Suppose $A$ be a finite dimensional $\mathbb K$ algebra. Let $P$ be a projective module in mod $A$. Let $P=Q\oplus \overline{Q}$ for indecomposable $A$ module $Q$. Let $\pi:P\to Q$ be the natural projection. Let $M$ be a submodule of $P$ such that $\pi(M)=Q.$ Now how can we prove that $M=Q\oplus \overline{M}$ for $\overline{M}$ a submodule of $\overline{Q}$?
My thought goes to prove that the short sequence $0\to \overline{M} \to M \to Q \to 0$ is exact and so splits, by using the exactness of the sequence $0\to \overline{Q} \to P \to Q \to 0$ and the functions between these sequences? Please let me know if I am wrong?
And please let me know how can we show this? It does not seem a difficult question! But I have no other idea!
Thanks!
I think you have to write $M=\bar M\oplus Q'$ where $Q'$ is a submodule isomorphic to $Q$. Suppose that $P,Q,\bar Q$ are vector spaces. Let $M$ be a vector space supplementary to $\bar Q$, $\pi_{\mid M}(M)=Q$, but $Q$ is not necessarily a subspace of $M$.
Since $\pi_{\mid M}:M\rightarrow Q$ is surjective and $P$ is a projective module, there exists $f:P\rightarrow M$ such that $\pi=\pi_{\mid M}\circ f$. For every $m\in M$, $\pi(m-f(\pi(m))=\pi(m)-\pi_{\mid M}(f(\pi(m))=\pi(m)-(\pi_{\mid M}\circ f)(\pi(m))=\pi(m)-\pi(m)=0$, this implies that $m-f(\pi(m))\in \bar Q$, $m=m-f(\pi(m))+f(\pi(m))$. This implies that $M=M\cap \bar Q+f(Q)$. Let $x\in M\cap\bar Q\cap f(Q)$, $x=f(q), q\in Q$, this implies that $\pi(x)=\pi(f(q))=\pi(q)=q$, since $x\in\bar Q,\pi(x)=0$, we deduce that $q=x=0$, and $M=\bar Q\cap M\oplus f(Q)$.