Prove there exists $l \geq 1$ such that $t^l M$ is free as a $\mathbb{C}[[t]]$-module

Question

Problem: Let $M$ be a finitely generated $\mathbb{C}[[t]]$-module. Prove that there exists a $l \geq 1$ such that $t^l M$ is free as a $\mathbb{C}[[t]]$-module.

Attempt: I let $B = \left\{m_1, \ldots, m_n \right\}$ generate $M$. Let $$l = \text{min} \left\{ v(m_i) \mid 1 \leq i \leq n \right\}$$ where $v$ is the valuation function, i.e. $$ v(\sum_{i=0}^{\infty} a_i t^{i} ) = \text{min} \left\{i \mid a_i \neq 0 \right\}. $$

I wanted to prove that $t^l M$ is free. But I think this is kind of obvious if $M$ is finitely generated? I would just take as basis $\left\{ t^l m_1, \ldots, t^l m_n \right\}$.

Help is appreciated!

Answer 1

Your attempt is meaningless since $v(m_i)$ is not defined (the $m_i$ are not elements of $\mathbb{C}[[t]]$), and even if it were meaningful I don't know why you would think $t^lM$ is free for that particular value of $l$. Instead, I would suggest you use the classification of finitely generated modules over a PID.

More details are hidden below.

The torsion submodule of $M$ is annihilated by $t^l$ for some $l$. So, for that value of $l$, $t^lM$ is a finitely generated torsion-free module. Since $\mathbb{C}[[t]]$ is a PID, this implies it is free.