My issue is the following:
Let $M$ be an $A$-module with $A$ a commutative unitary ring, $u: M\longrightarrow M $ an $A$-morphism.
I cannot solve or find the proof of the following assertions:
1- If $u(M)=u^{2}(M)$ then $M=\ker u + u(M)$
2- If $\ker(u) = \ker(u^2)$ then $\ker(u) \cap u(M) = \lbrace 0 \rbrace$
Can anyone help me with this please.
Thank you in advance.
For every $m \in M$, you have that $u(m) = u^2(n)$, for some $n \in M$ because $u(M) = u^2(M)$. Now $u(m-u(n)) = u(m) - u^2(n) = 0$, so $m - u(n) \in \operatorname{Ker} u$ and $m = u(n) + (m - u(n)) \in u(m) + \operatorname{Ker} u$. If $x \in \operatorname{Ker} u \cap u(M)$ then $x = u(y)$ for some $y \in M$, so $u(x) = u^2(y)$. However $0 = u(x) = u^2(y)$, since $x \in \operatorname{Ker} u$, so $y \in \operatorname{Ker} u^2 = \operatorname{Ker} u \Rightarrow x = u(y) = 0$.