶^{f_{i}} M_{i+1} ⟶ …

the sequence

… ⟶ T (M_{i−1}) ⟶^{T (f_{i−1})} T (M_{i}) ⟶^{T (f_{i})} T (M_{i+1}) ⟶ …

is also exact.

Lemma 10.14. Let P ∈ Mod-R. Then the functor P ⊗_{R} - : R-Mod ⟶ Ab preserves exact sequences of the form

M_{1} ⟶ M_{2} ⟶ M_{3} ⟶ 0,

i.e. the sequences

P ⊗_{R} M_{1} ⟶ P ⊗_{R} M_{2} ⟶ P ⊗_{R} M_{3} ⟶ 0

are exact. (The functor P ⊗_{R} - is right exact.)

Proof. This follows from Corollary 6.13, Exercise 5.2 (1) and Exercise 6.2. We give a direct proof. Let

M_{1} ⟶^{f} M_{2} ⟶^{g} M_{3} ⟶ 0

be exact. This is equivalent to g surjective, gf = 0 and Ke(g) ⊆ Im(f). The map P ⊗_{R} g is surjective, for ∑ p_{i} ⊗ m_{i3} = ∑ p_{i} ⊗ g(m_{i2}) for arbitrary m_{i3} ∈ M_{3} and suitable m_{i2} ∈ M_{2}. Furthermore (P ⊗_{R}g)(P ⊗_{R}f) = P ⊗_{R}gf = 0. It remains to show Ke(P ⊗_{R}g) ⊆ Im(P ⊗_{R}f). Since Im(P ⊗_{R} f) ⊆ Ke(P ⊗_{R} g), we obtain a homomorphism by the homomorphism theorem

ψ : (P ⊗_{R} M_{2})/ Im(P ⊗_{R} f) ⟶ P ⊗_{R} M_{3}

with ψ(\overline{p ⊗ m_{2}}) = p ⊗ g(m_{2}). Furthermore we define a homomorphism

φ : P ⊗_{R} M_{3} ⟶ (P ⊗_{R} M_{2})/ Im(P ⊗_{R} f)

with φ(p ⊗ m_{3}) := \overline{p ⊗ m_{2}} for an m_{2} ∈ M_{2} with g(m_{2}) = m_{3}. For this purpose we first define φ̃: P ×M_{3} ⟶ P ⊗_{R}M_{2}/ Im(P ⊗_{R}f) by φ̃(p, m_{3}) := \overline{p ⊗ m_{2}} for an m_{2} ∈ M_{2} with g(m_{2}) = m_{3}. If also g(m′_{2}) = m_{3} holds then we have g(m_{2} − m′_{2}) = 0, so there is an m_{1} ∈ M_{1} with m_{2}−m′_{2} = f(m_{1}). This implies \overline{p ⊗ m_{2}} = \overline{p ⊗ (m′_{2} + f(m_{1}))} = \overline{p ⊗ m′_{2}}+\overline{p ⊗ f(m_{1})} = \overline{p ⊗ m′_{2}}, i.e. φ̃ is well defined. It is easy to verify that φ̃ is R-bilinear and thus φ is a well defined homomorphism. Now φψ = id and ψφ = id hold since φψ(\overline{p ⊗ m_{2}}) = φ(p⊗g(m_{2})) = \overline{p ⊗ m_{2}} and ψφ(p⊗m_{3}) = ψ(\overline{p ⊗ m_{2}}) = p⊗g(m_{2}) = p⊗m_{3}. So we get Ke(P ⊗_{R} g) = Ke(φ(P ⊗_{R} g)) = Ke(ν : P ⊗_{R} M_{2} ⟶ P ⊗_{R} M_{2}/ Im(P ⊗_{R} f)) = Im(P ⊗_{R} f). Thus P ⊗_{R} M_{1} ⟶ P ⊗_{R} M_{2} ⟶ P ⊗_{R} M_{3} ⟶ 0 is exact. □

Definition 10.15. A module P_{R} is called R-flat, if P ⊗_{R} - is an exact functor.

Proposition 10.16. A module P_{R} is flat if and only if P ⊗_{R} - preserves monomorphisms, i.e. if for each monomorphism f : M ⟶ N the map P ⊗_{R} f : P ⊗_{R} M ⟶ P ⊗_{R} N is a monomorphism.

Proof. If P_{R} is flat and if f : M ⟶ N is a monomorphism then 0 ⟶ M ⟶^{f} N is exact. Consequently 0 ⟶ P ⊗_{R} M ⟶^{P ⊗_{R}f} P ⊗_{R} N is exact and thus P ⊗_{R} f : P ⊗_{R} M ⟶ P ⊗_{R} N is a monomorphism. Assume that P ⊗_{R} - preserves monomorphisms and that the sequence

… ⟶ M_{i−1} ⟶^{f_{i−1}} M_{i} ⟶^{f_{i}} M_{i+1} ⟶ …

is exact. Then the sequences

0 ⟶ Im(f_{i−1}) ⟶ M_{i} ⟶ Im(f_{i}) ⟶ 0

are exact. Since P ⊗_{R} - preserves monomorphisms, the sequences

0 ⟶ P ⊗_{R} Im(f_{i−1}) ⟶ P ⊗_{R} M_{i} ⟶ P ⊗_{R} Im(f_{i}) ⟶ 0

are exact. The canonical map P ⊗_{R} Im(f) ⟶ Im(P ⊗_{R} f) is surjective, since each element ∑ p_{i} ⊗ f(m_{i}) ∈ Im(P ⊗_{R} f) is in the image of this map. Observe, however, that this map is in general not injective. The maps Im(f) ⟶ N and thus also P ⊗_{R} Im(f) ⟶ P ⊗_{R} N are, however, by hypothesis injective hence P ⊗_{R} Im(f) ⟶ Im(P ⊗_{R} f) is injective and thus bijective. From the isomorphism P ⊗_{R} Im(f) ≅ Im(P ⊗_{R} f) we thus get the exactness of

0 ⟶ Im(P ⊗_{R} f_{i−1}) ⟶ P ⊗_{R} M_{i} ⟶ Im(P ⊗_{R} f_{i}) ⟶ 0.

So the sequence

… ⟶ P ⊗_{R} M_{i−1} ⟶^{P ⊗_{R}f_{i−1}} P ⊗_{R} M_{i} ⟶^{P ⊗_{R}f_{i}} P ⊗_{R} M_{i+1} ⟶ …

is also exact. □

Proposition 10.17. S^{−1}R is a flat R-module.

Proof. Let f : M ⟶ N be a monomorphism and let S^{−1}f(\frac{m}{s}) = 0 = \frac{f(m)}{s}. Then there is a t ∈ S with tf(m) = 0 = f(tm), so with tm = 0. Then \frac{m}{s} = 0, hence S^{−1}f is a monomorphism. □

Recall from Basic Algebra:

(1) An ideal 𝔭 ⊆ R is called a prime ideal if and only if 𝔭 ≠ R and (rs ∈ 𝔭 ⟹ r ∈ 𝔭 ∨ s ∈ 𝔭).

(2) If 𝔪 ⊆ R is a maximal ideal, then 𝔪 is a prime ideal.

(3) 𝔭 ∈ R is a prime ideal if and only if the residue class ring R/𝔭 is an integral domain.

Lemma 10.18. Let 𝔭 ⊆ R be an ideal. The following are equivalent (1) 𝔭 is a prime ideal.

(2) R \ 𝔭 is a multiplicatively closed set.

Proof. follows immediately from the definition. □

Definition 10.19. Let 𝔭 ⊆ R be a prime ideal and M be an R-module. Then M_{𝔭} := S^{−1}M with S = R \ 𝔭 is called the localization of the module M at 𝔭. The set Spec(R) := {𝔭 ⊆ R|𝔭 prime ideal} is called the spectrum of the ring R. The set Specm(R) := {𝔪 ⊆ R|𝔪 maximal ideal} is called the maximal spectrum of the ring R.

Proposition 10.20. Let M be an R-module, such that M_{𝔪} = 0 for all 𝔪 ∈ Spec(R). Then M = 0.

Proof. Assume there is an m ∈ M with m ≠ 0. Then I := Ke(R ∋ r ↦ rm ∈ M) ⫋ R is an ideal. Since R is finitely generated there is a maximal ideal 𝔪 with I ⊆ 𝔪 ⫋ R. Since M_{𝔪} = 0, we have \frac{m}{s} = 0 in M_{𝔪}, hence there is a t ∈ R \ 𝔪 with tm = 0. This, however, gives t ∈ I ⊆ 𝔪, a contradiction. □

Corollary 10.21. Let f : M ⟶ N be given. The following are equivalent (1) f is a mono-(epi- resp. iso-)morphism.

(2) For all 𝔪 ∈ Spec(R) the localization f_{𝔪} is a mono-(epi- resp. iso-)morphism.

Proof. (1) ⟹ (2): follows from 10.17 and 10.12.

(2) ⟹ (1): The sequence 0 ⟶ Ke(f) ⟶ M ⟶^{f} N ⟶ Cok(f) ⟶ 0 is exact. Consequently

0 ⟶ Ke(f)_{𝔪} ⟶ M_{𝔪} ⟶^{f_{𝔪}} N_{𝔪} ⟶ Cok(f)_{𝔪} ⟶ 0

is exact. Thus we get in particular Ke(f)_{𝔪} ≅ Ke(f_{𝔪}) and Cok(f)_{𝔪} ≅ Cok(f_{𝔪}). Now if f_{𝔪} is a monomorphism for all 𝔪 ∈ Specm(R), then we have Ke(f)_{𝔪} = 0 for all 𝔪, hence Ke(f) = 0 and f is a monomorphism. An analogous argument can be used for epimorphisms with Cok(f). Taken together these two results give the claim for isomorphisms. □

Proposition 10.22. Let R be a commutative ring and 𝔭 ⊆ R be a prime ideal. Then R_{𝔭} is a local ring.

Proof. Since 0 ⟶ 𝔭 ⟶ R ⟶ R/𝔭 ⟶ 0 is exact and R/𝔭 ≠ 0, the sequence 0 ⟶ 𝔭_{𝔭} ⟶ R_{𝔭} ⟶ (R/𝔭)_{𝔭} ⟶ 0 is exact and (R/𝔭)_{𝔭} ≠ 0 since there is no t ∈ R \ 𝔭 with t · \overline{r} = 0 for any \overline{r} ≠ 0 (10.9). So 𝔭_{𝔭} ⫋ R_{𝔭} is a proper ideal. If \frac{r}{s} ∉ 𝔭_{𝔭}, then r ∉ 𝔭 and s ∉ 𝔭 hence \frac{s}{r} \frac{r}{s} = 1 and thus \frac{r}{s} is a unit. So the non units of R_{𝔭} form an ideal 𝔭_{𝔭}, i.e. R_{𝔭} is local and 𝔭_{𝔭} is the maximal ideal. □

Corollary 10.23. Let 𝔭 ⊆ R be a prime ideal. Then the quotient field Q(R/𝔭) is isomorphic to R_{𝔭}/𝔭_{𝔭}.

Proof. As in the preceding proof (R/𝔭)_{𝔭} ≅ R_{𝔭}/𝔭_{𝔭}. Furthermore R_{𝔭}/𝔭_{𝔭} is a field, because 𝔭_{𝔭} is the maximal ideal of R_{𝔭}. Furthermore we have

(R/𝔭)_{𝔭} = S^{−1}(R/𝔭) = \{\frac{\overline{r}}{s}|\overline{r} ∈ R/𝔭, s ∉ 𝔭\} ≅ \{\frac{\overline{r}}{\overline{s}}|\overline{r} ∈ R/𝔭, \overline{s} ∈ R/𝔭, \overline{s} ≠ 0\} = Q(R/𝔭).

□

Proposition 10.24. Let _{R}M be a finitely generated module. Let M/𝔪M = 0 for all maximal ideals 𝔪 ⊆ R. Then M = 0.

Proof. M/𝔪M ≅ R/𝔪 ⊗_{R} M ≅ R_{𝔪}/𝔪_{𝔪} ⊗_{R_{𝔪}} R_{𝔪} ⊗_{R} M ≅ M_{𝔪}/𝔪_{𝔪}M_{𝔪}. Since R_{𝔪} is local and M_{𝔪} is finitely generated, it follows that M_{𝔪} = 0 for all maximal ideals 𝔪 ⊆ R. So we get M = 0. □

Corollary 10.25. Let f : M ⟶ N be an R-homomorphism and let N be finitely generated. Let f/𝔪f : M/𝔪M ⟶ N/𝔪N be an epimorphism for all maximal ideals 𝔪 ⊆ R. Then f is an epimorphism.

Proof. M ⟶^{f} N ⟶ Q ⟶ 0 is exact and thus Q is finitely generated. We apply the functor R/𝔪 ⊗_{R} - and get the exact sequence M/𝔪M ⟶ N/𝔪N ⟶ Q/𝔪Q ⟶ 0. Since f/𝔪f is an epimorphism, we get Q/𝔪Q = 0, hence Q = 0. So f is an epimorphism. □

11. Monoidal Categories

For our further investigations it is useful to introduce a generalized version of a tensor product. This shall be done in this section. With this generalized notion of a tensor product we also obtain generalizations of the notion of an algebra and of a representation.

Definition 11.1. A monoidal category (or tensor category) consists of a category C, a functor ⊗ : C × C ⟶ C, called tensor product, an object I ∈ C, called unit, natural isomorphisms

α(A, B, C) : (A ⊗ B) ⊗ C ⟶ A ⊗ (B ⊗ C),
λ(A) : I ⊗ A ⟶ A,
ρ(A) : A ⊗ I ⟶ A,

called associativity, left unit and right unit, such that the following diagrams, called coherence diagrams or constraints, commute:

[figure: ((A ⊗ B) ⊗ C) ⊗ D; (A ⊗ (B ⊗ C)) ⊗ D; α(A,B,C)⊗1; A ⊗ ((B ⊗ C) ⊗ D); α(A,B⊗C,D); α(A⊗B,C,D); 1⊗α(B,C,D); (A ⊗ B) ⊗ (C ⊗ D); A ⊗ (B ⊗ (C ⊗ D)); α(A,B,C⊗D)]

[figure: (A ⊗ I) ⊗ B; A ⊗ (I ⊗ B); α(A,I,B); A ⊗ B; ρ(A)⊗1; 1⊗λ(B)]

A monoidal category is called strict, if the morphisms α, λ, ρ are identities.

Remark 11.2. We define A_{1} ⊗ … ⊗ A_{n} := (… (A_{1} ⊗ A_{2}) ⊗ …) ⊗ A_{n}. The coherence theorem of S. MacLane says that all diagrams whose morphisms are formed using α, λ, ρ, identities, inverses, tensor products, and compositions thereof commute. We will not prove this theorem. It implies that each monoidal category can be replaced by (is monoidally equivalent to) a strict monoidal category, that is in all diagrams we may omit the morphisms α, λ, ρ, i. e. replace them by identities. In particular on A_{1} ⊗ … ⊗ A_{n} there is only one automorphism formed with coherence morphisms, the identity.

Remark 11.3. For each monoidal category C one can construct the monoidal category C^{symm} symmetric to C which coincides with C as a category, which has the tensor product A ⊠ B := B ⊗ A, and coherence morphisms

α(C, B, A)^{−1} : (A ⊠ B) ⊠ C ⟶ A ⊠ (B ⊠ C),

ρ(A) : I ⊠ A ⟶ A,
λ(A) : A ⊠ I ⟶ A.

Then the coherence diagrams commute again, so that C^{symm} becomes a monoidal category.

Example 11.4. (1) Let R be an arbitrary ring. The category _{R}M_{R} of R-R- bimodules with the tensor product M ⊗_{R} N is a monoidal category. In particular the 𝕂-modules form a monoidal category. (2a) Let G be a monoid. A vector space V together with a family of subspaces (V_{g}|g ∈ G) is called G-graded, if V = ⊕_{g∈G}V_{g} holds. Let V and W be G-graded vector spaces. A linear map f : V ⟶ W is called of degree e ∈ G, if for all g ∈ G f(V_{g}) ⊆ W_{g} holds.

The G-graded vector spaces and linear maps of degree e ∈ G form the category M^{G} of G-graded vector spaces. M^{G} carries a monoidal structure with the tensor product V ⊗ W where the subspaces (V ⊗ W)_{g} are defined by

(V ⊗ W)_{g} := ⊕_{h,k∈G,hk=g}V_{h} ⊗ W_{k}.

If G is a group, this can also be written as (V ⊗ W)_{g} := ⊕_{h∈G}V_{h} ⊗ W_{h^{−1}g}. (2b) Let G be a monoid. A family of vector spaces (V_{g}|g ∈ G) is called a G-family of vector spaces. Let (V_{g}) and (W_{g}) be G-families of vector spaces. A family of linear maps (f_{g} : V_{g} ⟶ W_{g} is called a G-family of linear maps. The G-families of vector spaces and G-families of linear maps form the category (M)^{G} of G-families of vector spaces. (M)^{G} carries a monoidal structure with the tensor product (V_{g}) ⊗ (W_{g}) where the subspaces (V ⊗ W)_{g} are defined by

(V_{g}) ⊗ (W_{g}) := ((⊕_{h,k∈G,hk=g}V_{h} ⊗ W_{k})_{g}).

(3) A (chain) complex of R-modules over a ring R

M = (… ⟶^{∂_{3}} M_{2} ⟶^{∂_{2}} M_{1} ⟶^{∂_{1}} M_{0})

consists of a family of R-modules M_{i} and a family of homomorphisms ∂_{n} : M_{n} ⟶ M_{n−1} with ∂_{n−1}∂_{n} = 0. (This chain complex is indexed with ℕ_{0}. One can also consider chain complexes, that are indexed with ℤ. See also Section 1.6.) Let M and N be two chain complexes. A homomorphism f : M ⟶ N of chain complexes consists of a family of homomorphisms of R-modules f_{n} : M_{n} ⟶ N_{n}, such that f_{n}∂_{n+1} = ∂_{n+1}f_{n+1} for all n ∈ ℕ_{0}. The chain complexes of R-modules with these homomorphisms form the category Comp-R of chain complexes.

Lemma 11.5. The following diagrams in a monoidal category commute

[figure: (I ⊗ A) ⊗ B; I ⊗ (A ⊗ B); α; A ⊗ B; λ(A)⊗1_{B}; λ(A⊗B); (A ⊗ B) ⊗ I; A ⊗ (B ⊗ I); α; A ⊗ B; ρ(A⊗B); 1_{A}⊗ρ(B)]

and λ(I) = ρ(I) holds.

Proof. We first observe that the identity functor Id_{C} and the functor I ⊗ - are isomorphic by the natural isomorphism λ. In particular we have I ⊗ f = I ⊗ g ⟹ f = g. In the diagram all subdiagrams commute, except for the right hand trapezoid. Since the morphisms are isomorphisms, also the right hand trapezoid commutes, hence the whole diagram commutes. The commutativity of the second diagram follows by analogous conclusions. Furthermore the following diagram commutes

[figure: ((I ⊗ I) ⊗ A) ⊗ B; (I ⊗ (I ⊗ A)) ⊗ B; α⊗1; I ⊗ ((I ⊗ A) ⊗ B); α; (ρ⊗1)⊗1; (1⊗λ)⊗1; 1⊗(λ⊗1); (I ⊗ A) ⊗ B; I ⊗ (A ⊗ B); α; α; 1⊗α; α; 1; I ⊗ (A ⊗ B); I ⊗ (A ⊗ B); 1; (I ⊗ I) ⊗ (A ⊗ B); I ⊗ (I ⊗ (A ⊗ B)); α; ρ⊗(1⊗1); 1⊗λ]

[figure: I ⊗ (I ⊗ I); (I ⊗ I) ⊗ I; α; I ⊗ (I ⊗ I); α; I ⊗ I; 1⊗ρ; ρ; I ⊗ I; ρ⊗1; 1⊗λ; I; ρ; ρ]

Here the left hand triangle commutes because of the property shown before, the right hand triangle is given through the axiom. Finally the lower square commutes, since ρ is a natural transformation. In particular we get ρ(1 ⊗ ρ) = ρ(1 ⊗ λ). Since ρ is an isomorphism and I ⊗ - ≅ Id_{C}, it follows ρ = λ. □

Problem 11.1. For morphisms f : I ⟶ M and g : I ⟶ N in a monoidal category C we define (f ⊗ 1 : N ⟶ M ⊗ N) := (f ⊗ 1_{I})ρ(I)^{−1} and (1 ⊗ g : M ⟶ M ⊗ N) := (1 ⊗ g)λ(I)^{−1}. Show that the diagram

[figure: N; M ⊗ N; f ⊗ 1; I; I; f; g; 1 ⊗ g]

commutes.

Definition 11.6. Let (C, ⊗) and (D, ⊗) be monoidal categories. A functor

F : C ⟶ D

together with a natural transformation

ξ(M, N) : F(M) ⊗ F(N) ⟶ F(M ⊗ N)

and a morphism

ξ_{0} : I_{D} ⟶ F(I_{C})

is called weakly monoidal, if the following diagrams commute:

[figure: (F(M) ⊗ F(N)) ⊗ F(P); F(M ⊗ N) ⊗ F(P); ξ⊗1; F((M ⊗ N) ⊗ P); ξ; α; F(α); F(M) ⊗ (F(N) ⊗ F(P)); F(M) ⊗ F(N ⊗ P); 1⊗ξ; F(M ⊗ (N ⊗ P)); ξ]

[figure: I ⊗ F(M); F(I) ⊗ F(M); ξ_{0}⊗1; F(I ⊗ M); ξ; F(M); F(λ); λ]

[figure: F(M) ⊗ I; F(M) ⊗ F(I); 1⊗ξ_{0}; F(M ⊗ I); ξ; F(M).; F(ρ); ρ]

In addition if ξ and ξ_{0} are isomorphisms then the functor is called monoidal. The functor is called strict monoidal, if ξ and ξ_{0} are identity morphisms. A natural transformation ζ : F ⟶ F′ between weakly monoidal functors is called monoidal, if the diagrams

[figure: F(I); ξ_{0}; I; F′(I); ξ′_{0}; ζ; F(M) ⊗ F(N); F(M ⊗ N); ξ; ζ; ζ⊗ζ; F′(M) ⊗ F′(N); F′(M ⊗ N); ξ′]

commute.

In monoidal categories one can generalize notions like algebra and coalgebra. For this purpose we define

Definition 11.7. Let C be a monoidal category. An algebra or a monoid in C is an object A together with a multiplication ∇ : A ⊗ A ⟶ A, that is associative:

[figure: A ⊗ A ⊗ A; A ⊗ A; id ⊗∇; ∇⊗id; ∇; A ⊗ A; A; ∇]

and a unit element η : I ⟶ A, for which the following diagram commutes

[figure: I ⊗ A ≅ A ≅ A ⊗ I; A ⊗ A; id ⊗η; η⊗id; ∇; A ⊗ A; A.; ∇; id]

Let A and B algebras in C. A morphism of algebras f : A ⟶ B is a morphism in C, such that the following diagrams commute:

and Remark 11.8. Obviously the composition of two morphisms of algebras is again a morphism of algebras. Also the identity morphism is a morphism of algebras. Thus we obtain the category Alg(C) of algebras in C.

[figure: A ⊗ A; B ⊗ B; f⊗f; ∇_{A}; ∇_{B}; A; B; f]

[figure: I; A; η_{A}; B; η_{B}; f]

Definition 11.9. A coalgebra or a comonoid in a monoidal category C is an object C together with a comultiplication Δ : C ⟶ C ⊗ C, that is coassociative:

[figure: C; C ⊗ C; Δ; Δ; Δ⊗id; C ⊗ C; C ⊗ C ⊗ C; id ⊗Δ]

and a counit ϵ : C ⟶ I, for which the diagram

[figure: C; C ⊗ C; Δ; Δ; id ⊗ϵ; C ⊗ C; I ⊗ C ≅ C ≅ C ⊗ I; ϵ⊗id; id]

commutes. Let C and D be coalgebras. A morphism of coalgebras f : C ⟶ D is a morphism in C, such that

[figure: C; D; f; Δ_{C}; Δ_{D}; C ⊗ C; D ⊗ D; f⊗f]

and

[figure: C; D; f; I; ϵ_{C}; ϵ_{D}]

commute.

Remark 11.10. Obviously the composition of two morphisms of coalgebras is again a morphism of coalgebras. Also the identity morphism is a morphism of coalgebras. Thus we obtain the category Coalg(C) of coalgebras in C.

12. Bialgebras and Hopf Algebras

12.1. Bialgebras.

Definition 12.1. (1) A bialgebra (B, ∇, η, Δ, ϵ) consists of an algebra (B, ∇, η) and a coalgebra (B, Δ, ϵ) such that the diagrams

[figure: B ⊗ B; B ⊗ B ⊗ B ⊗ B; Δ⊗Δ; 1⊗τ⊗1; ∇; B; B ⊗ B; Δ; B ⊗ B ⊗ B ⊗ B; ∇⊗∇]

and

[figure: 𝕂; B; η; η⊗η; B; B ⊗ B; Δ; B ⊗ B; B; ∇; 𝕂; ϵ⊗ϵ; ϵ; 𝕂; 𝕂; id; B; η; ϵ]

commute, i.e. Δ and ϵ are homomorphisms of algebras resp. ∇ and η are homomorphisms of coalgebras.

(2) Given bialgebras A and B. A map f : A ⟶ B is called a homomorphism of bialgebras if it is a homomorphism of algebras and a homomorphism of coalgebras.

(3) The category of bialgebras is denoted by 𝕂- Bialg.

Problem 12.1. (1) Let (B, ∇, η) be an algebra and (B, Δ, ε) be a coalgebra. The following are equivalent:

a) (B, ∇, η, Δ, ε) is a bialgebra.

b) Δ : B ⟶ B ⊗ B and ε : B ⟶ 𝕂 are homomorphisms of 𝕂-algebras.

c) ∇ : B ⊗ B ⟶ B and η : 𝕂 ⟶ B are homomorphisms of 𝕂-coalgebras.

(2) Let B be a finite dimensional bialgebra over field 𝕂. Show that the dual space B^{∗} is a bialgebra.

One of the most important properties of bialgebras B is that the tensor product over 𝕂 of two B-modules or two B-comodules is again a B-module.

Proposition 12.2. (1) Let B be a bialgebra. Let M and N be left B-modules. Then M ⊗_{𝕂} N is a B-module by the map

B ⊗ M ⊗ N ⟶^{Δ⊗1} B ⊗ B ⊗ M ⊗ N ⟶^{1⊗τ⊗1} B ⊗ M ⊗ B ⊗ N ⟶^{μ⊗μ} M ⊗ N.

(2) Let B be a bialgebra. Let M and N be left B-comodules. Then M ⊗_{𝕂} N is a Bcomodule by the map

M ⊗ N ⟶^{δ⊗δ} B ⊗ M ⊗ B ⊗ N ⟶^{1⊗τ⊗1} B ⊗ B ⊗ M ⊗ N ⟶^{∇⊗1} B ⊗ M ⊗ N.

(3) 𝕂 is a B-module by the map B ⊗ 𝕂 ≅ B ⟶^{ε} 𝕂.

(4) 𝕂 is a B-comodule by the map 𝕂 ⟶^{η} B ≅ B ⊗ 𝕂.

Proof. We give a diagrammatic proof for (1). The associativity law is given by

[figure: B ⊗ B ⊗ M ⊗ N; B ⊗ B ⊗ B ⊗ M ⊗ N; 1⊗Δ⊗1⊗1; B ⊗ B ⊗ M ⊗ B ⊗ N; 1⊗1⊗τ⊗1; B ⊗ M ⊗ N; 1⊗μ⊗μ; B ⊗ B ⊗ B ⊗ B ⊗ M ⊗ N; B ⊗ B ⊗ B ⊗ M ⊗ B ⊗ N; 1⊗1⊗1⊗τ⊗1; B ⊗ B ⊗ M ⊗ N; 1⊗1⊗μ⊗μ; B ⊗ B ⊗ B ⊗ B ⊗ M ⊗ N; B ⊗ B ⊗ M ⊗ B ⊗ B ⊗ N; 1⊗1⊗τ(B⊗B,M)⊗1; B ⊗ M ⊗ B ⊗ N; 1⊗μ⊗1⊗μ; B ⊗ M ⊗ N; B ⊗ B ⊗ M ⊗ N; Δ⊗1⊗1; B ⊗ M ⊗ B ⊗ N; 1⊗τ⊗1; M ⊗ N; μ⊗μ; ∇⊗1⊗1; Δ⊗1⊗1⊗1⊗1; Δ⊗1⊗1⊗1⊗1; Δ⊗1⊗1; 1⊗τ⊗1⊗1⊗1; 1⊗τ(B,B⊗M)⊗1⊗1; 1⊗τ⊗1; ∇⊗∇⊗1⊗1; ∇⊗1⊗∇⊗1; μ⊗μ]

The unit law is the commutativity of

[figure: M ⊗ N ≅ 𝕂 ⊗ M ⊗ N; B ⊗ M ⊗ N; η⊗1⊗1; 𝕂 ⊗ 𝕂 ⊗ M ⊗ N; B ⊗ B ⊗ M ⊗ N; η⊗η⊗1⊗1; M ⊗ N ≅ 𝕂 ⊗ M ⊗ 𝕂 ⊗ N; B ⊗ M ⊗ B ⊗ N; η⊗1⊗η⊗1; ≅; 1⊗τ⊗1; Δ⊗1⊗1; 1⊗τ⊗1; μ⊗μ; =; M ⊗ N; 1]

The corresponding properties for comodules follows from the dualized diagrams. The module and comodule properties of 𝕂 are easily checked. □

Problem 12.2. (1) Let B be a bialgebra and M_{B} be the category of right B- modules. Show that M_{B} is a monoidal category.

(2) Let B a bialgebra and M^{B} be the category of right B- comodules. Show that M^{B} is a monoidal category.

Definition 12.3. (1) Let (B, ∇, η, Δ, ϵ) be a bialgebra. Let A be a left B-module with structure map μ : B ⊗ A ⟶ A. Let furthermore (A, ∇_{A}, η_{A}) be an algebra such that ∇_{A} and η_{A} are homomorphisms of B-modules. Then (A, ∇_{A}, η_{A}, μ) is called a B-module algebra.

(2) Let (B, ∇, η, Δ, ϵ) be a bialgebra. Let C be a left B-module with structure map μ : B ⊗C ⟶ C. Let furthermore (C, Δ_{C}, ε_{C}) be a coalgebra such that Δ_{C} and ε_{C} are homomorphisms of B-modules. Then (C, Δ_{C}, ε_{C}, μ) is called a B-module coalgebra.

(3) Let (B, ∇, η, Δ, ϵ) be a bialgebra. Let A be a left B-comodule with structure map δ : A ⟶ B ⊗ A. Let furthermore (A, ∇_{A}, η_{A}) be an algebra such that ∇_{A} and η_{A} are homomorphisms of B-comodules. Then (A, ∇_{A}, η_{A}, δ) is called a B-comodule algebra.

(4) Let (B, ∇, η, Δ, ϵ) be a bialgebra. Let C be a left B-comodule with structure map δ : C ⟶ B ⊗ C. Let furthermore (C, Δ_{C}, ε_{C}) be a coalgebra such that Δ_{C} and ε_{C} are homomorphisms of B-comodules. Then (C, Δ_{C}, ε_{C}, δ) is called a B-comodule coalgebra.

Remark 12.4. If (C, Δ_{C}, ε_{C}) is a 𝕂-coalgebra and (C, μ) is a B-module, then (C, Δ_{C}, ε_{C}, μ) is a B-module coalgebra iff μ is a homomorphism of 𝕂-coalgebras. If (A, ∇_{A}, η_{A}) is a 𝕂-algebra and (A, δ) is a B-comodule, then (A, ∇_{A}, η_{A}, δ) is a B-comodule algebra iff δ is a homomorphism of 𝕂-algebras.

Similar statements for module algebras or comodule coalgebras do not hold.

Problem 12.3. (1) Let B be a bialgebra. Describe what an algebra A and a coalgebra C are in the monoidal category M_{B} (in the sense of section 11).

(2) Let B be a bialgebra. Describe what an algebra A and a coalgebra C are in the monoidal category M^{B} (in the sense of section 11).

Remark 12.5. The notions of a bialgebra, a comodule algebra, and a Hopf algebra cannot be generalized in the usual way to an arbitrary monoidal category, since we need the multiplication on the tensor product of two algebras. To define this we need the commutation, exchange morphism, or flip of two tensor factors. Such exchange morphisms are known under the name of symmetry or quasisymmetry (braiding). They will be discussed later on.

12.2. Hopf Algebras. The difference between a monoid and a group lies in the existence of an additional map S : G ∋ g ↦ g^{−1} ∈ G for a group G that allows forming inverses. This map satisfies the equation S(g)g = 1 or in a diagrammatic form

[figure: G; {1}; ε; G; 1; Δ; G × G; G × G; S×id; mult]

We want to carry this property over to bialgebras B instead of monoids. An “inverse map” shall be a morphism S : B ⟶ B with a similar property. This will be called a Hopf algebra.

Definition 12.6. A left Hopf algebra H is a bialgebra H together with a left antipode S : H ⟶ H, i.e. a 𝕂-module homomorphism S such that the following diagram commutes:

[figure: H; 𝕂; ε; H; η; Δ; H ⊗ H; H ⊗ H; S⊗id; ∇]

Symmetrically we define a right Hopf algebra H. A Hopf algebra is a left and right Hopf algebra. The map S is called a (left, right, two-sided) antipode.

Using the Sweedler notation (2.20) the commutative diagram above can also be expressed by the equation

∑ S(a_{(1)})a_{(2)} = ηε(a)

for all a ∈ H. Observe that we do not require that S : H ⟶ H is an algebra homomorphism.

Problem 12.4. (1) Let H be a bialgebra and S ∈ Hom(H, H). Then S is an antipode for H (and H is a Hopf algebra) iff S is a two sided inverse for id in the algebra (Hom(H, H), ∗, ηε) (see 2.21). In particular S is uniquely determined.

(2) Let H be a Hopf algebra. Then S is an antihomomorphism of algebras and coalgebras i.e. S “inverts the order of the multiplication and the comultiplication”.

(3) Let H and K be Hopf algebras and let f : H ⟶ K be a homomorphism of bialgebras. Then fS_{H} = S_{K}f, i.e. f is compatible with the antipode.

Definition 12.7. Because of Problem 12.4 (3) every homomorphism of bialgebras between Hopf algebras is compatible with the antipodes. So we define a homomorphism of Hopf algebras to be a homomorphism of bialgebras. The category of Hopf algebras will be denoted by 𝕂- Hopf.

Proposition 12.8. Let H be a bialgebra with an algebra generating set X. Let S : H ⟶ H^{op} be an algebra homomorphism such that ∑ S(x_{(1)})x_{(2)} = ηε(x) for all x ∈ X. Then S is a left antipode of H.

Proof. Assume a, b ∈ H such that ∑ S(a_{(1)})a_{(2)} = ηε(a) and ∑ S(b_{(1)})b_{(2)} = ηε(b). Then

\begin{aligned}∑ S((ab)_{(1)})(ab)_{(2)} & = ∑ S(a_{(1)}b_{(1)})a_{(2)}b_{(2)} = ∑ S(b_{(1)})S(a_{(1)})a_{(2)}b_{(2)} \\ & = ∑ S(b_{(1)})ηε(a)b_{(2)} = ηε(a)ηε(b) = ηε(ab). \end{aligned}

Since every element of H is a finite sum of finite products of elements in X, for which the equality holds, this equality extends to all of H by induction. □

Example 12.9. (1) Let V be a vector space and T (V) the tensor algebra over V. We have seen in Problem 2.2 that T (V) is a bialgebra and that V generates T (V) as an algebra. Define S : V ⟶ T (V)^{op} by S(v) := −v for all v ∈ V. By the universal property of the tensor algebra this map extends to an algebra homomorphism S : T (V) ⟶ T (V)^{op}. Since Δ(v) = v ⊗ 1 + 1 ⊗ v we have ∑ S(v_{(1)})v_{(2)} = ∇(S ⊗ 1)Δ(v) = −v + v = 0 = ηε(v) for all v ∈ V, hence T (V) is a Hopf algebra by the preceding proposition.

(2) Let V be a vector space and S(V) the symmetric algebra over V (that is commutative). We have seen in Problem 2.3 that S(V) is a bialgebra and that V generates S(V) as an algebra. Define S : V ⟶ S(V) by S(v) := −v for all v ∈ V. S extends to an algebra homomorphism S : S(V) ⟶ S(V). Since Δ(v) = v ⊗ 1 + 1 ⊗ v we have ∑ S(v_{(1)})v_{(2)} = ∇(S ⊗ 1)Δ(v) = −v + v = 0 = ηε(v) for all v ∈ V, hence S(V) is a Hopf algebra by the preceding proposition.

Example 12.10. (Group Algebras) For each algebra A we can form the group of units U(A) := {a ∈ A|∃a^{−1} ∈ A} with the multiplication of A as composition of the group. Then U is a covariant functor U : 𝕂-Alg ⟶ Gr. This functor leads to the following universal problem. Let G be a group. An algebra 𝕂G together with a group homomorphism ι : G ⟶ U(𝕂G) is called a (the) group algebra of G, if for every algebra A and for every group homomorphism f : G ⟶ U(A) there exists a unique homomorphism of algebras g : 𝕂G ⟶ A such that the following diagram commutes

[figure: G; U(𝕂G); ι; f; U(A).; g]

The group algebra 𝕂G is (if it exists) unique up to isomorphism. It is generated as an algebra by the image of G. The map ι : G ⟶ U(𝕂G) ⊆ 𝕂G is injective and the image of G in 𝕂G is a basis. The group algebra can be constructed as the free vector space 𝕂G with basis G and the algebra structure of 𝕂G is given by 𝕂G ⊗ 𝕂G ∋ g ⊗ h ↦ gh ∈ 𝕂G and the unit η : 𝕂 ∋ α ↦ αe ∈ 𝕂G. The group algebra 𝕂G is a Hopf algebra. The comultiplication is given by the diagram

[figure: G; 𝕂G; ι; f; 𝕂G ⊗ 𝕂G; Δ]

with f(g) := g ⊗g which defines a group homomorphism f : G ⟶ U(𝕂G⊗𝕂G). The counit is given by

[figure: G; 𝕂G; ι; f; 𝕂; ε]

where f(g) = 1 for all g ∈ G. One shows easily by using the universal property, that Δ is coassociative and has counit ε. Define an algebra homomorphism S : 𝕂G ⟶ (𝕂G)^{op} by

[figure: G; 𝕂G; ι; f; (𝕂G)^{op}; S]

with f(g) := g^{−1} which is a group homomorphism f : G ⟶ U((𝕂G)^{op}). Then one shows with Proposition 12.8 that 𝕂G is a Hopf algebra.

Proposition 12.11. The following three monoidal categories are monoidally equivalent (1) the category M^{G} of G-graded vector spaces M^{G},

(2) the category of G-families of vector spaces (M)^{G},

(3) the monoidal category of 𝕂G-comodules M^{𝕂G}.

Proof. We only indicate the construction for the equvalence between (1) and (3). For a G-graded vector space V one constructs the 𝕂G-comodule V with the structure map δ : V ⟶ V ⊗ 𝕂G, δ(v) := v ⊗ g for all v ∈ V_{g} and for all g ∈ G. Conversely let V, δ : V ⟶ V ⊗ 𝕂G be a 𝕂G-comodule. Then one constructs the graded vector space V with graded (homogenous) components V_{g} := {v ∈ V |δ(v) = v ⊗ g}. It is easy to verify, that this is an equivalence of categories. Since 𝕂G is a bialgebra, the category of 𝕂G-comodules is a monoidal category by Exercise 12.2 (2). One checks that under the equivalence between M^{G} and M^{𝕂G} tensor products are mapped into corresponding tensor products so that we have a monoidal equivalence. □

Example 12.12. The following is a bialgebra B = 𝕂⟨x, y⟩/I, where I is generated by x^{2}, xy + yx. The diagonal is Δ(y) = y ⊗ y, Δ(x) = x ⊗ y + 1 ⊗ x and the counit is ϵ(y) = 1, ϵ(x) = 0.

Proposition 12.13. The monoidal category Comp-𝕂 of chain complexes over 𝕂 is monoidally equivalent to the category of B-comodules M^{B} with B as in the preceding example.

Proof. We use the following construction. A chain complex M is mapped to the B-comodule M = ⊕_{i∈ℕ}M_{i} with the structure map δ : M ⟶ M ⊗ B, δ(m) := ∑ m ⊗ y^{i} + ∂_{i}(m) ⊗ xy^{i−1} for all m ∈ M_{i} and for all i ∈ ℕ resp. δ(m) := m ⊗ 1 for m ∈ M_{0}. Conversely if M, δ : M ⟶ M ⊗ B is a B-comodule, then one associates with it the vector spaces M_{i} := {m ∈ M|∃m′ ∈ M[δ(m) = m ⊗ y^{i} + m′ ⊗ xy^{i−1}} and the linear maps ∂_{i} : M_{i} ⟶ M_{i−1} with ∂_{i}(m) := m′ for δ(m) = m ⊗ y^{i} + m′ ⊗ xy^{i−1}. One checks that this is an equivalence of categories. By Exercise 12.5 this is a monoidal equivalence. □

Problem 12.5. (1) Give a detailed proof that M^{G} and M^{𝕂G} are equivalent as monoidal categories.

(2) Give a detailed proof that Comp-𝕂 and M^{B} with B as in the preceding Proposition 12.13 are equivalent categories. Since M^{B} is a monoidal category, the tensor product can be transported to Comp-𝕂. Describe the tensor product in the category Comp-B.

You may use the following arguments: Let m ∈ M ∈ M^{B}. Since y^{i}, xy^{i} form a basis of B we have δ(m) = ∑_{i} m_{i} ⊗y^{i} +∑_{i} m′_{i} ⊗xy^{i}. Apply (δ ⊗ 1)δ = (1 ⊗ Δ)δ to this equation and compare coefficients then δ(m_{i}) = m_{i} ⊗ y^{i} + m′_{i−1} ⊗ xy^{i−1}, δ(m′_{i}) = m′_{i} ⊗ y^{i}. Hence for each m_{i} ∈ M_{i} there is exactly one ∂(m_{i}) ∈ M_{i−1}, so that

δ(m_{i}) = m_{i} ⊗ y^{i} + ∂(m_{i}) ⊗ xy^{i−1}, δ(m′_{i}) = m′_{i} ⊗ y^{i}.

Apply furthermore (ϵ⊗1)δ(m) = m then you get m = ∑ m_{i} with m_{i} ∈ M_{i}, so M = ⨁_{i∈ℕ} M_{i}. Thus define ∂ : M_{i} ⟶ M_{i−1} by the above equation. Furthermore one has ∂^{2} = 0. The converse construction can be found in the proof of the proposition.

(3) A cochain complex over 𝕂 has the form

M = (M^{0} ⟶^{∂_{0}} M^{1} ⟶^{∂_{1}} M^{2} ⟶^{∂_{2}} …)

with ∂_{i+1}∂_{i} = 0. Show that the category 𝕂-Cocomp of cochain complexes is equivalent to BM, where B is chosen as in Example 12.12.

(4) Show that the bialgebra B from Example 12.12 is not a Hopf algebra.

(5) Find a bialgebra B′ such that the category of complexes … ⟶ M_{1} ⟶ M_{0} ⟶ M^{1} ⟶ M^{2} ⟶ … and M^{B′} are monoidally equivalent. Show that B′ is a Hopf algebra.

The example 𝕂G of a Hopf algebra gives rise to the definition of particular elements in arbitrary Hopf algebras, that share certain properties with elements of a group. We will use and study these elements later on in the course on Non Commutative Geometry and Quantum Groups.

Definition 12.14. Let H be a Hopf algebra. An element g ∈ H, g ≠ 0 is called a group-like element if

Δ(g) = g ⊗ g.

Observe that ε(g) = 1 for each group-like element g in a Hopf algebra H. In fact we have g = ∇(ε ⊗ 1)Δ(g) = ε(g)g ≠ 0 hence ε(g) = 1. If the base ring is not a field then one adds this property to the definition of a group-like element.

Problem 12.6. (1) Let 𝕂 be a field. Show that an element x ∈ 𝕂G satisfies Δ(x) = x ⊗ x and ε(x) = 1 if and only if x = g ∈ G.

(2) Show that the group-like elements of a Hopf algebra form a group under multiplication of the Hopf algebra.

Example 12.15. (Universal Enveloping Algebras) A Lie algebra consists of a vector space 𝔤 together with a (linear) multiplication 𝔤 ⊗ 𝔤 ∋ x ⊗ y ↦ [x, y] ∈ 𝔤 such that the following laws hold:

[x, y] = −[y, x],
[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 (Jacobi identity).

A homomorphism of Lie algebras f : 𝔤 ⟶ 𝔥 is a linear map f such that f([x, y]) = [f(x), f(y)]. Thus Lie algebras form a category 𝕂- Lie. An important example is the Lie algebra associated with an associative algebra (with unit). If A is an algebra then the vector space A with the Lie multiplication

[x, y] := xy − yx

is a Lie algebra denoted by A^{L}. This construction of a Lie algebra defines a covariant functor -^{L} : 𝕂-Alg ⟶ 𝕂- Lie. This functor leads to the following universal problem. Let 𝔤 be a Lie algebra. An algebra U(𝔤) together with a Lie algebra homomorphism ι : 𝔤 ⟶ U(𝔤)^{L} is called a (the) universal enveloping algebra of 𝔤, if for every algebra A and for every Lie algebra homomorphism f : 𝔤 ⟶ A^{L} there exists a unique homomorphism of algebras g : U(𝔤) ⟶ A such that the following diagram commutes

[figure: 𝔤; U(𝔤)^{L}; ι; f; A^{L}.; g]

The universal enveloping algebra U(𝔤) is (if it exists) unique up to isomorphism. It is generated as an algebra by the image of 𝔤. The universal enveloping algebra can be constructed as U(𝔤) = T (𝔤)/(x ⊗ y − y ⊗ x − [x, y]) where T (𝔤) = 𝕂 ⊕ 𝔤 ⊕ 𝔤 ⊗ 𝔤 … is the tensor algebra. The map ι : 𝔤 ⟶ U(𝔤)^{L} is injective. The universal enveloping algebra U(𝔤) is a Hopf algebra. The comultiplication is given by the diagram

[figure: 𝔤; U(𝔤); ι; f; U(𝔤) ⊗ U(𝔤); Δ]

with f(x) := x⊗1+1⊗x which defines a Lie algebra homomorphism f : 𝔤 ⟶ (U(𝔤)⊗U(𝔤))^{L}. The counit is given by

[figure: 𝔤; U(𝔤); ι; f; 𝕂; ε]

with f(x) = 0 for all x ∈ 𝔤. One shows easily by using the universal property, that Δ is coassociative and has counit ε. Define an algebra homomorphism S : U(𝔤) ⟶ (U(𝔤))^{op} by

[figure: 𝔤; U(𝔤); ι; f; (U(𝔤))^{op}; S]

with f(x) := −x which is a Lie algebra homomorphism f : 𝔤 ⟶ (U(𝔤)^{op})^{L}. Then one shows with Proposition 12.8 that U(𝔤) is a Hopf algebra. (Observe, that the meaning of U in this example and the previous example (group of units, universal enveloping algebra) is totally different, in the first case U can be applied to an algebra and gives a group, in the second case U can be applied to a Lie algebra and gives an algebra.)

The preceding example of a Hopf algebra gives rise to the definition of particular elements in arbitrary Hopf algebras, that share certain properties with elements of a Lie algebra. We will use and study these elements later on in the course on Non Commutative Geometry and Quantum Groups.

Definition 12.16. Let H be a Hopf algebra. An element x ∈ H is called a primitive element if

Δ(x) = x ⊗ 1 + 1 ⊗ x.

Let g ∈ H be a group-like element. An element x ∈ H is called a skew primitive or gprimitive element if

Δ(x) = x ⊗ 1 + g ⊗ x.

Problem 12.7. Show that the set of primitive elements P (H) = {x ∈ H|Δ(x) = x⊗1+1⊗x} of a Hopf algebra H is a Lie subalgebra of H^{L}.

Proposition 12.17. Let H be a Hopf algebra with antipode S. The following are equivalent:

(1) S^{2} =id.

(2) ∑ S(a_{(2)})a_{(1)} = ηε(a) for all a ∈ H.

(3) ∑ a_{(2)}S(a_{(1)}) = ηε(a) for all a ∈ H.

Proof. Let S^{2} =id. Then

\begin{aligned}∑ S(a_{(2)})a_{(1)} & = S^{2}(∑ S(a_{(2)})a_{(1)}) = S(∑ S(a_{(1)})S^{2}(a_{(2)})) \\ & = S(∑ S(a_{(1)})a_{(2)}) = S(ηε(a)) = ηε(a) \end{aligned}

by using Problem 12.4. Conversely assume that (2) holds. Then

\begin{aligned}S ∗ S^{2}(a) & = ∑ S(a_{(1)}S^{2}(a_{(2)}) = S(∑ S(a_{(2)})a_{(1)} \\ & = S(ηε(a)) = ηε(a). \end{aligned}

Thus S^{2} and id are inverses of S in the convolution algebra Hom(H, H), hence S^{2} = id. Analogously one shows that (1) and (3) are equivalent. □

Corollary 12.18. If H is a commutative Hopf algebra or a cocommutative Hopf algebra with antipode S, then S^{2} =id.

Remark 12.19. Kaplansky: Ten conjectures on Hopf algebras In a set of lecture notes on bialgebras based on a course given at Chicago university in 1973, made public in 1975, I. Kaplansky formulated ten conjectures on Hopf algebras that have been the aim of intensive research.

(1) If C is a Hopf subalgebra of the Hopf algebra B then B is a free left C-module.

(Yes, if H is finite dimensional [Nichols-Zoeller]; No for infinite dimensional Hopf algebras [Oberst-Schneider]; B : C is not necessarily faithfully flat [Schauenburg])

(2) Call a coalgebra C admissible if it admits an algebra structure making it a Hopf algebra. The conjecture states that C is admissible if and only if every finite subset of C lies in a finite-dimensional admissible subcoalgebra.

(Remarks.

(a) Both implications seem hard.

(b) There is a corresponding conjecture where “Hopf algebra” is replaced by “bialgebra”.

(c) There is a dual conjecture for locally finite algebras.)

(No results known.)

(3) A Hopf algebra of characteristic 0 has no non-zero central nilpotent elements.

(First counter example given by [Schmidt-Samoa]. If H is unimodular and not semisimple, e.g. a Drinfel’d double of a not semisimple finite dimensional Hopf algebra, then the integral Λ satisfies Λ ≠ 0, Λ^{2} = ε(Λ)Λ = 0 since D(H) is not semisimple, and aΛ = ε(a)Λ = Λε(a) = Λa since D(H) is unimodular [Sommerhäuser].)

(4) (Nichols). Let x be an element in a Hopf algebra H with antipode S. Assume that

for any a in H we have

∑ b_{i}xS(c_{i}) = ε(a)x

where Δa = ∑ b_{i} ⊗ c_{i}. Conjecture: x is in the center of H.

(Yes, since ax = ∑ a_{(1)}xε(a_{(2)}) = ∑ a_{(1)}xS(a_{(2)})a_{(3)}) = ∑ ε(a_{(1)})xa_{(2)} = xa.) In the remaining six conjectures H is a finite-dimensional Hopf algebra over an

algebraically closed field.

(5) If H is semisimple on either side (i.e. either H or the dual H^{∗} is semisimple as an

algebra) the square of the antipode is the identity.

(Yes if char(𝕂) = 0 [Larson-Radford], yes if char(𝕂) is large [Sommerhäuser])

(6) The size of the matrices occurring in any full matrix constituent of H divides the

dimension of H.

(Yes if Hopf algebra is defined over ℤ [Larson]; in general not known; work by [Montgomery-Witherspoon], [Zhu], [Gelaki])

(7) If H is semisimple on both sides the characteristic does not divide the dimension. (Larson-Radford)

(8) If the dimension of H is prime then H is commutative and cocommutative. (Yes in characteristic 0 [Zhu: 1994]) Remark. Kac, Larson, and Sweedler have partial results on 5 – 8. (Was also proved by [Kac]) In the two final conjectures assume that the characteristic does not divide the

dimension of H.

(9) The dimension of the radical is the same on both sides.

(Counterexample by [Nichols]; counterexample in Frobenius-Lusztig kernel of U_{q}(sl(2)) [Schneider])

(10) There are only a finite number (up to isomorphism) of Hopf algebras of a given

dimension.

(Yes for semisimple, cosemisimple Hopf algebras: Stefan 1997) (Counterexamples: [Andruskiewitsch, Schneider], [Beattie, others] 1997)

Quickies in Advanced Algebra 101

13. Quickies in Advanced Algebra

I. Allgemeine Modultheorie. (1) Sei R ein Ring. Dann ist _{R}R ein R-Links-Modul.

(2) Sei M eine abelsche Gruppe und End(M) der Endomorphismenring von M. Dann ist M ein End(M)-Modul.

(3) {(¯1, ¯0), (¯0, ¯1)} ist eine Erzeugendenmenge für den ℤ-Modul ℤ/(2) × ℤ/(3).

(4) {(¯1, ¯1)} ist eine Erzeugendenmenge für den ℤ-Modul ℤ/(2) × ℤ/(3).

(5) _{ℤ}ℤ/(n) besitzt als Modul keine Basis, d.h. dieser Modul ist nicht frei.

(6) Sei V = ⨁_{i=0}^{∞} Kb_{i} ein abzählbar unendlich dimensionaler Vektorraum über dem Körper K. Seien p, q, a, b ∈ Hom(V, V) definiert durch

p(b_{i}) := b_{2i},
q(b_{i}) := b_{2i+1},
a(b_{i}) := \begin{cases}b_{i/2}, & wenn i gerade ist, und \\ 0, & wenn i ungerade ist. \end{cases}
b(b_{i}) := \begin{cases}b_{i−1/2}, & wenn i ungerade ist, und \\ 0, & wenn i gerade ist. \end{cases}

Zeige pa + qb = id_{V}, ap = bq = id, aq = bp = 0.

Zeige, daß für R = End_{K}(V) gilt _{R}R = Ra ⊕ Rb und R_{R} = pR ⊕ qR.

(7) Sind {(0, …, a, …, 0)|a ∈ K_{n}} und {(a, 0, …, 0)|a ∈ K_{n}} isomorph als M_{n}(K)- Moduln?

(8) Zu jedem Modul P gibt es einen Modul Q mit P ⊕ Q ≅ Q.

(9) Welche der folgenden Aussagen ist wahr?

(a) P_{1} ⊕ Q = P_{2} ⊕ Q ⟹ P_{1} = P_{2}?

(b) P_{1} ⊕ Q = P_{2} ⊕ Q ⟹ P_{1} ≅ P_{2}?

(c) P_{1} ⊕ Q ≅ P_{2} ⊕ Q ⟹ P_{1} ≅ P_{2}?

(10) ℤ/(2) ⊕ ℤ/(6) ⊕ ℤ/(6) ⊕ … ≅ ℤ/(6) ⊕ ℤ/(6) ⊕ ℤ/(6) ⊕ . . ..

(11) ℤ/(2) ⊕ ℤ/(4) ⊕ ℤ/(4) ⊕ … ≅ ℤ̸/(4) ⊕ ℤ/(4) ⊕ ℤ/(4) ⊕ . . ..

(12) Man finde zwei abelsche Gruppen P und Q, so daß P isomorph zu einer Untergruppe von Q ist und Q isomorph zu einer Untergruppe von P ist und P ≅ Q̸ gilt.

II. Tensorprodukte

(1) In ℂ ⊗_{ℂ} ℂ gilt 1 ⊗ i − i ⊗ 1 = 0.

In ℂ ⊗_{ℝ} ℂ gilt 1 ⊗ i − i ⊗ 1 ≠ 0.

(2) Für jeden R-Modul gilt R ⊗_{R} M ≅ M.

(3) Sei der ℚ-Vektorraum V = ℚ^{n} gegeben.

(a) Bestimme dim_{ℝ}(ℝ ⊗_{ℚ} V).

(b) Gib explizit einen Isomorphismus ℝ ⊗_{ℚ} V ≅ ℝ^{n} an.

(4) Sei V ein ℚ-Vektorraum und W ein ℝ-Vektorraum.

(a) Hom_{ℝ}(.ℝ_{ℚ}, .W) ≅ W in ℚ-Mod.

(b) Hom_{ℚ}(.V, .W) ≅ Hom_{ℝ}(.ℝ ⊗_{ℚ} V, .W).

(c) Sei dim_{ℚ}V < ∞ und dim_{ℝ}W < ∞. Wie kann man verstehen, daß in 4b links unendliche Matrizen und rechts endliche Matrizen stehen?

102 Advanced Algebra – Pareigis

(d) Hom_{ℚ}(.V, Hom_{ℝ}(.ℝ, .W) ≅ Hom_{ℝ}(.ℝ ⊗_{ℚ} V, .W).

(5) ℤ/(18) ⊗_{ℤ} ℤ/(30) ≠ 0.

(6) m : ℤ/(18) ⊗_{ℤ} ℤ/(30) ∋ \overline{x} ⊗ \overline{y} ↦ \overline{xy} ∈ ℤ/(6) ist ein Homomorphismus und m ist bijektiv.

(7) Für ℚ-Vektorräume V und W gilt V ⊗_{ℤ} W ≅ V ⊗_{ℚ} W.

(8) Für jede endliche abelsche Gruppe M gilt ℚ ⊗_{ℤ} M = 0.

(9) ℤ/(m) ⊗_{ℤ} ℤ/(n) ≅ ℤ/(ggT(m, n)).

(10) ℚ ⊗_{ℤ} ℤ/(n) = 0.

(11) Hom_{ℤ}(ℚ, ℤ/(n)) = 0.

(12) Gib explizit Isomorphismen an für

\begin{matrix}ℤ ⊗_{ℤ} ℚ ≅ ℚ, \\ 3ℤ ⊗_{ℤ} ℚ ≅ ℚ. \end{matrix}

Zeige, daß das Diagramm kommutiert:

[figure: ℚ; ℚ; 3·; 3ℤ ⊗_{ℤ} ℚ; ℤ ⊗_{ℤ} ℚ; ≅; ≅]

(13) Der Homomorphismus 2ℤ⊗_{Z} ℤ/(2) ⟶ ℤ⊗_{ℤ}ℤ/(2) ist der Nullhomomorphismus,

beide Moduln sind aber von Null verschieden. III. Projektive Moduln

(1) Bestimme die Dual-Basis von ℝ^{2} im Sinne der Vorlesung.

(2) Zeige, daß die Spur eines Homomorphismus f : V ⟶ V gegeben ist durch

End_{K}(V) ≅ V ⊗ V^{∗} ⟶^{ev} K.

(3) Bestimme die Dual-Basis von _{R×S}R × 0 ⊆ R × S.

(4) K_{n} ist ein projektiver M_{n}(K)-Modul.

(5) Sei R := K × K mit einem Körper K.

(a) Zeige: P := {(a, 0)|a ∈ K} ist ein endlich erzeugter projektiver R-Modul.

(b) Sind die R-Moduln P und Q := {(0, a)|a ∈ K} isomorph?

(c) Man finde eine Dual-Basis für P.

(6) Zeige für R := M_{n}(K), daß P = K_{n} endlich erzeugt projektiv ist, und finde eine Dual-Basis.

(7) Zu jedem projektiven Modul P gibt es einen freien Modul F mit P ⊕ F ≅ F. IV. Kategorien und Funktoren

(1) In R-Mod gilt:

f : M ⟶ N Monomorphismus ⟺ f injektiver Homomorphismus.

(2) (a) Wenn f : M ⟶ N surjektiv ist, dann ist Hom_{R}(f, P) : Hom_{R}(N, P)

⟶ Hom_{R}(M, P) injektiv.

(b) ℤ ⟶ ℤ/(n) induziert eine injektive Abbildung

Hom_{ℤ}(ℤ/(n), M) ⟶ Hom_{ℤ}(ℤ.M) ≅ M. Warum kann man Hom_{ℤ}(ℤ/(n), M) mit {x ∈ M|nx = 0} ⊆ M identifizieren?

Quickies in Advanced Algebra 103

(c) T_{n}(M) := {x ∈ M|nx = 0} ist ein Funktor Ab ⟶ Ab.

(d) Die Einbettung T_{n}(M) ⟶ M ist ein funktorieller Homomorphismus.

(3) In R-Mod gilt:

f : M ⟶ N Epimorphismus ⟺ f surjektiv.

(4) Wenn F ein kovarianter darstellbarer Funktor ist und f : M ⟶ N ein Monomorphismus ist, dann ist F(f) : F(M) ⟶ F(N) ebenfalls ein Monomorphismus.

(5) Der Funktor F : M ↦ ℤ/(n) ⊗_{ℤ} M ist nicht darstellbar.

(6) Der Funktor F : V ↦ ℚ^{n} ⊗_{ℚ} V ist darstellbar.

(7) Der Funktor T_{n} : Ab ⟶ Ab mit T_{n}(M) := {x ∈ M|nx = 0} ist darstellbar.

(8) Jeder additive Funktor F : R-Mod ⟶ S-Mod erhält endliche direkte Summen,

d.h. F (M ⊕ N) ≅ F (M) ⊕ F (N). V. Morita-Äquivalenz

(1) Zeige, daß (K × K)-Mod nicht äquivalent zu K-Mod ist.

(2) Sei K ein Körper, B := M_{n}(K), _{K}P_{B} := K^{n} die Menge der Zeilenvektoren, _{B}Q_{K} die Menge der Spaltenvektoren. Finde f : P ⊗_{B} Q ⟶ K und g : Q ⊗_{K} P ⟶ B, so daß (K, B, P, Q, f, g) einen Morita- Kontext bildet. Ist dieser Morita-Kontext strikt?

(3) Zeige ℝ-Mod ≅ ℂ̸-Mod.

(4) Bestimme das Bild der Abbildungen f und g im kanonischen Morita-Kontext

(A, B, P, Q, f, g) für (a) A := ℤ/(6) und P := ℤ/(2),

(b) A := ℤ/(4) und P := ℤ/(4) ⊕ ℤ/(2),

(c) A := ℤ/(6) und P := ℤ/(6) ⊕ ℤ/(2).

VI. Halbeinfache Moduln (1) Finde alle einfachen Moduln über K, ℤ, K[x].

(2) Finde alle einfachen Moduln über ℂ[x], M_{2}(K), ℚ[x]/(x^{2}).

(3) Finde alle einfachen Moduln über

( \begin{matrix}K & K \\ 0 & K \end{matrix} ) .

(4) Stelle End_{K[x]}(K[x]/(x) ⊕ K[x]/(x − 1)) als Ring von Matrizen dar. VII. Radikal und Sockel

(1) Radikal und Sockel endlich erzeugter abelscher Gruppen. Bestimme (a) Rad(_{ℤ}ℤ/(p)), Soc(_{ℤ}ℤ/(p)).

(b) Rad(ℤ/(p^{n})), Soc(ℤ/(p^{n})).

(c) Rad(ℤ/(p^{n}) ⊕ ℤ/(p^{m})), Soc(ℤ/(p^{n}) ⊕ ℤ/(p^{m})).

(d) Für welche n ∈ ℕ ist Rad(_{ℤ}ℤ/(n)) = 0?

(2) Bestimme Radikal und Sockel der abelschen Gruppen (a) ℤ,

(b) ℚ,

(c) ℚ/ℤ.

VIII. Lokale Ringe (1) Sei R ein lokaler Ring. Dann ist R/𝔪 ein Schiefkörper.

104 Advanced Algebra – Pareigis

(2) Der Ring der formalen Potenzreihen K[[x]] ist ein lokaler Ring.

(3) Der Polynomring K[x] ist kein lokaler Ring. IX. Lokalization

(1) S := 2ℤ \ {0} ist multiplikativ abgeschlossen. S^{−1}ℤ ⫋ ℚ.

(2) (a) Wenn S ⊆ T multiplikativ abgeschlossene Mengen sind, dann wird dadurch

ein Homomorphismus ψ : S^{−1}M ⟶ T^{−1}M induziert.

(b) Finde eine hinreichende Bedingung dafür, daß ψ injektiv ist.

(c) Für S := ℤ \ (p) und T := ℤ \ {0} beschreibe man den Homomorphismus ψ.

(d) Für S ⊂ T zeige man S^{−1}T^{−1}M = T^{−1}S^{−1}M = T^{−1}M.

(e) Wenn S, T multiplikativ abgeschlossen sind, dann ist auch S ∩ T multiplikativ abgeschlossen. Wie drückt sich das für (S ∩ T)^{−1}M aus?

(f) Sei T := (ℤ \ (2)) ∩ (ℤ \ (3)). Bestimme T^{−1}ℤ.

(g) Ist ℤ/(6) ⟶ T^{−1}(ℤ/(6)) injektiv? surjektiv?

XII. Bialgebras and Hopf algebras

(1) Let H be a bialgebra and A be an H-left-module algebra. On A ⊗ H define a

multiplication

(a ⊗ h)(a′ ⊗ h′) := a(h_{(1)} · a′) ⊗ h_{(2)}h′.

Show that this defines a structure of an algebra on A⊗H. This algebra is usually denoted by A#H and the elements are denoted by a#h := a ⊗ h.

(2) Let A be a G-Galois extension field of the base field 𝕂. Define an homomorphism φ : A#𝕂G ⟶ End_{𝕂}(A) by φ(a#g)(b) := ag(b). Show that φ is a homomorphism of algebras.

(3) Let G := C_{2} be the cyclic group with two elements., A := ℂ, and 𝕂 := ℝ. Show that φ : ℂ#ℝC_{2} ⟶ End_{ℝ}(ℂ) is an isomorphism of algebras.