For simple $R$-modules, $A$,$B$, and $C$ if $A\oplus C\cong B\oplus C$ as $R$-modules then $A\cong B$ as $R$-modules.

Question

I'm supposing that there exists some $R$-module isomorphism $\varphi:A\oplus C\to B\oplus C$ and considering the following inclusion and projection maps:
$$\iota:A\to A\oplus C$$ and $$\pi:B\oplus C\to B,$$ which are both $R$-module homomorphisms. It follows that $F:=(\pi\circ \varphi)\circ\iota\in \text{Hom}_R(A,B)$. By Schur's Lemma, $F$ is either $0$ or an isomorphism.

Suppose we have $F(a)=0$ for all $a\in A$. $$\pi(\varphi(\iota(a)))=\pi(\varphi((a,0_C)))=0 \implies \varphi((a,0_C))=(0,c)$$ for some $c\in C$. As $a$ was arbitrary, $$\varphi(A\oplus \{0\})\subseteq \{0\}\oplus C\cong C.$$ $C$ is simple $\implies$ $\varphi(A\oplus \{0\})=\{0\}$ or $C$. As $\varphi$ is an isomorphism we must have $\varphi(A)=C$ .

Not sure how to proceed from here.

Answer 1

There are three ways you can do this.

Method 1

Consider these compositions

$$f:A\to A\oplus C\cong B\oplus C\to B$$ $$g:A\to A\oplus C\cong B\oplus C\to C$$ $$h:C\to A\oplus C\cong B\oplus C\to B$$

Each of them is zero or an isomorphism by Schur's lemma. If $f$ is an isomorphism we are done. If not then $f=0$. In this case $g$ is an isomorphism. Then $h$ cannot be $0$, otherwise all of $A\oplus C$ goes to the $C$ component of $B\oplus C$ under the isomorphism, which is absurd. Hence $h$ is an isomorphism, and $hg:A\cong B$.

Method 2

Consider the composition series $$A\oplus C\supset C\supset0$$ $$B\oplus C\supset C\supset0$$ and conclude with the Jordan-Holder theorem.

Method 3

Note that simple modules are indecomposable and that the direct sum of two simple modules is Noetherian and Artinian. Use the Krull-Remak-Schmidt theorem.

Answer 2

It is possible for $\phi$ to be an isomorphism while $F = 0$. Take $A = B = C$ and let $\phi: A \oplus A \longrightarrow A \oplus A$ be the twist isomorphism, i.e. $(a_1, a_2) \mapsto (a_2, a_1)$. Then let $\iota: A \longrightarrow A \oplus A$ via $a \mapsto (a, 0)$ and $\pi: A \oplus A \longrightarrow A$ via $(a_1, a_2) \mapsto a_1$. These are the inclusion to and projection onto the first coordinate, as in your definition. Then $F = \pi \circ \phi \circ \iota: A \longrightarrow A$ takes $a \mapsto (a, 0) \mapsto (0, a) \mapsto 0$, so $F = 0$ despite $\phi$ being an isomorphism.