Let $k$ be a field, $k[X]$ the polynomial ring in one variable and $M$ a torsion $k[X]$-module (not necessarily finitely generated).
Consider the submodules
\begin{equation*}
M_1 = \{a \in M \mid X^n a = 0 \text{ for some } n \in \mathbb{N}\},
\end{equation*}
\begin{equation*}
M_2 = \{a \in M \mid f a = 0 \text{ for some } f \in k[X], \text{ relatively coprime to } X\}.
\end{equation*}
Is it true that $M_1 + M_2 = M$?
The question arose while trying to understand this paper I unfortunately can not find on the internet. In the paper, $k[X]$ acts on $M$ via a vector space endomorphism $\phi \in \operatorname{End}_k(M)$ and $\phi$ has the property $\dim \phi^n M < \infty$ for some $n \in \mathbb{N}$ (that's why $M$ is a torsion $k[X]$-module). From the wording of the proof in question, I expect the proposition to be true in general but I don't even understand it in this special case.
This is true in a more general setting. Let $R$ be a principal ideal domain, $M$ a torsion $R$-module, and let $p$ be a prime element of $R$. Define $$M_1 = \{a\in M \ | \ p^n a = 0 \textrm{ for some } n\in \mathbb{N} \};$$ $$M_2 = \{a\in M \ | \ qa=0 \textrm{ for some } q\in R \textrm{ coprime with } p \}.$$
Now, let $a\in M$. Since $M$ is a torsion module, there exists a non-zero $z\in R$ such that $za=0$. Let $n$ be the greatest integer such that $z=p^nq$, with $p$ coprime with $q$. By Bézout's identity, there exist $\alpha, \beta\in R$ such that $\alpha p^n + \beta q = 1$. Then $$ a = \alpha p^n a + \beta q a,$$ with $\alpha p^n a \in M_2$ and $\beta q a\in M_1$. This finishes the proof.
You can then apply this to the special case $R=k[X]$ and $p=X$.