Q:

Questions about Lowey length

Let $\Lambda$ be an artin algebra. 

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 $P/radP \cong M/radM$)
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? 

A:

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