# Why $\operatorname{Hom}_A(e_jA, e_iA) \cong \operatorname{Hom}_A(e_jA, e_i\text{rad}A)$?

On line 6 of the proof of Corollary 3.4 of page 62 of the book Elements of Representation Theory of Associative Algebras, Volume 1, it is said that $\operatorname{Hom}_A(e_jA, e_iA) \cong \operatorname{Hom}_A(e_jA, e_i\text{rad}A)$. I think that it is possible that $e_iA \not\cong e_i\text{rad}A$. Why $\operatorname{Hom}_A(e_jA, e_iA) \cong \operatorname{Hom}_A(e_jA, e_i\text{rad}A)$? Thank you very much.

• I notice that you wrote "$e_i\mathrm{rad}(A)$" where the text wrote "$\mathrm{rad}(e_iA)$". Could that be part of the confusion, or is it just a typo? And what does $I.4.2$ say? – rschwieb Apr 5 '13 at 14:19
• @rschwieb I.4.2 gives the first isomorphism, hence has nothing to do with the question. – Julian Kuelshammer Apr 5 '13 at 15:54

It is always the case for finite dimensional algebras that $e_iA\ncong e_i\operatorname{rad} A\cong \operatorname{rad} e_iA$. This is not the argument for the isomorphism.
The argument is that every non-isomorphism $e_jA\to e_iA$ has an image in the radical of $e_i A$. This is because $e_i A$ has one-dimensional top, hence every morphism $e_jA\to e_iA$ whose image does not lie in $\operatorname{rad} e_iA$ must be surjective. Thus, because we are talking about indecomposable projectives, we have an isomorphism.
• thank you very much. We know that $\operatorname{rad}e_iA$ is the unique maximal ideal of $e_iA$. But how to show that $e_iA/ \operatorname{rad}e_iA$ is one dimensional? – LJR Apr 6 '13 at 6:59