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Let $A$ be a k-algebra,where k is a fixed field. {$x_1,x_2, \cdots,x_n $} is a complete set of primitive orthogonal idempotents of $A$. $M$ is a left $A$-module such that $x_iM=0$ for $i=1,2,\cdots,n-1$.

I want to know whether there exists a simple module $S$ which is a direct summand of $soc(M)$,such that its projective cover $P(S) \in add(_AAe_n)$?

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  • $\begingroup$ If $M$ is non-zero, $x_n M \neq 0$ and then also $x_n soc(M) \neq 0$ but $x_i soc(M)=0$ for all other $i$, so $soc(M)$ is a direct sum of of the same simple module $S$ having projective cover $Ae_n$ or? $\endgroup$
    – Mare
    Commented Sep 22, 2016 at 7:50
  • $\begingroup$ @Mare Could you tell me how to get $x_nsoc(M) \neq 0$? $\endgroup$
    – Xiaosong Peng
    Commented Sep 22, 2016 at 11:02
  • $\begingroup$ I assume A is finite dimensional? Then the socle of a nonzero module is nonzero and $x_i soc(M)=0$ since $x_i M=0$. So since the socle is nonzero one has to have $x_n soc(M) \neq 0$ since it is zero for all other idempotents. $\endgroup$
    – Mare
    Commented Sep 22, 2016 at 11:15
  • $\begingroup$ @Mare Get it. But then coule we get the projective cover of $soc(M)$ is the direct sum of $Ae_n$? I hope so, but I can't prove it. $\endgroup$
    – Xiaosong Peng
    Commented Sep 22, 2016 at 11:20
  • $\begingroup$ $x_n S \neq 0$ means that $S$ is the simple module $Ae_n / J e_n$ so it has projective cover A$e_n$ and soc(M) is just a direct sum of such $S$. $\endgroup$
    – Mare
    Commented Sep 22, 2016 at 11:25

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