Let $(X,\mathcal{O}_X)$ be a locally ringed space and let $\mathcal{F}$ be an $\mathcal{O}_X$-module. For a section $s \in \mathcal{F}(X)$ and a point $x$, we say $s(x)=0$ if the stalk $s_x$ is zero modulo the (unique) maximal ideal of $\mathcal{O}_{X,x}$.

Is it true in general that $\{x \in X | s(x) = 0 \}$ is closed?

I think I proved it for locally free $\mathcal{O}_X$-modules: if one of the components of a section doesn't vanish, then this component is invertible (in particular non-zero) in an open neighbourhoud. But I'm not able to extend this proof to $\mathcal{O}_X$-modules in general.

  • $\begingroup$ Working with the espace étalé, remember that $0$ represents the zero germ, so $s(x)=0$ means that $s$ is actually $0$ in a neighborhood of $x$. $\endgroup$ Commented Feb 2, 2014 at 14:00
  • $\begingroup$ @TedShifrin Yeah, I got it. The $0$'s are different, sorry I commented before thinking. $\endgroup$
    – user40276
    Commented Feb 2, 2014 at 14:02
  • 1
    $\begingroup$ The definition is $\mathcal{F}(x) = \mathcal{F}_x / \mathfrak{m}_x \mathcal{F}_x$ and $s(x)$ is the class of $s_x$ in $\mathcal{F}(x)$. $\endgroup$ Commented Feb 2, 2014 at 14:08

1 Answer 1


When $X=\mathrm{Spec}(\mathbb{Z})$ and $\mathcal{F}$ is the quasi-coherent sheaf associated to $\mathbb{Z}/p \mathbb{Z} = \langle s \rangle$ for a prime $p$, prove that $\{x \in X : s(x)=0\} = X \setminus \{(p)\}$, which is not closed.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .