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I have the following question:

Let $A$ be an Artin algebra. Let $S_1$ and $S_2$ be simple modules in $\text{mod}(A)$ and let $P(S_1)$ be the projective cover of $S_1$. Let $f: P(S_1) \rightarrow S_2$ be module homomorphism with $f \neq 0$. Then $S_1 \cong S_2$.

Any help is highly appreciated!

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  • $\begingroup$ What is $\theta_S$ ? If I had to guess, I'd say it is the morphism $P_S \twoheadrightarrow S \hookrightarrow I_S$. Is that correct? $\endgroup$
    – Johannes Hahn
    Commented Dec 17, 2020 at 18:27
  • $\begingroup$ Yes, this is exactly what $\theta_s$ is. $\endgroup$
    – mathStudent
    Commented Dec 17, 2020 at 18:30
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    $\begingroup$ The head of a projective indecomposable module over an Artinian ring is simple, so there is exactly one isomorphism class of simples that $P_S$ can surject onto (the equality should really be an isomorphism, not an equality). I am sure this is in Alperin, but I don't have the book in front of me right now. $\endgroup$
    – Alex B.
    Commented Dec 17, 2020 at 19:47
  • $\begingroup$ I’m voting to close this question because it is quite basic and answered in a comment. $\endgroup$
    – Mark Wildon
    Commented Feb 17, 2021 at 20:24

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Here, $P_S$ is a projective cover of a simple module $S$. This means that there is a surjection $\sigma\colon P_S\to S$ from projective $P_S$ onto $S$, which has superfluous kernel $K$. The fact that $P_S/K\cong S$ is simple implies that $K$ is a maximal submodule of $P_S$, and the fact that $K$ is superfluous then implies that every proper submodule of $P_S$ is contained in $K$. In particular, the only simple quotient of $P_S$ up to isomorphism is $P_S/K\cong S$.

In the underlined part of Lemma 2.2, $g\colon P_s\to S_r$ is a nonzero homomorphism onto a simple module. The kernel must be $K$ and we must have $S_r\cong P_S/K\cong S$.

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