Let $A$ be a central, simple $K$-algebra and let $P$ be a finitely generated projective $A$-module.

I want to show that the endomorphism ring $\operatorname{End}_A(P)$ is also a central, simple $K$-algebra.

I know that $P$ is a semisimple $A$-module since $A$ is a simple, Artinian ring and I tried using Wedderburn's theorem on $\operatorname{End}_A(A^n)$ where $P\oplus Q\cong A^n$ but I couldn't get very far.

Would someone be able to provide a proof of this?

Many thanks.

  • $\begingroup$ Do you assume that $A$ is finite-dimensional over $K$ ? $\endgroup$ – Captain Lama Apr 26 '16 at 20:54
  • $\begingroup$ Yes I am assuming this. $\endgroup$ – eddie Apr 26 '16 at 20:55

Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective).

Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any module and any ring).

Now for any $K$-algebra $A$ and any simple module $I$, $\operatorname{End}_A(I)$ is a division algebra ; call it $D$.So $\operatorname{End}_A(P)\simeq M_n(D)$.

The fact that $D$ is central can be seen by taking $P=A$ : if $A\simeq I^r$, $A^{op} = \operatorname{End}_A(A) = M_r(D)$ is central and $Z\left( M_r(D)\right) = Z(D)$, so $D$ is central.

Then $\operatorname{End}_A(P)\simeq M_n(D)$ with $D$ a central division algebra over $K$ : it's easy to check that this implies $\operatorname{End}_A(P)$ is central simple.

  • $\begingroup$ No, it's actually not that obvious, and now that I think about it I don't see how to prove it without paraphrasing the demonstration of the Artin-Wedderburn structure theorem (from which it follows easily : if you have two non-isomorphic simple modules then $A$ is a direct product, which is of course impossible if it's simple). $\endgroup$ – Captain Lama Apr 26 '16 at 22:00
  • $\begingroup$ Is it because a simple $A$-module is always isomorphic to a minimal left $A$-ideal and if you have two non-isomorphic simple modules then by Wedderburn's theorem, $A$ would have at least two components (as a regular module over itself), hence it would not be a simple ring? I assume the proof then proceeds as above? Regards $\endgroup$ – eddie Apr 29 '16 at 15:30

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