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I have seen a few different definitions of an Azumaya algebra in the literature- for example, Wikipedia prefers the following one:

An Azumaya algebra over a commutative local ring $R$ is an $R$-algebra $A$ that is free and of finite rank $r$ as an $R$-module, such that the tensor product $A\otimes_R A^{op}$ (where $A^{op}$ is the opposite algebra) is isomorphic to the matrix algebra $\mathrm{End}_R(A) \sim M_r(R)$ via the map sending $a\otimes b$ to the endomorphism $x \mapsto axb$ of A.

On the other hand, for purposes of checking this definition, I have seen the following characterization used:

$A$ is a finitely generated projective $R$-module and for all maximal ideals $\mathfrak{m}\subset R$, $A/\mathfrak{m}A$ is a central simple $R/\mathfrak{m}$ algebra.

Is there an obvious way to show that these are equivalent?

EDIT: It has been noticed in the comments that the first definition specified "local ring"- this is too specific for the two definitions to be equivalent. I'm interested in what can be said when one removes local (and as Mariano points out, replaces free by projective) from the first definition and allows $R$ to be just a commutative ring.

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If you look on page 186 of Noncommutative algebra by Farb and Dennis, it is said that over a field, an algebra is central simple if and only if it is Azumaya. That's probably how one could prove your statement. –  M Turgeon May 6 '12 at 19:18
    
@MTurgeon: Thanks for the reference! Could you possibly expand on your point? Unfortunately, I'm not as well versed in the commutative algebra maximal ideal trickery as I ought to be. –  KReiser May 6 '12 at 19:28
    
Well, it's just an idea, but if you can check that $A$ is Azumaya if and only if $A/\mathfrak{m}A$ is Azuyama for all maximal ideals $\mathfrak{m}$, then you can use Frab & Dennis' result to conclude. –  M Turgeon May 6 '12 at 21:32
    
However, I don't know if this is true, and even if it is, I might not be easier than showing directly what you want to show. It's just something I'm suggesting. –  M Turgeon May 6 '12 at 21:33
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If you remove the local condition from the first definition, you should also replace freeness by projectivity... –  Mariano Suárez-Alvarez May 7 '12 at 3:30
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1 Answer

An Azumaya algebra is by definition an álgebra $A$ which is separable over its center, that is, the multiplication map $m\colon A\otimes_{C(A)}A\to A,$ is a $(A,A)$-bimodule epimorphism that splits.

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The map has to be split as an $(A,A)$-bimodule, rather, no? –  Mariano Suárez-Alvarez Jun 14 '13 at 18:11
    
Yes, i was mistaking. I already changed my answer. –  Hector Pinedo Jun 14 '13 at 21:31
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