Given a idempotent endomorphism, show that $M = \ker(f) \oplus \ker (f-\text{id}_M)$

Question

Let $M$ be an A-module and $f \in \text{End}(M)$ an idempotent endomorphism. Show that $M = \ker(f) \oplus \ker (f-\text{id}_M)$.

I don't know how to approach this exercise, using only properties of modules and direct sums, without projections.

Can you help me? Thanks in advance.

Answer 1

For any $m\in M$, write $m=\big(m-f(m)\big)+f(m)$. Now, $f\big(m-f(m)\big)=f(m)-f\circ f(m)=f(m)-f(m)=0$ i.e. $m-f(m)\in\ker (f)$. Similarly, $\big(f-\text{id}_M\big)\big(f(m)\big)=0$. So, $M=\ker f+ \ker (f-\text{Id}_M)$.

Now, let $x\in \ker f\cap \ker (f-\text{Id}_M)$, then, $f(x)=0=f(x)-\text{id}_M(x)=f(x)-x$, so that $x=0$. Hence, $M=\ker f\oplus \ker (f-\text{Id}_M)$.