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Given a locally ringed space $(X,\mathcal O_X)$ and a global section $s\in \Gamma (X,\mathcal O_X)$, which does not vanish at any point $x\in X$, meaning the image of $s$ in the stalk $\mathcal O_{X,x}$ does not belong to the maximal ideal of the local ring $\mathcal O_{X,x}$, does it mean that $s$ is invertible as an element of the ring $\Gamma (X,\mathcal O_X)$? In other words, if $s$ is invertible stalk-locally, does it mean that $s$ is invertible?

It is easy to see that if $s$ is locally-invertible in the usual sense, then it is invertible, since the uniqueness of the inverse guarantees that the local inverses can be patched together. It is also easy to see that for schemes, the answer is yes, since for an affine scheme this follows from the fact that an element of a ring that does not belong to any prime (or maximal for that matter) ideal is invertible. But is it true in general?

REMARK: I couldn't find batter tags. Feel free to edit.

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Yes. It suffices to prove that if $s_x \in \mathcal{O}_{X,x}$ is invertible, then there exists an open neighborhood $U$ of $x$ such that $s \vert_U \in \mathcal{O}_X(U)$ is invertible. But this is obvious from the definition of the stalk.

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I don't see how is this obvious from the definition of the stalk. Of course the inverse of $s_x$ is the image of some local section $t$ on some open $U$. But all you know is that $(st)_x = 1$, and it doesn't mean that $st = 1$ on $U$. – KotelKanim Nov 19 '11 at 12:06
Let $t \in \mathcal{O}_X(V)$ be a section such that $s_x t_x = 1$ in $\mathcal{O}_{X,x}$. By the definition of the stalk, there exists $U \subseteq V$ such that $s \vert_U \cdot t \vert_U = 1$, so $s \vert_U$ is invertible on $U$. – Andrea Nov 19 '11 at 12:10
hmm, you are right. It seems that this is true even in a general ringed space. thanks. – KotelKanim Nov 19 '11 at 15:26
You are welcome. In general, you must think that locally ringed spaces are not so bad; in fact, they unify in a single concept differentiable manifolds, complex manifolds, real/complex analytic spaces and schemes. If $s$ is a section on a locally ringed space, the question "is $s$ zero in a point?" has sense. On the contrary, this question is meaningless in ringed space which are not locally ringed. – Andrea Nov 19 '11 at 17:59
What I meant was that a section whose image is invertible in each stalk is invertible. Of course in general you can't talk about having a "zero in a point". Am I right? – KotelKanim Nov 20 '11 at 7:17

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