# Sheaf of rings with vanishing stalk?

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. – Hurkyl Jun 30 '12 at 18:45
Work with locally ringed spaces. – Martin Brandenburg Jun 30 '12 at 20:34

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.... – Hurkyl Jun 30 '12 at 18:42
@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, – Matt E Jun 30 '12 at 18:44
I agree with Matt E that a local ring is non-zero by definition, since maximal ideal is (probably universally) defined to be proper. – ashpool Jun 30 '12 at 18:50
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. – ashpool Jun 30 '12 at 18:55
@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, – Matt E Jun 30 '12 at 19:01