Section of a coherent algebraic sheaf being zero over a principal open I have a question on Proposition 6 of §43 of Serre's Coherent Algebraic Sheaves (page 51-52 in the link).
The proposition is stating that, given $X$ any irreducible algebraic variety, $Q$ a regular function on $X$, $\mathcal{F}$ a coherent algebraic sheaf on $X$ and $s$ a global section of $\mathcal{F}$ which restriction to $X_Q$ is zero, then for $n$ sufficiently large the section $Q^n s $ is zero on all $X$. (Recall that $X_Q$ is defined to be the open of $X$ given by thos $x$ such that $Q(x) \neq 0$).
This is my reasoning: the section $Q \cdot s$ is zero on $X$ since, at any point $x \in X$, either $x \in X_Q$, so $s(x)=0$, or $Q(x)=0$, so  $Qs(x)= Q(x)\cdot s(x)=0$ 
Following this reasoning, the whole proposition would be useless (for example the coherence condition wouldn't play any role). Where am I wrong?
 A: Given an element $f \in \mathcal O_X(X)$ and a point $x \in X$, there are two notions of $f$ being zero/vanishing at $x$.


*

*The germ $f_x$ is zero as an element of $\mathcal O_{X,x}$. This is meant, if you read $f_x=0$.

*The germ $f_x$ is contained in the maximal ideal of $\mathcal O_{X,x}$, i.e. it is zero in the residue field $k(x)$. This is meant, if you read $f(x)=0$.
In particular $X_f$ is defined via the second notion, i.e.
$$X_f = \{x \in X ~|~ f_x \in \mathcal O_{X,x}^* ~\}.$$
In particular, if $x \notin X_f$, the germ $f_x$ is contained in the maximal ideal of the local ring $\mathcal O_{X,x}$, but it might be a non-zero element. In particular $(fs)_x$ might be a non-zero element of $\mathcal F_x$, i.e. $fs \in \mathcal F(X)$ might be a non-zero global section.
As an example, take $X = \operatorname{Spec} k[x], \mathcal F = (k[x]/(x^2))^\sim, f=x \in k[x], s=1 \in k[x]/(x^2)$.
The localization of $s$ at all primes in $X_f = \{(x)\}^c$ is zero, since we have $\operatorname{Supp} k[x]/(x^2) = \{(x)\}$. Hence, by the proposition, $f^ns=0$ for some $n$. And here, we have to take $n=2$, since $f^1s=x$ is a non-zero element of $k[x]/(x^2)$.

Some notes to understand the meaning of the second notion: Let $X = \operatorname{Spec}k[x_1, \dotsc, x_n]$ the affine $n$-space. and $f \in \mathcal O_X(X)=k[x_1, \dotsc, x_n]$ a regular function.
Let $p=(p_1, \dotsc, p_n)$ be a root of $f$, i.e. $f(p)=0$ and let $x = \mathfrak m_p = (x_1-p_1, \dotsc, x_n-p_n)$ be the corresponding maximal ideal. Of course the localization $f_x \in k[x_1, \dotsc, x_n]_x$ is non-zero, since any localization map from an integral domain is injective.
Hence the first notion of $f$ being zero at $x$ is not the correct one if we want to interpret $x$ as a root of $f$.
The second notion however is: $f(p)=0$ is equivalent to $f \in \mathfrak m_p$ by the Nullstellensatz, hence $f(p)=0$ is equivalent to the statement that $f_x$ is contained in the maximal ideal of $\mathcal O_{X,x}=k[x_1, \dotsc, x_n]_{\mathfrak m_p}$.
