Algebras over a field and equivalence of module categories

Let $k$ be a field. Let $A$ be a finitely generated $k$-algebra. Consider the inclusion of the center $i:Z(A)\hookrightarrow A$. I'm interested in what general conditions there are for the pull-back functor $i^*: Z(A)\mathrm{-mod}\rightarrow A\mathrm{-mod}$ to be an equivalence of categories. Clearly, this is the identity functor if $A$ is commutative- but it seems like an interesting question to ask in the case that $A$ is non-commutative.

This seems like such an obvious question to ask that someone must have developed a theory about it- but I haven't been able to find out what the name of this theory is. I'd appreciate any answers, references, or reading suggestions that you can offer.

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Some possible keywords: ‘Morita equivalence’, ‘descent theory’... –  Zhen Lin Mar 10 '12 at 23:48
If by $i^*$ you mean the functor $i^*(M)=A\otimes_{Z(A)}M$, I believe that $i$ must be an isomorphism. –  Frank Murphy Mar 11 '12 at 1:42
What functor exactly do you mean? –  Mariano Suárez-Alvarez Mar 11 '12 at 1:57

As Frank notes in his comment, if the funtor $\newcommand\Mod{\mathrm{Mod}}A\otimes_Z(-):{}_Z\Mod\to{}_A\Mod$ is an equivalence, then $A$ is commutative. Indeed, that functor maps the regular left $Z$-module $Z$ to the regular $A$-module $A$, so it induces an isomorphism between the endomorphism algebras of these objects, that is, between $\hom_Z(Z,Z)$ and $\hom_A(A,A)$. Since these endomorphism algebras are $Z$ and $A^{\mathrm{op}}$, we see that $A$ is commutative.

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Dear Mariano, A typo: "regular $A$-module $Z$" should read "regular $A$-module $A$". Cheers, –  Matt E Mar 11 '12 at 4:24
Thank you for your answer. Do you have any suggestions on texts or papers that give a good treatment of basic notions of Morita equivalence? –  KReiser Mar 11 '12 at 18:32
I like Pierce's book on Associative algebras, and I've learned to appreciate the book by Anderson and Fuller, for example. –  Mariano Suárez-Alvarez Mar 11 '12 at 21:37
Thank you very much, I'll check those out. –  KReiser Mar 11 '12 at 23:37