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!

  • $\begingroup$ 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? $\endgroup$ Dec 17 '20 at 18:27
  • $\begingroup$ Yes, this is exactly what $\theta_s$ is. $\endgroup$ Dec 17 '20 at 18:30
  • 1
    $\begingroup$ 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.
    Dec 17 '20 at 19:47
  • $\begingroup$ I’m voting to close this question because it is quite basic and answered in a comment. $\endgroup$ 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 $S_r\cong P_S/K\cong S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.