Proving that a module can be decomposed as a direct sum of submodules

Letting $X$ be a ring and $K$ be an $X$-module, I need to show that if $K \cong A \times B$ for some $X$-modules $A,B$, then $\exists$ submodules $M'$ and $N'$ of $K$ such that:

$K=M' \oplus N'$

$M' \cong A$

$N' \cong B.$

I understand the concepts of internal and external direct sum of modules, and I showed that if $K = M \oplus N$ for $M,N$ submodules of $K$, then $K \cong M \times N.$ (I showed the isomorphism by defining a well-defined map, and then showing that the map is a surjective homomorphism, followed by the kernel being $\{0\}$ and applying the First Isomorphism Theorem.)

But I have tried doing this problem for hours now, and have not been able to crack it. How should I begin?

• If $\varphi:A\times B\longrightarrow K$ is such an isomorphism, just let $M=\varphi\bigl(A\times\{0\}\bigr)$ and $N=\varphi\bigl(\{0\}\times B\bigr)$ . – Bernard Mar 27 '15 at 20:41

Let $\phi : A\times B\to K$ be the isomorphism.
Look at the submodules of $K$ which are given by $\phi(A\times \{0\})$ and $\phi(\{0\}\times B)$.