Notes on Azumaya algebras

Yorck Sommerhäuser

Version 1

Contents

1 Central separable algebras 1 1.1 The definition of separability . . . . . . . . . . . . . . . . . . . . 1 1.2 The separability idempotent . . . . . . . . . . . . . . . . . . . . . 2 1.3 Centers of separable algebras . . . . . . . . . . . . . . . . . . . . 2 1.4 The Villamayor-Zelinsky theorem . . . . . . . . . . . . . . . . . . 3 1.5 Ideals in central separable algebras . . . . . . . . . . . . . . . . . 4 1.6 Tensor products of central separable algebras . . . . . . . . . . . 5 1.7 Generators and faithfulness . . . . . . . . . . . . . . . . . . . . . 8 1.8 Characterization of central separable algebras . . . . . . . . . . . 9 1.9 Bimodules over central separable algebras . . . . . . . . . . . . . 11

2 Azumaya algebras 14 2.1 Lifting idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2 Central idempotents . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.3 Criteria for Azumaya algebras . . . . . . . . . . . . . . . . . . . . 16 2.4 Local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.5 Matrix rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Bibliography 20

1 Central separable algebras

1.1 In the following, R denotes a commutative ring. Tensor products without an index are taken over R.

Definition Suppose that A is an R-algebra.

1. A is called central if the map

R → Z(A), r ↦ r1_{A}

is a ring isomorphism.

2. A is separable if it is projective as an A ⊗ A^{op}-module, where A is considered as an A ⊗ A^{op}-module via

A ⊗ A^{op} × A → A, (a ⊗ a′, a′′) ↦ aa′′a′

3. A is called an Azumaya algebra if it is faithful, finitely generated, and projective as an R-module and the mapping

A ⊗ A^{op} → End_{R}(A), a ⊗ b ↦ (c ↦ acb)

is bijective (cf. [3], Chap. 8, p. 186; [6], Chap. III, § 5, Par. (5.1), p. 134).

It is the purpose of this section to prove that A is an Azumaya algebra if and only if it is central separable. This goal will be achieved in Paragraph 1.8. In the definition of centrality, note that, in contrast to the case of fields, the injectivity of the map r ↦ r1_{A} is not automatic.

1.2 We need an equivalent characterization of separability:

Proposition For an R-algebra A, the following assertions are equivalent:

1. A is separable.

2. The exists an element e ∈ A ⊗ A such that μ_{A}(e) = 1_{A} and (a ⊗ 1_{A})e = e(1_{A} ⊗ a) for all a ∈ A.

Proof. The multiplication map μ_{A} : A ⊗ A → A is A ⊗ A^{op}-linear. If A is projective over A ⊗ A^{op}, there must be a map f : A → A ⊗ A satisfying μ_{A} ∘ f = id_{A}. The element e = f(1_{A}) satisfies the requirements. Therefore, the first assertion implies the second.

For the converse, define

f : A → A ⊗ A, a ↦ (a ⊗ 1_{A})e

This map is A ⊗ A^{op}-linear and satisfies μ_{A} ∘ f = id_{A}. Therefore, A is a direct summand of the free A ⊗ A^{op}-module A ⊗ A, and thus projective. □

Note that an element e with these properties is an idempotent in A ⊗ A^{op}: If e = ∑_{i} a_{i} ⊗ b_{i}, we have

e^{2} = (∑_{i} a_{i} ⊗ b_{i})e = (∑_{i} a_{i}b_{i} ⊗ 1_{A})e = e

It is therefore often called a separability idempotent.

The above characterization implies directly that separability is preserved under passage to quotients:

Corollary Suppose that I ⊂ A is a two-sided ideal. If A is a separable Ralgebra, then A/I is a separable R-algebra.

Proof. If e ∈ A ⊗ A is a separability idempotent for A and π : A → A/I is the quotient map, then (π ⊗ π)(e) is a separability idempotent for A/I (cf. [2], Chap. II, Prop. 1.11, p. 46). □

1.3 It is clear that an A ⊗ A^{op}-module is the same as an A-bimodule. For every A-bimodule M, we can consider the centralizer M^{A} of M:

M^{A} := {m ∈ M | am = ma for all a ∈ A}

If we consider A as an A ⊗ A^{op}-module as above, we have a map

Hom_{A⊗Aop}(A, M) → M^{A}, f ↦ f(1_{A})

which is bijective since the inverse map is

M^{A} → Hom_{A⊗Aop}(A, M), m ↦ (a ↦ am)

By definition, A is separable if it is projective as an A ⊗ A^{op}-module. This is equivalent to the requirement that Hom_{A⊗Aop}(A, −) is an exact functor. We therefore can say the following:

Proposition Suppose that A is a separable R-algebra. Then for each exact sequence

M → M′ → M′′

the sequence

M^{A} → M′^{A} → M′′^{A}

is also exact.

Moreover, we have seen that this property characterizes separable algebras (cf. [2], Chap. II, Lem. 1.2, p. 42, Cor. 1.5, p. 43, [6], § III.5, Lem. (5.1.2), p. 135).

We want to apply this proposition to determine the center of a quotient algebra:

Corollary Suppose that A is a separable R-algebra and that I ⊂ A is a twosided ideal. If

π : A → A/I

is the canonical projection, then Z(A/I) = π(Z(A)).

Proof. Consider A/I as an A-bimodule by pullback via π. Since π is surjective, it follows from the preceding proposition that the induced map

A^{A} → (A/I)^{A}

is also surjective. But this is precisely the assertion. (cf. [2], Chap. II, Prop. 1.11, p. 46 [6], § III.5, Lem. (5.1.4), p. 136). □

1.4 Next, we give the Villamayor-Zelinsky theorem:

Proposition A separable R-algebra A that is projective as an R-module is a finitely generated R-module.

Proof. Since A is a projective R-module, we have for every family (a_{i})_{i∈I} of generators corresponding linear functions f_{i} : A → R such that

a = ∑_{i∈I} f_{i}(a)a_{i}

where in this sum only finitely many coefficients are nonzero. Choose a separability idempotent

e = ∑_{j=1}^{n} x_{j} ⊗ y_{j}

We then have that the set

J := {i ∈ I | (id_{A} ⊗f_{i})(e) ≠ 0}

is finite. For a ∈ A, we have

a = a ∑_{j=1}^{n} x_{j}y_{j} = a ∑_{j=1}^{n} ∑_{i∈J} x_{j}f_{i}(y_{j})a_{i} = ∑_{j=1}^{n} ∑_{i∈J} x_{j}f_{i}(ay_{j})a_{i}

and therefore the elements x_{j}a_{i} for j = 1, …, n and i ∈ J generate A as an R-module (cf. [2], Chap. II, Prop. 2.1, p. 47, [14], Lem. 9.2.12, p. 311). □

1.5 For a central separable R-algebra A, there is a bijective correspondence between ideals of R and two-sided ideals of A, as we will see in Paragraph 1.9. Here, we begin with some preparation:

Lemma Suppose that A is a separable R-algebra. Then Z(A) is an R-direct summand of A.

Proof. Choose a separability idempotent e ∈ A ⊗ A^{op}. Since the map

f : A ⊗ A^{op} → End_{R}(A), a ⊗ a′ ↦ (a′′ ↦ aa′′a′)

is an algebra homomorphism, f(e) is a projection. It follows from the defining properties of e that f(e) maps A to its center and that f(e) is the identity on the center. Therefore, we have the decomposition

A = Z(A) ⊕ ker(f(e))

(cf. [2], Chap. II, Lem. 3.1, p. 51, [6], § III.5, Lem. (5.1.6), p. 136). □

The next proposition is still preparatory, although it already provides the essential part of the correspondence:

Proposition Suppose that A is a central separable R-algebra.

1. For an ideal I ⊂ R, we have I = R ∩ IA.

2. For a maximal ideal I ⊂ A, J := R ∩ I is a maximal ideal of R, and we have I = JA.

Proof. In the first assertion, it is clear that I ⊂ R ∩ IA. For the reverse inclusion, choose an R-submodule L of A such that A = R ⊕ L; this exists by the preceding lemma. We then have

IA = IR ⊕ IL ⊂ I ⊕ L

and therefore R ∩ IA ⊂ I (cf. [2], Chap. II, Cor. 3.2, p. 51).

The proof of the second statement is more complicated. The algebra A/I is simple by assumption. From Corollary 1.3, we know that the canonical projection induces a surjective map from the center of A to the center of the quotient. This shows that Z(A/I) ≅ R/J.

To show that J is maximal, we have to show that R/J is a field. If r ∈ R \ J, its equivalence class r̄ ∈ R/J is nonzero. Since A/I is simple, the nonzero ideal r̄(A/I) must coincide with the whole quotient algebra, which shows that r̄ is invertible in A/I. Choose an element a ∈ A that satisfies r̄ā = 1_{A/I}. From the lemma above, we know that R/J is a direct summand of A/I, i.e., we can find an R/J-submodule L such that

A/I = R/J ⊕ L

We write ā in the form ā = ā_{1} + ā_{2}, where ā_{1} ∈ R/J and ā_{2} ∈ L. We then have

1_{A/I} = r̄ā_{1} + r̄ā_{2} ∈ R/J ⊕ L

This implies 1_{A/I} = r̄ā_{1}, and therefore r̄ is invertible in R/J, which proves that R/J is a field.

The ideal JA is obviously contained in I. Consider the factor algebra A/JA. For the same reasons as above, this is a separable algebra. From the first statement, we have R ∩ JA = J, and therefore we get as above that the center of A/JA is isomorphic to R/J. But a separable algebra over a field is semisimple, and if the center is one-dimensional the algebra must be simple itself. Therefore, A/JA is simple, and the surjective algebra homomorphism

A/JA → A/I

is an isomorphism, which proves that I = JA (cf. [2], Chap. II, Lem. 3.5, p. 53, [6], § III.5, Lem. (5.1.7), p. 137). □

1.6 In this paragraph, we show that the tensor product of two central separable algebras is central separable. Again, we need some preparation: Suppose that A and B are two R-algebras and that M and M′ are two A-modules and that N and N′ are two B-modules. As usual, we have a canonical map

ψ : Hom_{A}(M, M′) ⊗ Hom_{B}(N, N′) → Hom_{A⊗B}(M ⊗ N, M′ ⊗ N′)

Lemma If M is a finitely generated projective A-module and N is a finitely generated projective B-module, then ψ is an isomorphism.

Proof. Choose generators m_{1}, …, m_{k} of M and generators n_{1}, …, n_{l} of N, as well as corresponding homomorphisms f_{i} ∈ Hom_{A}(M, A) and g_{j} ∈ Hom_{B}(N, B) such that

m = ∑_{i=1}^{k} f_{i}(m)m_{i} n = ∑_{j=1}^{l} g_{j}(n)n_{j}

for all m ∈ M and all n ∈ N. If φ : M ⊗ N → M′ ⊗ N′ is A ⊗ B-linear, we can, for all i = 1, …, k and j = 1, …, l, write

φ(m_{i} ⊗ n_{j}) = ∑_{p=1}^{q_{ij}} m′_{ijp} ⊗ n′_{ijp}

Now define

μ_{ijp} : M → M′, m ↦ f_{i}(m)m′_{ijp} ν_{ijp} : N → N′, n ↦ g_{j}(n)n′_{ijp}

We claim that

ψ(∑_{i=1}^{k} ∑_{j=1}^{l} ∑_{p=1}^{q_{ij}} μ_{ijp} ⊗ ν_{ijp}) = φ

This holds since

ψ(∑_{i=1}^{k} ∑_{j=1}^{l} ∑_{p=1}^{q_{ij}} μ_{ijp} ⊗ ν_{ijp})(m ⊗ n) = ∑_{i=1}^{k} ∑_{j=1}^{l} ∑_{p=1}^{q_{ij}} μ_{ijp}(m) ⊗ ν_{ijp}(n)

= ∑_{i=1}^{k} ∑_{j=1}^{l} ∑_{p=1}^{q_{ij}}(f_{i}(m) ⊗ g_{j}(n))(m′_{ijp} ⊗ n′_{ijp})

= ∑_{i=1}^{k} ∑_{j=1}^{l}(f_{i}(m) ⊗ g_{j}(n))φ(m_{i} ⊗ n_{j})

= φ(∑_{i=1}^{k} ∑_{j=1}^{l} f_{i}(m)m_{i} ⊗ g_{j}(n)n_{j}) = φ(m ⊗ n)

This proves that the map is surjective. To prove injectivity, we consider the map

T_{A} : Hom_{A}(M, A) ⊗ M′ → Hom_{A}(M, M′), f ⊗ m′ ↦ (m ↦ f(m)m′)

and the analogous map

T_{B} : Hom_{B}(N, B) ⊗ N′ → Hom_{B}(N, N′), g ⊗ n′ ↦ (n ↦ g(n)n′)

Now suppose that

ψ(∑_{p=1}^{q} μ_{p} ⊗ ν_{p}) = 0

This implies ∑_{p=1}^{q} μ_{p}(m_{i}) ⊗ ν_{p}(n_{j}) = 0. We have μ_{p}(m) = ∑_{i=1}^{k} f_{i}(m)μ_{p}(m_{i}) and ν_{p}(n) = ∑_{j=1}^{l} g_{j}(n)ν_{p}(n_{j}), which implies

∑_{p=1}^{q} μ_{p} ⊗ ν_{p} = ∑_{p=1}^{q} T_{A}(∑_{i=1}^{k} f_{i} ⊗ μ_{p}(m_{i})) ⊗ T_{B}(∑_{j=1}^{l} g_{j} ⊗ ν_{p}(n_{j}))

This shows that ∑_{p=1}^{q} μ_{p} ⊗ ν_{p} is the image of the tensor

∑_{p=1}^{q} ∑_{i=1}^{k} ∑_{j=1}^{l} f_{i} ⊗ g_{j} ⊗ μ_{p}(m_{i}) ⊗ ν_{p}(n_{j})

with respect to the map

Hom_{A}(M, A) ⊗ Hom_{B}(N, B) ⊗ M′ ⊗ N′ → Hom_{A}(M, M′) ⊗ Hom_{B}(N, N′)
f ⊗ g ⊗ m′ ⊗ n′ ↦ T_{A}(f ⊗ m′) ⊗ T_{B}(g ⊗ n′)

But this is zero, since already the tensor ∑_{p=1}^{q} μ_{p}(m_{i}) ⊗ ν_{p}(n_{j}) is zero. (cf. [2], Chap. I, Sec. 2.4, p. 14, [7], Lem. 1.4.1, p. 15). □

This lemma is used in the treatment of the center in the following context:

Proposition If A and B are central separable R-algebras, then A ⊗ B is a central separable R-algebra.

Proof. Let us first observe that the canonical map R → Z(A ⊗ B) is injective. This follows from the fact that R is a direct summand of A by Lemma 1.5, and similarly a direct summand of B, which implies that R ≅ R ⊗R is a direct summand of A⊗B. To see that A⊗B is separable, choose a separability idempotent e_{A} = ∑_{i=1}^{n} a_{i} ⊗a′_{i} for A and a separability idempotent e_{B} = ∑_{j=1}^{m} b_{j} ⊗b′_{j} for B. Recall from Paragraph 1.2 that this means that we have

∑_{i=1}^{n} aa_{i} ⊗ a′_{i} = ∑_{i=1}^{n} a_{i} ⊗ a′_{i}a  ∑_{i=1}^{n} a_{i}a′_{i} = 1_{A}

for all a ∈ A, and similarly for B. This implies that

e_{A⊗B} := ∑_{i=1}^{n} ∑_{j=1}^{m} a_{i} ⊗ b_{j} ⊗ a′_{i} ⊗ b′_{j}

is a separability idempotent for A ⊗ B. To see that A ⊗ B is central, note that we have from the above lemma that the canonical map

Hom_{A⊗Aop}(A, A) ⊗ Hom_{B⊗Bop}(B, B) → Hom_{A⊗Aop⊗B⊗Bop}(A ⊗ B, A ⊗ B)

is a bijection. But as we have discussed in Paragraph 1.3, the above homomorphisms are given by multiplication with central elements. Therefore, we see that Z(A ⊗ B) = Z(A) ⊗ Z(B), which implies the assertion (cf. [2], Chap. II, Prop. 3.3, p. 52, [6], § III.5, (5.1), p. 134). □

It may be noted in this context that for a central separable algebra A the algebra A^{op} is also central separable. Namely, the center of A^{op} obviously coincides with the center of A, and a separability idempotent for A^{op} is given by exchanging the tensor factors in a separability idempotent for A.

1.7 Before we proceed, we need another criterion when a module over a commutative ring is a generator.

Definition Suppose that M is a module over the commutative ring R.

1. The annihilator of M is the ideal A_{M} := {r ∈ R | am = 0 for all m ∈ M}

2. The trace ideal is the ideal T_{M} := ⟨{f(m) | m ∈ M, f ∈ Hom_{R}(M, R)}⟩

We need the following variant of Nakayama’s lemma:

Lemma Suppose that M is a finitely generated R-module. If the ideal I ⊂ R satisfies IM = M, we have I + A_{M} = R.

Proof. Suppose that M is generated by m_{1}, …, m_{n}. By assumption, we can find elements a_{ij} ∈ I such that

m_{i} = ∑_{j=1}^{n} a_{ij}m_{j}

Now the Cayley-Hamilton theorem (cf. [12], Thm. 2.44, p. 27) implies that det(E_{n} − A) ∈ A_{M}, where E_{n} is the unit matrix and A = (a_{ij})_{i,j≤n}. Expanding det(E_{n} − A) via the Leibniz formula, we see that det(E_{n} − A) ∈ 1_{R} + I, which proves the assertion. (For a different proof, see [2], Chap. I, Lem. 1.7, p. 6, [6], Chap. III, Lem. (5.1.11), p. 139). □

For projective module, this implies the following:

Corollary Suppose that P is a finitely generated projective R-module. Then we have T_{M} + A_{M} = R.

Proof. Suppose that P is generated by p_{1}, …, p_{n}. Since P is projective, we can find elements and linear forms f_{1}, …, f_{n} ∈ P^{∗} = Hom_{R}(P, R) such that

p = ∑_{i=1}^{n} f_{i}(p)p_{i}

for all p ∈ P. This shows that T_{M}M = M, and therefore the assertion follows from the preceding lemma (cf. [2], Chap. I, Prop. 1.9, p. 7). □

This provides us with the following criterion for generators:

Proposition For a finitely generated projective R-module P, the following assertions are equivalent:

1. P is a generator.

2. P is faithful.

Proof. If P is a generator, we know from [1], Lem. 1, p. 6 that R is a direct summand in a direct sum of copies of P. Therefore, an element vanishing on P vanishes on R, and is therefore zero. This shows that the first statement implies the second.

Conversely, if M is faithful, then we have A_{M} = {0}, and therefore T_{M} = R. By the reference above, this tells that M is a generator. (cf. [2], Chap. I, Cor. 1.10, p. 8). □

1.8 For an R-algebra A, we can look at the space A ⊗ A as an A-bimodule. The centralizer then is

(A ⊗ A)^{A} = {x ∈ A ⊗ A | (a ⊗ 1_{A})x = x(1_{A} ⊗ a) for all a ∈ A}

In the case of a separable algebra, the separability idempotents are contained in this space. Of course, we can also view this space as a subset of A ⊗ A^{op}, where it is obviously a right ideal. If A is a central separable algebra, we can say something more:

Lemma Suppose that the algebra A is central separable. Then the left ideal of A ⊗ A^{op} generated by (A ⊗ A)^{A} coincides with A ⊗ A^{op}.

Proof. Since (A⊗A)^{A} is already a right ideal, the generated left ideal is in fact a two-sided ideal. If it does not coincide with the whole algebra, it must be contained in a maximal two-sided ideal I ⊂ A⊗A^{op}. It follows from Proposition 1.6 that A ⊗ A^{op} is a central separable algebra. If we define J := I ∩ R, it follows from Proposition 1.5 that I = J(A⊗A^{op}). For a separability idempotent e of A, this implies e ∈ I = J(A ⊗ A^{op}). But if μ_{A} is the multiplication map of A, this would imply 1_{A} = μ_{A}(e) ∈ J, which is a contradiction. □

This fact can be restated as follows:

Corollary Suppose that the algebra A is central separable. Then A is an A ⊗ A^{op}-generator.

Proof. We have to show that the trace ideal coincides with A⊗A^{op}. The dual of the A⊗A^{op}-module A is Hom_{A⊗Aop}(A, A⊗A^{op}). As explained in Paragraph 1.3, this space is isomorphic to (A⊗A)^{A}. The trace ideal, i.e., the ideal generated by the values of elements from Hom_{A⊗Aop}(A, A ⊗ A^{op}) on elements of A, therefore coincides with A(A ⊗ A)^{A}. Using the definition of the centralizer, we see that this is exactly the left ideal generated by (A ⊗ A)^{A} in A ⊗ A^{op}, which is A ⊗ A^{op} by the above lemma. □

We can now give the promised characterization of central separable algebras. Recall that a module is called a progenerator if it is a finitely generated projective generator.

Proposition For an R-algebra A, the following assertions are equivalent:

1. A is central separable.

2. A is central and an A ⊗ A^{op}-progenerator.

3. A is an R-progenerator and the canonical map A ⊗ A^{op} → End_{R}(A) is an isomorphism.

4. A is an Azumaya algebra.

Proof. 2. ⇒ 1. This is obvious, since by definition an algebra is separable if it is projective as an A ⊗ A^{op}-module.

1. ⇒ 2. By assumption, we have that A is projective over A⊗A^{op}, and obviously finitely generated by the generating element 1_{A}. That it is a generator follows from the preceding corollary.

2. ⇒ 3. This follows from the first Morita theorem (cf. [1], p. 9), which asserts in particular that progenerator is also a progenerator as a module over its endomorphism ring, and that the endomorphism ring with respect to the new module structure is isomorphic to the original ring. Here, the endomorphism ring of the A ⊗ A^{op}-progenerator A is End_{A⊗Aop}(A) ≅ Z(A) ≅ R, so A is an R-progenerator, and the endomorphism ring with respect to the new module structure is End_{R}(A), which shows that A ⊗ A^{op} ≅ End_{R}(A).

3. ⇒ 2. This follows also from the first Morita theorem in a similar way. If A is an R-progenerator, this theorem asserts that

End_{End_{R}(A)}(A) ≅ R

and that A is an End_{R}(A)-progenerator. Under the assumption that A ⊗ A^{op} ≅ End_{R}(A), this implies End_{A⊗Aop}(A) ≅ R. But since End_{A⊗Aop}(A) ≅ Z(A), we see that R is central, and also that A is an A ⊗ A^{op}-progenerator.

3. ⇔ 4. This follows directly from Proposition 1.7 (cf. [2], Chap. II, Thm. 3.4, p. 52). □

1.9 If A is an A ⊗ A^{op}-progenerator and the space Hom_{A⊗Aop}(A, A) is isomorphic to R, we know from the first Morita theorem that the category of A-bimodules is isomorphic to the category of R-modules. To make this equivalence explicit, we need the following lemma:

Lemma Suppose that A and B are rings. Suppose that P is a finitely generated projective left A-module, that M is an A-B-bimodule, and that N is a left Bmodule. Then the map

θ : Hom_{A}(P, M) ⊗_{B} N → Hom_{A}(P, M ⊗_{B} N), f ⊗ n ↦ (p ↦ f(p) ⊗ n)

is an isomorphism.

Proof. The space P^{∗} := Hom_{A}(P, A) is a right A-module, and we can, similarly to the proof of Lemma 1.6, consider the map

T_{M} : P^{∗} ⊗_{A} M → Hom_{A}(P, M), f ⊗ m ↦ (p ↦ f(p)m)

which is right B-linear. If we choose generators p_{1}, …, p_{k} of P, as well as corresponding homomorphisms f_{i} ∈ P^{∗} such that p = ∑_{i=1}^{k} f_{i}(p)p_{i} for all p ∈ P, we see that the inverse of T_{M} is given as

T_{M}^{−1} : Hom_{A}(P, M) → P^{∗} ⊗_{A} M, g ↦ ∑_{i=1}^{k} f_{i} ⊗ g(p_{i})

Inserting the module M ⊗_{B} N instead of M, we have the following commutative diagram:

[figure: Hom_{A}(P, M) ⊗_{B} N; Hom_{A}(P, M ⊗_{B} N); θ; (P^{∗} ⊗_{A} M) ⊗_{B} N; P^{∗} ⊗_{A} (M ⊗_{B} N); T_{M} ⊗_{B} id_{N}; T_{M⊗N}]

This shows that θ is a composition of isomorphisms, and therefore itself an isomorphism (cf. [2], Chap. I, 2.7, p. 15). □

One may view the fact that T_{M} is an isomorphism as the special case of the above lemma in which M = A = B. The correspondence between R-modules and A-bimodules now takes the following form:

Proposition Suppose that A is a central separable R-algebra.

1. If M is an A-bimodule, then the map

A ⊗ M^{A} → M, a ⊗ m ↦ am

is a bimodule isomorphism.

2. If N is an R-module, then the centralizer of the A-bimodule A ⊗ N is isomorphic to N under the map

N → (A ⊗ N)^{A}, n ↦ 1_{A} ⊗ n

Proof. By assumption, A is an A ⊗ A^{op}-progenerator. Recall from the first Morita theorem that the category equivalence is given by tensoring with the progenerator, resp. by tensoring with the dual of the progenerator. The proof of the proposition now consists in the reinterpretation of the tensoring with the dual of the progenerator Hom_{A⊗Aop}(A, A ⊗ A^{op}). From the special case of the lemma discussed above and the discussion in Paragraph 1.3, we have for an A-bimodule M that

Hom_{A⊗Aop}(A, A ⊗ A^{op}) ⊗_{A⊗Aop} M ≅ Hom_{A⊗Aop}(A, M) ≅ M^{A}  (*)

which is given on elements by f ⊗ m ↦ f(1_{A})m.

Let us set up in detail the Morita context that arises from this situation. We let P := A, considered as a left A ⊗ A^{op}-module and a right R-module. Let Q := Hom_{A⊗Aop}(A, A⊗A^{op}), considered as a left R-module and a right A⊗A^{op}- module. We then have the mappings

(·, ·) : P ⊗_{R} Q = A ⊗_{R} Hom_{A⊗Aop}(A, A ⊗ A^{op}) → A ⊗ A^{op}, a ⊗ f ↦ f(a)

as well as

[·, ·] : Q⊗_{A⊗Aop} P = Hom_{A⊗Aop}(A, A⊗A^{op})⊗_{A⊗Aop} A → R, f ⊗a ↦ f(1_{A})(a)

The second map needs a little explanation. A priori, the square brackets map f ⊗ a to the A ⊗ A^{op}-endomorphism

A → A, a′ ↦ f(a′).a

Now, an A ⊗ A^{op}-endomorphism is given by multiplication with a central element, which in this case is f(1_{A})(a). Since A is central, this element is in fact in R.

For an A-bimodule M, we now get from the first Morita theorem that the map

A ⊗ Hom_{A⊗Aop}(A, A ⊗ A^{op}) ⊗_{A⊗Aop} M → M, a ⊗ f ⊗ m ↦ f(a).m = af(1_{A}).m

is an isomorphism. In view of the isomorphism (*), this is the composition of the map

A ⊗ Hom_{A⊗Aop}(A, A ⊗ A^{op}) ⊗_{A⊗Aop} M → A ⊗ M^{A}, a ⊗ f ⊗ m ↦ a ⊗ f(1_{A})m

considered above and the map

A ⊗ M^{A} → M, a ⊗ m ↦ am

given in the first statement. As the first map is an isomorphism, the second must be an isomorphism, too. This proves the first statement.

If M = A ⊗ N for an R-module N, the first Morita theorem also asserts that the map

Hom_{A⊗Aop}(A, A ⊗ A^{op}) ⊗_{A⊗Aop} A ⊗ N → R ⊗ N
f ⊗ a ⊗ n ↦ [f, a] ⊗ n = (f(1_{A}).a) ⊗ n = f(1_{A}).(a ⊗ n)

is an isomorphism. On the other hand, we get from the isomorphism (*) above that the image of the above map is (A ⊗ N)^{A}. (Note that the canonical map from R ⊗ N to A ⊗ N is injective, since R is a direct summand of A by Lemma 1.5.) The second assertion now follows by composition with the isomorphism N → R ⊗ N, n ↦ 1_{A} ⊗ n (cf. [2], Chap. II, Cor. 3.6, p. 54). □

From this correspondence, we can infer the relation between the ideals of A and the ideals of R that we alluded to in Paragraph 1.5. Denote the set of two-sided ideals of A by I_{A} and the set of (two-sided) ideals of R by I_{R}.

Corollary If A is a central separable R-algebra, the mappings

I_{A} → I_{R}, I ↦ R ∩ I

and

I_{R} → I_{A}, J ↦ JA

are mutually inverse bijections.

Proof. Note that a two-sided ideal in A is nothing else than a subbimodule of A, and similarly an ideal in the commutative ring R is nothing but a submodule. Since A is a projective R-module, it is in particular flat, and therefore we have for an ideal J ⊂ R that the induced homomorphism

J ⊗ A → R ⊗ A

is injective. This implies that the canonical isomorphism between R ⊗ A and A induces an isomorphism between J ⊗ A and JA.

From the preceding proposition, we now get the isomorphism

J → (J ⊗ A)^{A} → (JA)^{A} = Z(A) ∩ JA = R ∩ JA

which proves that R ∩ JA = J.

On the other hand, if I ⊂ A is a two-sided ideal, we know from the preceding proposition that the map

A ⊗ I^{A} → I, a ⊗ r ↦ ar

is an isomorphism. Since I^{A} = Z(A) ∩ I = R ∩ I, this proves (R ∩ I)A = I (cf. [2], Chap. II, Cor. 3.7, p. 54). □

In the preceding proof, we have use the R-projectivity of A to get the isomorphism between J ⊗ A and JA. It may be noted that one can alternatively use the A ⊗ A^{op}-projectivity of A to insure that the surjectivity of J ⊗ A → JA implies the surjectivity of

(J ⊗ A)^{A} = Hom_{A⊗Aop}(A, J ⊗ A) → (JA)^{A} = Hom_{A⊗Aop}(A, JA)

as pointed out in Proposition 1.3.

2 Azumaya algebras

2.1 The goal of this section is to find criteria which allows us to verify whether a given algebra is an Azumaya algebra. In the sequel, R denotes a commutative ring. We give complete proofs except for the following fact:

Proposition Suppose that R is a complete noetherian local ring with maximal ideal I, and that A is a finitely generated algebra over R. If p ∈ A/IA is an idempotent, then there is an idempotent e ∈ A whose residue class ē satisfies ē = p.

Proof. See [8], Chap. 7, Thm. (21.31), p. 330, and Prop. (21.34), p. 332. A more general result is stated in [2], Chap. II, § 7, p. 73. □

2.2 Of course, Proposition 2.1 applies in particular to central idempotents. However, we can say more about central idempotents: Their liftings are also central, and, more importantly, their liftings are unique. The first assertion follows from the following fact, which is known as Dade’s lemma:

Lemma Suppose that I is an ideal of R which is contained in the Jacobson radical. Suppose that A is a finitely generated algebra over R and that e ∈ A is an idempotent. Then e is central in A if and only if its equivalence class ē is central in A/IA.

Proof. It is obvious that ē is central in A/IA if e is central in A. Suppose that ē is central in A/IA. Then we have ē(A/IA)(1 − ē) = {0}, and therefore eA(1 − e) ∈ IA. Multiplying this with e on the left and 1 − e on the right, we see that

eA(1 − e) ∈ I(eA(1 − e))

Since eA(1 − e) is obviously finitely generated, we get from Nakayama’s lemma that eA(1 − e) = {0}. Now, for a ∈ A, the equation ea(1 − e) = 0 is equivalent to ea = eae. By the same reasoning, we get that (1 − e)Ae = {0}, which implies (1 − e)ae = 0 resp. ae = eae. Therefore, e is central in A. (cf. [8], Chap. 7, (22.10), p. 342). □

Denote by J_{c}(A) resp. J_{cp}(A) the set of central resp. centrally primitive idempotents of A. We then can say the following:

Proposition Suppose that R is a complete noetherian local ring with maximal ideal I and that A is a finitely generated algebra over R. Then the mappings

J_{c}(A) → J_{c}(A/IA), e ↦ ē and J_{cp}(A) → J_{cp}(A/IA), e ↦ ē

are bijective.

Proof. As the Jacobson radical is the intersection of all maximal ideals, I is the Jacobson radical of R in this situation. It follows from Proposition 2.1 and Dade’s lemma above that the first map is surjective. Let us see that the map is injective. If e and e′ are two central idempotents of A which have the same residue class modulo IA, then e′′ = e(1−e′) is a central idempotent with residue class zero. This implies that e′′ ∈ IA, and therefore e′′A ⊂ IA. As in the proof of Dade’s lemma, multiplication by e′′ gives e′′A = I(e′′A), and Nakayama’s lemma gives e′′A = {0}, which in turn implies e′′ = 0. By definition of e′′, this implies e = ee′. Interchanging the roles of e and e′, we see that e = e′, and therefore the first map is injective.

For the second map, the first question is whether it is well-defined. By definition (cf. [8], Chap. 7, § 22, p. 336), a nonzero idempotent is called centrally primitive if it cannot be written as the sum of two nonzero orthogonal central idempotents. Now suppose that e ∈ A is a centrally primitive idempotent. If ē could be written as the sum of two nonzero orthogonal central idempotents, we could lift these idempotents, so that we would have

ē = ē′ + ē′′

for two central idempotents e′, e′′ ∈ A. Since ē′ and ē′′ are orthogonal, their product is zero, and the same argument that we used above then shows that e′e′′ is also zero, which shows that e′ and e′′ are orthogonal. But then e′+e′′ is an idempotent, and therefore (1−e)(e′+e′′) and e(1−e′−e′′) are idempotents whose residue classes are zero, which implies as above that they are zero themselves. In different terms, these two equations assert that

e′ + e′′ = e(e′ + e′′)  and  e = e(e′ + e′′)

which implies e = e′ + e′′ in contradiction to the assumption. This establishes that the map is well-defined.

The injectivity of the second map follows from the injectivity of the first map, and it is also easy to see that the map is surjective: Given a centrally primitive idempotent in A/IA, we can lift it to a central idempotent of A, and the question is whether this idempotent is centrally primitive. But a decomposition of this idempotent into a sum of two nonzero orthogonal central idempotents would give a similar decomposition in A/IA, since we have seen a couple of times that nonzero idempotents have nonzero residue classes. Since such a decomposition does not exist by assumption, the lifted idempotent must be centrally primitive itself (cf. [8], Chap. 7, Prop. (21.22), p. 327, and Thm. (22.11), p. 342). □

This result implies that the Wedderburn decomposition for the finite dimensional algebra A/IA over the field F := R/I leads to a corresponding block decomposition of A. Note that, if A is projective over R, and therefore free, and if e ∈ A is a centrally primitive idempotent, then the corresponding block Ae is projective, and therefore also free, and in particular faithful.

2.3 We are getting closer to the main result of this section, a criterion for Azumaya algebras. First, we need an auxiliary result, which is analogous to the fact from linear algebra that

ker(f ⊗ g) = ker(f) ⊗ W + V ⊗ ker(g)

for two linear maps f : V → V′ and g : W → W′ between vector spaces.

Lemma Suppose that I is an ideal in the commutative ring R, and that M and N are two R-modules. Then the sequence

I(M ⊗ N) → M ⊗ N → (M/IM) ⊗ (N/IN)

is exact.

Proof. Since the tensor product is right exact (cf. [5], Prop. III.7.3, p. 110), we then have the exact sequence

(IM) ⊗ N → M ⊗ N → (M/IM) ⊗ N → 0

It is easy to see that the image of the map (IM) ⊗ N → M ⊗ N coincides with I(M ⊗ N). On the other hand, we have again by right exactness that the sequence

(M/IM) ⊗ IN → (M/IM) ⊗ N → (M/IM) ⊗ (N/IN) → 0

is exact. Since the left map is identically zero, we see that the canonical map (M/IM) ⊗ N → (M/IM) ⊗ (N/IN) is an isomorphism. This proves the assertion. □

The main piece of work now is the following proposition:

Proposition Suppose that R is a complete noetherian local ring with maximal ideal I and that A is a finitely generated algebra over R. Then A is separable if and only if the R/I-algebra A/IA is separable.

Proof. If A is separable, then A/IA is separable by Corollary 1.2. Suppose now that A/IA is separable. We proceed in steps, following [2], Chap. II, Thm. 7.1, p. 72:

(1) Let us first see that IA is the Jacobson radical J(A) of A. By definition, the Jacobson radical of A is the intersection of the annihilators of the simple A-modules. Now, if M is a simple A-module, the submodule IM must be zero or equal to M. As a simple module is generated by every nonzero element, and since A is a finitely generated R-module, we see that M is finitely generated as an R-module. Therefore, the case M = IM is impossible by Nakayama’s lemma, and we see that I annihilates M. Therefore, IA ⊂ J(A). On the other hand, A/IA is separable by assumption; therefore, it is semisimple and has vanishing Jacobson radical. Since J(A/IA) = J(A)/IA (cf. [8], Chap. 2, Prop. (4.6), p. 55, [3], Chap. 2, Exerc. 8, p. 68), we get J(A) = IA.

(2) Choose a separability idempotent

ē = ∑_{i=1}^{n} ā_{i} ⊗ b̄_{i}

From the above lemma, we know that the canonical map from

(A ⊗ A^{op})/I(A ⊗ A^{op}) → (A/IA) ⊗ (A/IA)^{op}

is an isomorphism, and therefore we can use Proposition 2.1 to lift ē to an idempotent of A ⊗ A^{op}, i.e., we can assume that the element

e := ∑_{i=1}^{n} a_{i} ⊗ b_{i}

is itself an idempotent. However, we do not yet have x := ∑_{i=1}^{n} a_{i}b_{i} = 1_{A}, and therefore we want to modifiy this idempotent.

(3) As we have ∑_{i=1}^{n} ā_{i}b̄_{i} = 1_{A/IA}, it follows that

1_{A} − x = 1_{A} − ∑_{i=1}^{n} a_{i}b_{i} ∈ IA = J(A)

Therefore x = 1_{A} − (1_{A} − x) is invertible (cf. [8], Chap. 2, Lem. (4.3), p. 54, [3], Prop. 2.8, p. 63), and we can look at the element

e′ := ∑_{i=1}^{n} x^{−1}a_{i}x ⊗ b_{i}

Obviously, this is again an idempotent in A ⊗ A^{op}, but furthermore, using that e ∈ A ⊗ A^{op} is an idempotent, we get for the product of its tensor components

∑_{i=1}^{n} x^{−1}a_{i}xb_{i} = ∑_{i,j=1}^{n} x^{−1}a_{i}a_{j}b_{j}b_{i} = ∑_{i=1}^{n} x^{−1}a_{i}b_{i} = x^{−1}x = 1_{A}

In addition, since 1_{A} − x ∈ IA, we have ē′ = ē.

(4) We are still lacking the second property of the separability idempotent, namely that e′ ∈ (A ⊗ A)^{A}. For this, we need a little preparation. Consider the multiplication map μ_{A} : A ⊗ A → A, and denote its kernel by L. It is a subbimodule of A ⊗ A which is a direct summand as a left A-module, since the map

π : A ⊗ A → L, a ⊗ a′ ↦ a ⊗ a′ − aa′ ⊗ 1_{A}

is a projection onto L that is left A-linear. If M is the kernel of π, we therefore have A ⊗ A = L ⊕ M.

Subbimodules of A ⊗ A can be considered alternatively as left ideals in A ⊗ A^{op}, and it is this viewpoint that we will adopt now. From this viewpoint, right multiplication with e′ is left A ⊗ A^{op}-linear. Moreover, since μ_{A}(e′) = 1_{A}, we have

π((a ⊗ a′)e′) = (a ⊗ a′)e′ − aa′ ⊗ 1_{A}

and therefore π(le′) = le′ for l ∈ L. This implies that the map

(A ⊗ A^{op})e′ → Le′, b ↦ π(b)e′

is a projection onto Le′. If M′ is its kernel, we get

(A ⊗ A^{op})e′ = Le′ ⊕ M′

(5) Still taking the above viewpoint, we note that e′ is a separability idempotent if and only if Le′ = {0}. To see this, note that a ⊗ 1_{A} − 1_{A} ⊗ a ∈ L, so Le′ = {0} implies (a ⊗ 1_{A} − 1_{A} ⊗ a)e′ = 0, which is precisely the missing condition for a separability idempotent. Conversely, if

(1_{A} ⊗ a′ − a′ ⊗ 1_{A})e′ = −(a′ ⊗ 1_{A} − 1_{A} ⊗ a′)e′ = 0

we also have

(a ⊗ 1_{A})(1_{A} ⊗ a′ − a′ ⊗ 1_{A})e′ = (a ⊗ a′ − aa′ ⊗ 1_{A})e′ = 0

As a ⊗ a′ − aa′ ⊗ 1_{A} is the general element of L, this shows Le′ = {0}.

(6) To see that this is really the case, denote the kernel of the multiplication map A/IA ⊗ A/IA → A/IA by L̄. Since ē′ = ē is a separability idempotent, we have L̄ē′ = {0}. From the above lemma, we have that the kernel of the canonical mapping from A ⊗ A to A/IA ⊗ A/IA is I(A ⊗ A), and therefore we have Le′ ⊂ I(A ⊗ A), and even Le′ ⊂ I(A ⊗ A^{op})e′. By the above, we get

Le′ ⊂ I(A ⊗ A^{op})e′ = ILe′ ⊕ IM′

and therefore Le′ = ILe′. As L is the image of A ⊗ A under the projection π, it is a finitely generated R-module, and this implies that also Le′ is a finitely generated R-module. But then Nakayama’s lemma implies that Le′ = {0}, which finally shows that e′ is a separability idempotent and A is separable. □

A more general version of this proposition can be found in [2], Chap. II, Thm. 7.1, p. 72, [13], Chap. 5, § 5, Thm. 7, p. 122, [6], Chap. III, § 5, Thm. (5.1.1), p. 135 and [7], Chap. III, Prop. 2.6, p. 80, although the above argument contains an essential part of the proof.

We finally have the following criterion for Azumaya algebras:

Corollary Suppose that R is a complete noetherian local ring with maximal ideal I and that A is a finitely generated algebra over R. Then A is an Azumaya algebra if and only if it is faithful and finitely generated as an R-module and the R/I-algebra A/IA is central simple.

Proof. Recall that we know from Proposition 1.8 that Azumaya algebras are the same as central separable algebras. If A is an Azumaya algebra, it is by definition faithful and finitely generated over R. Moreover, we know from Proposition 1.2 that the quotient A/IA is separable, and therefore semisimple. From Proposition 1.3, we know that the center of A/IA is the image of the center of A under the canonical epimorphism. The kernel of the restriction of this epimorphism to R is R∩IA, which is equal to I by Proposition 1.5. Therefore, we have Z(A/IA) ≅ R/I. But a semisimple algebra with a one-dimensional center must be simple itself, which proves the easier direction in the asserted equivalence.

For the converse, it follows from the above proposition that A is separable, and it remains to be shown that A is central. Since we have assumed that A is faithful, the canonical map R → Z(A) is injective. Since Z(A) is an R-direct summand of A by Lemma 1.5, Z(A) is a finitely generated R-module. Consider the diagram

[figure: R/I; Z(A)/IZ(A); R; Z(A)]

It is a consequence of Nakayama’s lemma that the upper map is surjective if the lower map is surjective (cf. [3], Cor. 2.12, p. 66; note that the statement there is incorrect; it must be assumed that M′ is finitely generated). This shows that the canonical map R → Z(A) is bijective, i.e., A is central. □

2.4 Suppose that R is a local ring with maximal ideal I and residue field F.

Proposition Suppose f : P → Q is an R-linear map between finitely generated projective R-modules. If the induced F -linear map

f̄: P/IP → Q/IQ

is an isomorphism, then f is an isomorphism.

Proof. Note that, in this situation, the projective modules are in fact free (cf. [10], § 2, Thm. 2.5, p. 9). Choose a basis v_{1}, …, v_{n} of P and w_{1}, …, w_{m} of Q. Then the residue classes v̄_{1}, …, v̄_{n} form a basis of P/IP and the residue classes w̄_{1}, …, w̄_{m} form a basis of Q/IQ. If the induced map f̄ is an isomorphism, we must have n = m. Suppose that d ∈ R is the determinant of (the matrix representation of) f. Then there is an element u ∈ R that satisfies ūd̄ = 1. Therefore, we have 1 − ud ∈ I. By [3], Chap. 2, Prop. 2.8, p. 63, this implies that ud is invertible, and therefore d is invertible, which shows that f is an isomorphism (cf. [3], Chap. 2, Exerc. 37, p. 75, [11], § 1, p. 6). □

2.5 Suppose that R is a complete noetherian local ring with maximal ideal I and residue field F.

Proposition If A is a finitely generated projective R-algebra such that

A/IA ≅ M(n × n, F),

then A ≅ M(n × n, R).

Proof. Suppose that e_{11} ∈ A/IA corresponds to the first matrix unit in the above isomorphism. By Proposition 2.1, we can lift e_{11} to an idempotent e ∈ A. Ae is a left ideal in A, and therefore a left A-module, which means that we have an R-algebra homomorphism

f : A → End_{R}(Ae)

As stated above, since Ae is projective and R is local, Ae is free. By passing to the quotient, we see that its rank must be n. Therefore, End_{R}(Ae) is also free of rank n^{2}. The induced map f̄ is an isomorphism by construction, and therefore Proposition 2.4 shows that f is also an isomorphism. □

This proof is taken from [6], Chap. III, § 5, Lem. (5.1.16), p. 142. We get as a corollary the result stated there:

Corollary If A is an Azumaya algebra over R such that

A/IA ≅ M(n × n, F),

then A ≅ M(n × n, R).

Note that it was not necessary in the proposition to assume that A is faithful; this follows from the fact that it is projective, and therefore free.

References

[1] H. Bass: The Morita theorems, Lect. Notes, Univ. Oregon, 1962

[2] F. DeMeyer/E. Ingraham: Separable algebras over commutative rings, Lect. Notes Math., Vol. 181, Springer, Berlin, 1971

[3] B. Farb/R. K. Dennis: Noncommutative algebra, Grad. Texts Math., Vol. 144, Springer, Berlin, 1993

[4] M. Gerstenhaber: On the deformation of rings and algebras, Ann. Math., II. Ser. 79 (1964), 59-103 [5] P. J. Hilton/U. Stammbach, A course in homological algebra, Grad. Texts

Math., Vol. 4, Springer, Berlin, 1971

[6] M. A. Knus: Quadratic and hermitian forms over rings, Grundlehren Math. Wiss., Vol. 294, Springer, Berlin, 1991

[7] M. A. Knus/M.. Ojanguren: Théorie de la descente et algèbres d’Azumaya, Lect. Notes Math., Vol. 389, Springer, Berlin, 1974

[8] T. Y. Lam: A first course in noncommutative rings, Grad. Texts Math., Vol. 131, Springer, Berlin, 1991

[9] S. Mac Lane: Homology, Grundlehren Math. Wiss., Vol. 114, Springer, Berlin, 1963

[10] H. Matsumura: Commutative ring theory, Camb. Stud. Adv. Math., Vol. 8, Camb. Univ. Press, Cambridge, 1986

[11] J. Milnor: Introduction to algebraic K-theory, Ann. Math. Stud., Vol. 72, Princeton University Press and University of Tokyo Press, Princeton, 1971

[12] C. Peskine: An algebraic introduction tio complex projective geometry, I. Commutative algebra, Camb. Stud. Adv. Math., Vol. 47, Camb. Univ. Press, Cambridge, 1996

[13] G. Renault: Algèbre non commutative, Collection ‘Varia mathematica’, Gauthier-Villars, Paris, 1975

[14] C. A. Weibel: An introduction to homological algebra, Camb. Stud. Adv. Math., Vol. 38, Camb. Univ. Press, Cambridge, 1994