Idempotent endomorphisms generate a direct sum decomposition Show that there is a one-to-one correspondence between indempotents $e\in\operatorname{End}_R(V)$ and direct sum decompositions $V=X\dotplus Y$.
Attempt:
We have $e^2 = e$.  Therefore, we can write an element $x\in V$ as $x = e(x) + (1-e)(x)$. Write $X = Re$ and $Y =  R(1-e)$.  Then $V = X + Y$.  Note that if $a$ belongs to  $X \cap Y$, then $a = e(x) = (1-e)(y)$. So, $e(a) = e^2(x) = e(x)= a$ and  $e(a) = e(1-e)(y) = -a$. Therefore $a= 0$. Hence  $V$ decomposes into a direct sum.

To finish a little more cleanly, if $e(r)=(1-e)(s)$, applying $e$ to both sides gets you $e(r)=e^2(r)=(e\circ(1-e))(s)$, but $e\circ(1-e)$ is the zero map, so $e(r)=0$.  This proves $e(R)\cap (1-e)(R)=\{0\}$.
In the other direction, if $V=X\dotplus Y$, let $e$ be the map that sends $x+y\mapsto x$ for every $x+y\in V$ (remember the elements of $V$ have a unique representation as a sum, so this map is well-defined.) 
You can confirm $e\in End_R(V)$, and further that $e\circ e=e$.
