# Question on simple modules and projective covers

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!

• 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? Dec 17 '20 at 18:27
• Yes, this is exactly what $\theta_s$ is. Dec 17 '20 at 18:30
• 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. Dec 17 '20 at 19:47
• I’m voting to close this question because it is quite basic and answered in a comment. Feb 17 at 20:24

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 $$S_r\cong P_S/K\cong S$$.