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How common is that a sheaf of rings has a vanishing stalk? To define the rank of a locally free sheaf of $\mathscr{O}$-modules, for instance, $\mathscr{O}_x=0$ may cause some problem, since the rank of a free $A$-module is not well-defined if $A=0$. It would make life easier if $\mathscr{O}_x\neq0\ \forall x\in X$, but is this condition somehow incorporated in the definition of sheaf of rings?

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    I believe that you can define the rank by continuously extending it from the support of $\mathcal{O}$, so long as $\mathcal{O}$ doesn't vanish on an entire connected component.2012-06-30
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    Work with locally ringed spaces.2012-06-30

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I will presume that by "a sheaf of rings" you mean "a sheaf of rings with $1$" (since this is what is usually so meant). If the stalk of $\mathcal O_X$ vanishes at $x$, then this means that $1 = 0$ in the stalk, and hence in $\mathcal O_X(U)$ for some neighbourhood $U$ of $x$. Thus the stalk of $\mathcal O_X$ vanishes at $x$ if and only if the sheaf $\mathcal O_X$ restricts to the zero sheaf in some n.h of $x$.

If e.g. $X$ is not only ringed, but locally ringed, then the stalks of $\mathcal O_X$ (which are then local rings by definition) never vanish at a point (since local rings are non-zero, again by definition).

Added: If $U$ is an open subset of $X$ with complement $Z$, and $i: Z \to X$ is the inclusion, then for any sheaf of rings $\mathcal O_Z$ on $Z$, the pushforward $i_* \mathcal O_Z$ will be a sheaf of rings on $X$ whose stalks vanish on $U$. Thus we can always find examples realizing the discussion of the first paragraph. More generally, if we let $Z$ be the support of any sheaf of rings $\mathcal O_X$ on $X$, this will coincide with the support of the identity section $1 \in \mathcal O_X(X)$, and hence will be a closed subset of $X$, and we will have that $\mathcal O_X = i_* i^{-1} \mathcal O_X,$ with $i^{-1}\mathcal O_X$ a sheaf of rings on $Z$, none of whose stalks vanish. Consequently, any sheaf of rings on $X$ can be obtained by a sheaf of rings with non-vanishing stalks on a closed subspace by pushing forward.

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    Local rings are non-zero by definition only if you adopt the convention that local rings are non-zero by definition. That convention is not universal: e.g. such a condition is not present in the definition of local ring used in *Sheaves in Geometry and Logic*. The problems with habit of excluding degenerate cases tends to be amplified in the context of sheaf theory anyways....2012-06-30
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    @Hurkyl: Dear Hyrkyl, In my experience, the definition of "locally ringed space", as it is used in algebraic geometry and related areas, presupposes that local rings are taken to be non-zero. (Otherwise the notion would behave quite differently.) There may be other conventions in different areas of mathematics; I am answering the question from the perspective of an algebraic geometer. Regards,2012-06-30
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    I agree with Matt E that a local ring is non-zero by definition, since maximal ideal is (probably universally) defined to be proper.2012-06-30
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    I don't think the first paragraph is answering the question, but I'm happy to restrict my attention to locally ringed spaces at present, so thanks for pointing that out.2012-06-30
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    @ashpool: Dear ashpool, I have added a paragraph expanding on the first one, which explains that every sheaf of rings can be obtained from one on a closed subset with no non-vanishing stalks via extension by zero. Regards,2012-06-30