# How to get Im $\overline{x}\not\subset e_i(\mathrm{rad}^2A)$

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!

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 $$x$$.