Let $\Lambda$ be an artin algebra.

  1. If $M$ is a finitely generated $\Lambda$-module with Loewy length 2 and finite projective dimension. How to get the exact sequence $$0 \rightarrow A \rightarrow P/rad^2P \rightarrow M \rightarrow 0$$ where $P$ is the projective cover of $M$ and $A$ is semisimple?(I just know that $P/radP \cong M/radM$)
  2. Suppose that $(rad \Lambda)^3=0$, for any $\Lambda$-module $M$, $\Omega M$ is the first syzygy of $M$. Then how to get $\Omega M$ has Loewy length at most 2?

1 Answer 1

  1. You have the projective cover $f:P \rightarrow M$ and since $M$ has Loewy length 2, rad^2(P) is in the kernel of f. Now the kernel A is a submodule of rad(P)/rad^2(P), which is semisimple.

  2. Use the $\Omega(M)$ is always a submodule of the radical of a projective module (property of minimal projective cover).

  • $\begingroup$ Why the kernel $A$ is a submodule of $rad(P)/rad^2(P)$? $\endgroup$ Nov 17, 2016 at 9:55
  • $\begingroup$ In general the kernel of a projective cover is contained in the radical of the projective module. it doesnt matter now that you factor out rad^2. $\endgroup$
    – Mare
    Nov 17, 2016 at 10:23

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