I try to show the following exercise from Group Representation Theory Book from Peter Webb:
Let $A$ be a ring with $1$. Let $V$ be a semisimple $A$-module with finitely many simple summands and $e,f \in End_A(V)$ idempotent. Show that $e(V) \cong f(V)$ if and only if there exists invertible $\alpha \in End_A(V)$ such that $f=\alpha e \alpha^{-1}$.
I can see how $f=\alpha e \alpha^{-1}$ implies $e(V) \cong f(V)$, but I dont get the other part.
I know that $e$ is idempotent if and only if $e$ is a projection and more, we have a bijection between decomposition of $V=W_1 \oplus W_2$ with orthogonal idempotents $e_1, e_2$ with $e_1+e_2=1$. I tried to use the semisimplicity of $V$ together with the above facts but it did not work.
Best Answer
Since $e$ is idempotent, it defines a direct sum decomposition $V = e(V) \oplus e(V)^\perp$ of $A$-modules, where $e(V)^\perp = (1-e)V$. Similarly for $f(V)$ and $f(V)^\perp$. Then, supposing that $V$ is semisimple and of finite length (if it is finite-dimensional and $A$ is over a field, for example), then $e(V)^\perp$ and $f(V)^\perp$ must be isomorphic (a semisimple module of finite length is classified by its list of irreducible submodules with multiplicity, and if we "subtract off" a component of type $e(V) \cong f(V)$ we must be left with whatever the type of $e(V)^\perp$ is, which is the same as $f(V)^\perp$).
After this, there must exist isomorphisms $\varphi \colon e(V) \to f(V)$ and $\psi \colon e(V)^\perp \to f(V)^\perp$ of $A$-modules, and we can join them together into an isomorphism $\alpha \colon V \to V$ defined by $\alpha|_{e(V)} = \varphi$ and $\alpha|_{e(V)^\perp} = \psi$. We have then that $ \alpha e = f \alpha$ by checking the equation on $e(V)$ and $e(V)^\perp$ separately.
Perhaps there is a way of weakening the finite length condition, but I think something similar needs to be used to deduce that $e(V)^\perp \cong f(V)^\perp$.