Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading this paper and have the following questions.

Let $A$ be a finite-dimensional algebra over a fixed field $k$.

Let the finitely-generated $A$-module $M$ be a generator–cogenerator for $A$, which means that all projective indecomposable $A$-modules and all injective indecomposable $A$-modules occur as direct summands of $M$.

On page 3 in the paper it says

"The identity of $End_A(M)$ is the sum of the “identity maps” on the indecomposable direct summands of $M$. Hence we have primitive idempotents of $End_A(M)$ corresponding to the summands of $M$. For any indecomposable summand $T$ of $M$ we denote the corresponding simple $End_A(M)$-module by $E_T$".

My questions are:

  • How can I prove this correspondence and
  • Why is $E_T$ simple?.

I would be grateful for references concerning literature or every other kinds of hints.

Thank you very much.

share|improve this question
2  
If $A$ is a direct summand of $M$, then $\pi:M \to M: (a,b) \mapsto (a,0)$ is an idempotent in $\operatorname{End}(M)$. If $\pi \in \operatorname{End}(M)$ is an idempotent, then $M$ is the direct sum of the kernel and image of $\pi$ (which are reversed for $1-\pi$). $E_T$ need not be simple: take $A=k[x]/(x^2)$. –  Jack Schmidt Jul 20 '12 at 15:14
    
@Jack:$\ $Thank you for your comment, your explanations and your counter-example. :-) $\ $ If we assume that $A=k[x,y]/(x^2,y^n)$, whereat $n\in \mathbb{N}$, why is $E_T$ simple then? –  Bernhard Boehmler Jul 20 '12 at 15:41
    
I looked more closely at the paper, and it does not seem it uses that $E_T$ is simple (though the paper claims it is many, many times). If you know what the injective modules look like over his $A$, let me know, and we can check them, but $E_T$ is very unlikely to be simple as $T$ is likely to have characteristic nonzero proper submodules like its socle, and I think this will be reflected in $E_T$. –  Jack Schmidt Jul 20 '12 at 16:52
2  
This is a general fact for finite dimensional algebra $A$ and fin. gen. modules $M$. $M=\oplus M_i$ be decomposition into (finitely many) indecomposable module, the "identity map" $M_i \to M_i$ is the primitive idempotents of the algebra $End_A(M)$. You guys seems to have misread the paper, $E_T$ is the notation for the simple corresponding to $T$ (note: primitive idem <-> proj indecomposable <-> simple), rather than $End_A(T,T)$ itself. –  Aaron Jul 21 '12 at 22:00
    
@Aaron: Thank you for the clarification.$\ $:-)$\ $It is clear to me now. –  Bernhard Boehmler Jul 22 '12 at 19:47
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.