Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A \cong B$.

Question

Let $R$ be a ring, $M$ an $R$-module, and $A, B ≤ M$ two submodules of $M$ such that $M = A ⊕ B$. Prove that $M/A \cong B$.

Answer 1

Saying that $M=A\oplus B$, means

1. $M=A+B$, and
2. $A\cap B=\{0\}$.

Thus the second isomorphism theorem yields $$ M/A=(A+B)/A\cong B/(A\cap B)=B/\{0\}\cong B $$