There exist $n \in \mathbb{N}$ such that $\mathrm{Im} f^{n} + \mathrm{Ker} f^{n} = M$ if $M$ satisfies DCC Let $M$ be an $R$-module, and let $f$ be an $R$-automorphism on $M$. Prove that if $M$ satisfies the descending chain condition, then there exist $n \in \mathbb{N}$ such that $\operatorname{Im} f^{n} + \operatorname{Ker} f^{n} = M$.

Note that $ \operatorname{im}(f) \supseteq \operatorname{im}(f^2) \supseteq \operatorname{im}(f^3) \supseteq \cdots$ so by the DCC, there is an integer $n$ such that $\operatorname{im}(f^n)=\operatorname{im}(f^{2n}).$ Let $m\in M.$ Since $\operatorname{im}(f^n)=\operatorname{im}(f^{2n}),$ there is some $t\in M$ such that $f^n(m)=f^{2n}(t).$ Write $m=(m-f^n(t)) + f^n(t)$ and we're done.
