Show that for all $m\in M$ there is $a\in A$ such that $f(m)=am$.

Question

The problem is following:

Let $M$ be a completely reducible $A$-module, $B=End_A(M)$ and let $f\in End_B(M)$. Show that for all $m\in M$ there is $a\in A$ such that $f(m)=am$.

My approach:

$M$ is a completely reducible $A$-module, so $M=M_1\oplus \cdots \oplus M_n$, where $M_i$ is an irreducible $A$-module for every $i$. Let $m\in M$. Note that for $m_i\in M_i$, $Am_i$ is a submodule of $M_i$, but since $M_i$ is irreducible it follows that $M_i=Am_i$. Then $m$ can be uniquely written as $m=a_1m_1\oplus \cdots \oplus a_nm_n$, where $m_i\in M_i$ and $a_i\in A$ for every $i$. Now, $f(m)= f(a_1m_1\oplus \cdots \oplus a_nm_n)=f(a_1m_1)\oplus \cdots \oplus f(a_nm_n)$.

But I don't see how I can finish this. Is this approach ok and how to proceed?

Answer 1

I think your approach doesn't work in general (but I might be wrong).

To make this kind an argument you need a slightly different method. Let $m\in M$, look at $Am\leq M$, this is a submodule and so from the complete reducibility, we get a projection (*) $\pi \in B$. $$f(m)=f(\pi(m))=\pi(f(m))\in Am$$ And so there is $a\in A$ such that $f(m)=am$.

It's possible that there is easier way to prove this, but I think this proof is elegant.

(*) A projection from of a submodule $N\subseteq M$ is a map $\pi:M\to N$ such that $\forall n\in N:\pi(n)=n$. If $\pi$ is a projection then $\ker(\pi)$ is a complement of $N$, and if $M=N\oplus N'$ we get a projection.