Section of a coherent sheaf vanishing outside a point I need some help in understanding an argument, probably basic, about coherent sheaves, which I've read in a paper, and as far as I understand can be described as follows:

Let $\mathcal F$ be a coherent sheaf of $\mathcal O_S$-modules, where $S = \{z\in \mathbb C: |z|< 1\}$. Let $\sigma$ be a section of $\mathcal F$ over the whole of $S$. Suppose $\sigma$ is "null" $^{\ast}$ outside of $0\in S$. Then (because of the coherence of $\mathcal F$) the germ $\sigma_0$ of $\sigma$ at $0$ is a torsion element in $\mathcal F_0$.
(${\ast}$) Here  "null", I guess, should mean that $\sigma_x = 0 \in \mathcal F_x$ for each $x \in S-0$, but please correct me if I'm wrong.

Is this assertion correct as it is stated, and why? Thank you very much.

Yes, the assertion is true.
The section $\sigma$ defines a morphism $\Sigma:\mathcal O\to \mathcal F$ sending $f$ to $f\sigma$.
If $\sigma_0\in \mathcal F_0$ were not a torsion element then the morphism $\Sigma_0:\mathcal O_0\to \mathcal F_0$ would be injective and thus for a suitably small open neighbourhood $U$  of $0$ the morphism $\Sigma\vert U:\mathcal O\vert U\to \mathcal F\vert U$ would be injective too (by coherence of  $\ker \Sigma)$.
Then $\Sigma_x:\mathcal O_x\to \mathcal F_x$ would be injective for all $x\in U$, which implies $\sigma_x\neq 0$ for all $x\in U$: contradiction.   
