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Suppose that $R$ is a finite dimensional $k$-algebra. I say that $R$ is Frobenius if it is locally bounded (see this question for a definition) and indecomposable projectives and injectives coincide. Could you help me to prove the following:

$R$ is Frobenius if and only if $_RR$ is injective as an $R$-module.

Of course if $R$ is Frobenius then $_RR$ is injective as an $R$-module, how can I prove the converse?

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Where does this notation come from? It seems non-standard to call $R$ Frobenius and not self-injective or quasi-Frobenius. – Julian Kuelshammer Nov 5 '12 at 17:33
it is from the book of happel "triangulated categories in the representation theory of finite dimensional algebras" – Nick Nov 6 '12 at 4:54

Of course a finite dimensional algebra is locally bounded. (If you need a proof of that, comment on this answer.) Now let $P$ be an indecomposable projective. Then (as noted in the answer to your previous question $P$ is a direct summand of $R$. Hence $P$ is injective (as $R$ is). EDIT: Now you get an injective map from isoclasses of indecomposable projectives to isoclasses of indecomposable injectives by mapping each projective to itself. Since there are equally many indecomposable projectives and indecomposable injectives (as many as simples) you have that this is a bijective map, so every indecomposable injective is projective. This is related to the Nakayama permutation. You can read this and similar things in Lam's Lectures on Modules and rings.

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Could you give me more detail on why injectives are projectives, please? – Nick Nov 6 '12 at 5:06
why are there equally many indecomposable projectives and indecomposable injectives? (if that is a theorem could you give me at least a reference?) – Nick Nov 6 '12 at 22:54
In the answer to your previous question… I told you how to set up a bijection between isoclasses of indecomposable projectives and isoclasses of simples by taking $P\mapsto P/\operatorname{rad}(P)$. As I wrote there, by similar arguments, you can prove that there is a bijection between isoclasses of indecomposable injectives and simples by taking $I\mapsto \operatorname{soc}(I)$. – Julian Kuelshammer Nov 6 '12 at 23:05
ah right. I the mentioned question I was asking also why the set of simple modules is finite, I don't know if this is evident, could you explain me why, please? – Nick Nov 7 '12 at 2:57
Each simple corresponds to an indecomposable projective, each indecomposable projective is a direct summand of the algebra, the algebra is finite-dimensional, so finitely many direct summands. Thus finitely many simples. – Julian Kuelshammer Nov 7 '12 at 7:12

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