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If we have $Z\subset X$ a closed irreducible subscheme of an integral scheme $X$ (which you can take to have various further niceness properties if you want), one can take its generic point $\eta_Z$ and localize at it, to get a local ring with quotient field the quotient field of $Z$.

However, intuitively, I'd like to be able to "take a small neighborhood of $Z$" and not just of $\eta_Z$, i.e. I'd like a canonically constructed subscheme of $X$ whose "special fiber" is $Z$ and doesn't contain any points of $X$ besides the ones which specialize to points of $Z$. Certainly in general, I wouldn't expect this to be possible, so I don't expect a general functorial construction like usual localizations. But are there known cases when something like this exists?

I'm particularly interested in when $\pi:X\to S$ is a flat family or something similarly nice, and $Z$ is a section of $\pi$, so that what I'm asking for is basically like the "relative local ring of the point $Z$ over the base $S$". Then I'd also like to ask that the pullback of my "localization at $Z$" to any fiber of $\pi$ over a closed point of $S$ yields the localization of that fiber at the point where it intersects $Z$.

In this case, if $X$ is a trivial bundle over $S$, like if $X=S\times_k Y$ for some $Y$, then obviously this is possible; I'm wondering if it's possible more generally.

(I'd also appreciate an explicit counterexample to it being possible in general flat families, since I don't expect it to be possible in that generality, but find it difficult to prove it's impossible in any particular case.)

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    $\begingroup$ Just as a local ring is not by definition a subscheme, I don't think you could expect a subscheme (not finitary enough), but is the thing you're seeking something like inverting all functions whose zero locus is disjoint from $Z$? It seems like it would be something like the coproduct in the category of $X$-schemes of the local ring at $\eta_Z$ and the formal completion of $X$ at $Z$. When $Z$ is a section, I would be thinking about relative effective Cartier divisors instead of arbitrary functions. In any case, I myself am very much interested in seeing a good answer. $\endgroup$
    – Somatic Custard
    Commented Aug 22, 2022 at 2:47
  • $\begingroup$ I imagine that you would like your "small neighborhood of $Z$ in $X$" to be the same as the corresponding "small neighborhood of $Z$ in $U$" if $U$ is an open subset of $X$ that contains $Z$. But this seems to mean that you do want to cut out every closed subset of $X$ that is disjoint from $Z$. This does not coincide with what you say you expect in the case when $X=Z\times Y$. $\endgroup$
    – Tom Goodwillie
    Commented Aug 22, 2022 at 11:16
  • $\begingroup$ A naive thing to do is take the limit of all open neighbourhoods $U \subseteq X$ containing $Z$. This exists as a locally ringed space (and agrees with the same limit as ringed space), but is not always a scheme (e.g. take $Z$ to be a line in $\mathbf P^2$). When $X$ is affine and $Z$ is closed, this is just a usual localisation of rings, but this procedure does not even commute with open immersions $U \hookrightarrow X$ (let alone with taking fibres). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 22, 2022 at 12:33
  • $\begingroup$ @R.vanDobbendeBruyn A native question. Why in the affine case it is just a localization? The process your are describing is just removing all closed subsets disjoint from $Z$. These closed subsets can have higher codimensions. Is it true that any higher codimension closed subset that is disjoint from $Z$ can be covered by codimension $1$ divisors that are disjoint from $Z$? $\endgroup$
    – user127776
    Commented Aug 23, 2022 at 18:31
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    $\begingroup$ @user127776 In the affine case, yes. If $I + J = R$ for ideals $I, J \subseteq R$ (i.e. $V(I) \cap V(J) = \varnothing$), then $I + (f) = R$ for some $f \in J$ (i.e. $V(I) \cap V(f) = \varnothing$). By the way, my construction agrees with Johan's, which is most easily seen if you already know it's also a limit of ringed spaces (as opposed to locally ringed spaces). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 23, 2022 at 22:37

1 Answer 1

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Let $W \subset X$ denote the set of points specializing to $Z$ with the induced topology. Denote $\mathcal{O}_W$ the pullback of $\mathcal{O}_X$. Then $W = (W, \mathcal{O}_W)$ is a locally ringed space and we can ask:

Question: is $W$ a scheme?

Usually not: for example if $X = \mathbf{P}^2_k$ and $Z$ is a line, then this is false. (Hint: show that there aren't any nonempty affine opens.)

Sometimes yes: if $Z = \{x\}$ consists of a closed point, then $W$ is the spectrum of the local ring $\mathcal{O}_{X, x}$.

Here is a criterion: suppose that we can write $W = \bigcap_{i \in I} U_i$ as the intersection of quasi-compact opens with the following properties: (a) $\forall i, j \in I, \exists k \in I$ such that $U_k \subset U_i \cap U_j$ and (b) $\forall i, j \in I$ if $U_i \subset U_j$ then $U_i \to U_j$ is an affine morphism. Then $W = \lim U_i$ is a scheme.

If $X$ is a smooth projective surface then (a) and (b) hold if $Z$ is the fibre of a nonconstant morphism from $X$ to a curve or if $Z$ can be blown down (to a point on another algebraic surface -- does not work if the contraction gives an algebraic space).

If $X$ is a smooth projective surface over the algebraic closure of a finite field and $f : X \to S$ is a nonconstant morphism to a smooth projective curve and $Z$ is the image of a section of $f$, then it is often the case that the normal bundle of $Z$ in $X$ is negative (think about elliptic surfaces for example). Then by Artin you can contract $Z$ and hence $W$ is a scheme.

Anyway, it is fun to play around with these ideas, but I do not know a good way to work with $W$ like this. Many people would, I think, instead consider the formal completion of $X$ along $Z$ and work with that instead.

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