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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.2012-05-06
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    @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.2012-05-06
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    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.2012-05-06
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    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.2012-05-06
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    Second definition does not seem to be assuming $R$ is local, so I'd venture they are not equivalent.. but the second may be generalizing the first.2012-05-06
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    @rschwieb You're right! I didn't see that!2012-05-06
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    Good catch! I've edited the original post.2012-05-07
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    If you remove the local condition from the first definition, you should also replace freeness by projectivity...2012-05-07
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    Edit added. Thanks for your observation.2012-05-07

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