Let $R$ be a ring and $A,B,C$ be left modules over $R$, if $A\oplus B\cong A\oplus C$ and $A$ is simple (i.e. it has exactly two submodules), does it follow that $B\cong C$ ?
I believe the conclusion follows if any of $B$ or $C$ is completely reducible, my question is does it follow without any complete reduciblity assumption?
Thank you.
The deeper and more general statement described by rschwieb is certainly interesting, but there's also a fairly elementary proof for the particular case in the question.
Suppose $A\oplus B\to A\oplus C$ is an isomorphism, with $A$ simple. Composing with projection $A\oplus C\to A$ we get a split epimorphism $\varphi:A\oplus B\to A$ whose kernel is isomorphic to $C$.
Let $\varphi(a,b)=\varphi_A(a)+\varphi_B(b)$. Since $A$ is simple, there are two possibilities:
(1) $\varphi_A$ is an isomorphism. In this case the map $B\to\ker(\varphi)$ given by $b\mapsto (\varphi_A^{-1}\varphi_B(b),-b)$ is an isomorphism. So $C\cong B$.
(2) $\varphi_A=0$. Then $\ker(\varphi)=A\oplus\ker(\varphi_B)$, and since $\varphi_B$ is a split epimorphism, also $B\cong A\oplus\ker(\varphi_B)$. So again $B\cong C$.