Bourbaki

Exercise 11, § 14

Tag 00HM, permanent. Cite it as [Bourbaki, Tag 00HM].

Algebra, VIII, p. 271

Bourbaki marks this one as harder than the rest.

Let A be a commutative integral domain, K its field of fractions, and S an Azumaya A-algebra (Exercise 8).

a) Prove that the K-algebra S(K)S_{(K)} is central simple. Identify S with a subalgebra of S(K)S_{(K)}. b) Assume from now on that A is the only A-subalgebra of K that is a finitely generated A-module, ∗*that is, that the ring A is integrally closed, cf. Comm. Alg., V,

§1, No. 1,2∗1, 2*. Prove that S is maximal (for the inclusion) among the A-subalgebras of S(K)S_{(K)} that are finitely generated as A-modules (apply Exercise 9, d)).

c) Suppose, moreover, that A is Noetherian and that the K-algebra S(K)S_{(K)} is isomorphic to EndK_K(V), where V is a vector space over K. Prove that there exists a finitely generated A-module M without torsion such that the A-algebra S is isomorphic to EndA(M)_A(M) (let xx be a nonzero vector of V; take M to be the set of elements s(x)s(x) as ss runs through S).

∗*d) Prove that in the previous construction, we may assume M reflexive (replace M

with its bidual). If, moreover, the ring A is regular (AC, X, §4, no2^o2, p. 55, définition 1), then the A-module M is projective (apply AC, X, §4, p. 164, exercice 1212).∗.*

Solution

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Cites

  • AC, X, §4, n$^o2$, p. 55 not in this corpus: AC X
  • AC, X, §4, p. 164 not in this corpus: AC X
  • Exercise 8
  • Exercise 9
Transcribed from the printed volume. Not translated.