Followup to my previous question, $M = \text{Image}(u^\infty) \oplus \text{Ker}(u^\infty)$. See my previous question here, Intersection of images and union of kernels.


  Let $A$ be a ring (not necessarily commutative), let $M$ be an $A$-module, and let $u: M \to M$ be an $A$-module morphism. Put $$\text{Image}(u^\infty) := \bigcap_{k=1}^\infty \text{Image}(u^k),\text{ resp. }\text{Ker}(u^\infty) = \bigcup_{k=1}^\infty \text{Ker}(u^k).$$Are $\text{Image}(u^\infty)$ and $\text{Ker}(u^\infty)$ necessarily $A$-submodules in $M$?


My question this time around is, do we moreover have that$$M = \text{Image}(u^\infty) \oplus \text{Ker}(u^\infty)?$$
Since we have$$M \supseteq \text{Im}(u) \supseteq \text{Im}(u^2) \supseteq \dots$$and $M$ is of finite length, we must have some $k \in \mathbb{N}$ such that  for all $k' \ge k$,$$\text{Im}(u^{k'}) = \text{Im}(u^k) = \text{Im}(u^\infty).$$Since we also have$$\text{Ker}(u) \subseteq \text{Ker}(u^2) \subseteq \dots \subseteq M,$$we get that for all $k' \ge k$,$$\text{Ker}(u^{k'}) = \text{Ker}(u^k) = \text{Ker}(u^\infty).$$


Proof that $M = \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty)$. Assume $a \in M$, and let $a_1 = u^k(a) \in \text{Im}(u^\infty)$. So$$u^k(a - a_1) = a_1 - u^k(a_1) = a- u^{2k}(a).$$Since $a_1 \in \text{Im}(u^\infty)$, there is $a_2 \in M$ such that $a = u^{2k}(a_2)$. Hence,$$u^k(a - a_1 - u^k(a_2 - a)) = u^k(a - u^k(a_2)) = 0.$$So$$b_1 = a- u^k(a_2) \in \text{Ker}(u^k) = \text{Ker}(u^\infty),\text{ }b_2 = u^k(a_2) \in \text{Im}(u^k) = \text{Im}(u^\infty).$$Thus,$$a = b_1 \oplus b_2 \in \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty),$$and so$$M = \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty).$$
Proof that $\text{Ker}(u^\infty) \cap \text{Im}(u^\infty) = \{0\}$. Assume that $\lambda \in \text{Ker}(u^\infty) \cap \text{Im}(u^\infty)$. So $u^k(\lambda) = 0$, and there is $\lambda_1 \in M$ such that $\lambda = u^k(\lambda_1)$. Therefore, $$u^{2k}(\lambda_1) = 0 \text{ and }\lambda_1 \in \text{Ker}(u^{2k}) = \text{Ker}(u^\infty) = \text{Ker}(u^k).$$So $\lambda = 0$.
Since we have$$M \supseteq \text{Im}(u) \supseteq \text{Im}(u^2) \supseteq \dots$$and $M$ is of finite length, we must have some $k \in \mathbb{N}$ such that  for all $k' \ge k$,$$\text{Im}(u^{k'}) = \text{Im}(u^k) = \text{Im}(u^\infty).$$Since we also have$$\text{Ker}(u) \subseteq \text{Ker}(u^2) \subseteq \dots \subseteq M,$$we get that for all $k' \ge k$,$$\text{Ker}(u^{k'}) = \text{Ker}(u^k) = \text{Ker}(u^\infty).$$


Proof that $M = \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty)$. Assume $a \in M$, and let $a_1 = u^k(a) \in \text{Im}(u^\infty)$. So$$u^k(a - a_1) = a_1 - u^k(a_1) = a- u^{2k}(a).$$Since $a_1 \in \text{Im}(u^\infty)$, there is $a_2 \in M$ such that $a = u^{2k}(a_2)$. Hence,$$u^k(a - a_1 - u^k(a_2 - a)) = u^k(a - u^k(a_2)) = 0.$$So$$b_1 = a- u^k(a_2) \in \text{Ker}(u^k) = \text{Ker}(u^\infty),\text{ }b_2 = u^k(a_2) \in \text{Im}(u^k) = \text{Im}(u^\infty).$$Thus,$$a = b_1 \oplus b_2 \in \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty),$$and so$$M = \text{Ker}(u^\infty) \oplus \text{Im}(u^\infty).$$
Proof that $\text{Ker}(u^\infty) \cap \text{Im}(u^\infty) = \{0\}$. Assume that $\lambda \in \text{Ker}(u^\infty) \cap \text{Im}(u^\infty)$. So $u^k(\lambda) = 0$, and there is $\lambda_1 \in M$ such that $\lambda = u^k(\lambda_1)$. Therefore, $$u^{2k}(\lambda_1) = 0 \text{ and }\lambda_1 \in \text{Ker}(u^{2k}) = \text{Ker}(u^\infty) = \text{Ker}(u^k).$$So $\lambda = 0$.