I need to show that if M is an A-module and $A\times A$ is isomorphic to $M\times M$ as A-modules then M is a direct summand of $A\times A$.
In other words that $A\times A$ = $N\times M$ for some A-module N.
I'm not sure how to approach proving this result, but I need it as a part of a homological algebra problem I'm working on. Any suggestions?
It's certainly not going to be true that $A\times A=N\times M$. Indeed, that would only be possible if $A=N$ and $A=M$, in order for $A\times A$ and $N\times M$ to be equal as sets.
But, that's not what "is a direct summand" means. If a module $X$ is the direct sum of submodules $Y$ and $Z$, that doesn't mean that $X=Y\times Z$, but that the canonical map $Y\times Z\to X$ is an isomorphism (so in particular $X\cong Y\times Z$). Moreover, in this context it is certainly not being asserted that $M$ is literally a submodule of $A\times A$ (there is absolutely no reason that $M$ would be a subset of $A\times A$), so what "$M$ is a direct summand of $A\times A$" means is that $M$ is isomorphic to a direct summand.
In other words, what you are trying to prove here is just that there exists a module $N$ such that $A\times A\cong N\times M$. This is now completely trivial: can you see an $N$ that works?