I'm attempting to prove the following statement:
"Let $R$ be a ring, let $M$ be an $R$-module, and let $\phi:M \to M$ be an $R$-module homomorphism. Prove that if $\phi$ is an idempotent in the ring $\text{End}_R(M)$, that is, $\phi^2=\phi$, then $M=\text{Ker}(\phi) \oplus \text{Im}(\phi)$"
This is what I have so far:
"We have that $\text{Ker}(\phi) + \text{Im}(\phi) \subseteq M$ because $\text{Ker}(\phi) \subseteq M$ and $\text{Im}(\phi) \subseteq M$.
To see the other inclusion, suppose that $x \in M$.
Then we can write $x=(x-\phi(x))+\phi(x)$.
We have that $\phi(x) \in \text{Im}(\phi)$ by definition.
We claim that $x-\phi(x) \in \text{Ker}(\phi)$.
We have the following:
$\phi(x-\phi(x))=\phi(x)-\phi^2(x)$
$=\phi(x)-\phi(x)$
$=0$.
Therefore, $x- \phi(x) \in \text{Ker}(\phi)$, meaning that $x=(x-\phi(x))+\phi(x) \in \text{Ker}(\phi)+ \text{Im}(\phi)$.
Therefore, $M \subseteq \text{Ker}(\phi) + \text{Im}(\phi)$.
Thus, by double containment, $M=\text{Ker}(\phi)+\text{Im}(\phi)$.
We claim that $\text{Ker}(\phi)$ and $\text{Im}(\phi)$ intersect trivially.
Suppose that $x \in \text{Ker}(\phi) \cap \text{Im}(\phi)$.
Then $x \in \text{Ker}(\phi)$ and $x \in \text{Im}(\phi)$.
Since $x \in \text{Im}(\phi)$, then we have that $\phi(x')=x$ for some $x' \in M$."
Is what I have so far correct? If so, how should I show that $\text{Ker}(\phi)$ and $\text{Im}(\phi)$ intersect trivially?
Everything in the OP's argument looks fine to me and indeed, it is very nearly completely complete. I would finish things off like this:
Let
$x \in \ker \phi \cap \text{Im} \phi; \tag 1$
then
$x \in \ker \phi \Longrightarrow \phi(x) = 0; \tag 2$
also,
$x \in \text{Im}(\phi) \Longrightarrow \exists y, \; x = \phi(y); \tag 3$
so we have
$x = \phi(y) = \phi^2(y) = \phi(\phi(y)) = \phi(x) = 0, \tag 4$
whence
$\ker \phi \cap \text{Im}(\phi) = \{0\}. \tag 5$
It is probably worth noting that this whole exercise focuses on a central result of idempotnent endomorphisms on modules, and the arguments are in fact ubiquitous.