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

[kalish1]

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?

This question has been crossposted here: abstract algebra - Proving that a module can be decomposed as a direct sum of submodules - Mathematics Stack Exchange

 

[Euge]

Hi kalish,

Let $\phi$ be is isomorphism from $A\times B$ onto $K$. Set $M' = \phi(A \times 0)$ and $N' = \phi(0 \times B)$. Since $M'$ and $N'$ are homomorphic images of submodules of $A\times B$, $M'$ and $N'$ are submodules of $K$. Given $x\in K$, there exists a unique element $(a,b) \in A\times B$ such that $x = \phi(a,b)$. Since $(a,b) = (a,0) + (0,b)$ and $\phi$ is a homomorphism, $\phi(a,b) = \phi(a,0) + \phi(0,b) \in M' + N'$. The representation of $x$ as $\phi(a,0) + \phi(0,b)$ is unique by uniqueness of $(a,b)$. Thus $x \in M' \oplus N'$. Since $x$ was an arbitrary point of $K$, $K \subseteq M' \oplus N'$ and thus $K = M' \oplus N'$.