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I am reading the book “Elements of the Representation Theory of Associative Algebras 1”. I have a question about the lemma 3.3 in the chapter 2.

Here is the question.

Assume that $A$ is a finite dimensional $K$-algebra with a complete set of primitive orthogonal idempotents$\{e_1,e_2,\dots, e_n\}$ and we consider the right $A$-modules.

If we choose $x\in e_i~(\mathrm{rad}A)~e _ j$ is nonzero, by the isomorphism $e_i~(\mathrm{rad}A)~e _ j\cong \mathrm{Hom}_A(e_jA, e_i~(\mathrm{rad}A))$, we can define $\overline{x}:e_j A\rightarrow e_i~(\mathrm{rad}A) $ by the formula $\overline{x}(e_j a)=xe_ja$.

By this formula , I have get that $\overline{x}(e_j)=x$ and Im $\overline{x}\subset e_i(\mathrm{rad}~A)$.

But the author also says that Im $\overline{x}\not\subset e_i(\mathrm{rad}^2A)$. I don’t know how to get it.

Firstly, I want to prove that $x\notin e_i(\mathrm{rad}^2A)$, but I don’t know how to prove it.

Any help and references are greatly appreciated.

Thanks!

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1 Answer 1

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From what you have written, you can't deduce what you are asking for. However, the book has one additional assumption on $x$, namely that $x$ is part of a set of elements whose residue classes with respect to $\operatorname{rad}^2 A$ form a basis of $e_i(\operatorname{rad} A/\operatorname{rad}^2A)e_j$. In particular, $x$ can't lie in $\operatorname{rad}^2 A$ since otherwise $x+\operatorname{rad}^2 A$ would be zero in $e_i(\operatorname{rad} A/\operatorname{rad}^2A)e_j$, so it can't be part of a basis. This means that the image of the corresponding map $\bar{x}$ can't lie in $e_i\operatorname{rad}^2 A$, since $e_j$ is mapped to $x$.

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  • $\begingroup$ Thanks! I forget it. $\endgroup$
    – fusheng
    Commented Dec 5, 2023 at 6:17

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