# Torsion in module over the ring of convergent power series

I need to understand a passage from a paper which I don't quite understand.

Let $M$ be a module over the ring $\mathbb C\{t\}$ of convergent power series. We want to show that $M$ is torsion-free, i.e., if $\psi(t)\cdot m = 0$ for some non-zero $\psi \in \mathbb C\{t\}$ then $m = 0$.

What the author is actually showing, if I understand correctly, is that $t\cdot m = 0$ implies $m = 0$, for every $m$. This is of course necessary, but is it sufficient?

So the question is:

Let $M$ be a $\mathbb C\{t\}$-module. Suppose $t\cdot m = 0$ implies $m=0$. Is then $M$ torsion-free?

• Are there conditions on the radius of convergence, or you just consider those power series with a nonzero radius of convergence? – egreg May 2 '16 at 22:44
• Just series with a nonzero radius of convergence – user336494 May 2 '16 at 22:49

## 1 Answer

The ring $\mathbb{C}\{t\}$ is local and its maximal ideal is generated by $t$, see Ring of Convergent Power Series in R and C is a Local Ring

In other words, power series $f(t)$ with $f(0)\ne0$ are invertible. So every nonzero convergent power series can be written as $$t^k g(t)$$ for some integer $k\ge0$ and $g(0)\ne0$. Therefore $t^kg(t)m=0$ if and only if $t^km=0$.

Suppose $tm=0$ implies $m=0$. Now, if $t^km=0$, with $k>0$, we have $t^{k-1}m=0$ and, by induction, $m=0$.