21
$\begingroup$

Motivation:

Say $F: D \to Sch$ is a diagram in the category of schemes, and we're interested in whether it has a colimit (gluings, pushouts, and "categorical" quotients are all examples of colimits). Its colimit $Q$ in the category of locally ringed spaces always exists, and a scheme $Y$ is the scheme-colimit of $F$ iff there is a map $Q\to Y$ that is "initial among maps to schemes", i.e. any other map from $Q$ to a scheme factors through it.

Thus the problem of when colimits of schemes exists can be answered if we know when a locally ringed space admits a "schemification".

Examples:

1) The process of turning a classical variety $V$ into a "variety with generic points" $V^s$ is a schemification. Proving this boils down the classical affine case. If $R$ is a finite-type reduced $k=\overline{k}$-algebra, and we map $V=mSpec(R)$ to a scheme $Y$, first cover $Y$ by affines $Spec(B_i)$, and then pull back this cover to $V$ and refine it by principal opens $mSpec(R_{f_{ij}})$, so we have maps $B_i \to R_{f_{ij}}$ that determine the map $V\to Y$ (by the adjunction of $Spec({\cal O}(-))$ to the inclusion $AffSch\hookrightarrow LRS)$. But these define the desired map $V^s\to Y$ because $V^s$ can be obtained by gluing $Spec(R_{f_{ij}})$.

2) For $k=\overline{k}$, if we let $\mathbb{G_m}=\mathbb{A}^1_k\setminus 0$ act on $\mathbb{A}^1_k$, then the coequalizer in schemes of the action and the projection $G_m \times_k \mathbb{A}^1_k \rightrightarrows \mathbb{A}^1_k$ (i.e. the "categorical quotient") is $Spec(k)$, hence the locally ringed space coequalizer $Q$, which has two points and is not a scheme, has $Spec(k)$ as its schemification.

3) Some locally ringed spaces have no schemification; for example, two affine lines glued along their generic points. This is precisely because the gluing diagram has no coequalizer in schemes, as per BCnrd and Anton Geraschenko's answer here.

When does a locally ringed space $X$ admit a "schemification", i.e. a map to a scheme $X^s$ through which any other map to a scheme must factor?

EDIT: (Response to Martin's answer)

4) Any locally ringed space $X$ with exactly one closed point $x$ has its "affinization", $X^a:=Spec({\cal O}_X(X))$, as a schemification. This is because in a map $F:X\to Y$ with $Y$ a scheme, every point maps to a generization of $f(x)$, so and since open sets are closed under generization, $f$ factors through an affine neighborhood of $f(x)$. But any map from $X$ to an affine factors uniquely through $X^a$, so we're done.

5) Schemification commutes with disjoint unions, in that if $X= \coprod X_i$ then $X^s$ exists iff all $X_i^s$ exist, and in that case $X^s = \coprod X_i^s$. So if $X$ is locally connected, since it then decomposes into a disjoint union of connected clopen components, we might as well assume it is connected.

$\endgroup$
5
  • 2
    $\begingroup$ HAve you read CAP.IV of "Topos Annelés et schemas relatifs" by M. Hakim , Ergebn. Math. Grenzgeb. , 64 ? $\endgroup$ Commented Apr 4, 2011 at 7:07
  • $\begingroup$ I haven't. Should I? I just skimmed through a copy and didn't see anything about turning espaces annelés en anneaux locaux into schemas. $\endgroup$ Commented Apr 4, 2011 at 7:34
  • $\begingroup$ @Andrew: There is some stuff in the next chapter (chapter 5) that might be useful to you (if you can show that your locally ringed topos is a relative scheme for some other ringed topos $(X,A)$, you have a universal scheme associated to it by way of a crazy adjunction induced by the universal map $(X,A)\to (Set,Z)$), but I don't think it really addresses your question in any sort of concrete way. $\endgroup$ Commented Apr 4, 2011 at 7:41
  • 2
    $\begingroup$ @Buschi, Chapter IV seems to be about forming locally ringed topoi (or spaces) from ringed topoi (or spaces). If I understand, this construction is right adjoint to the inclusion $LRT \to RT$. @Harry, Could you say more specifically what result you're talking about? I'm dubious, because I'd expect any ringed topos to be a relative scheme over itself... $\endgroup$ Commented Apr 4, 2011 at 8:01
  • $\begingroup$ @Andrew: I'm thinking of something like a "generalized GAGA", but I'm not really sure what conditions you need on $(X,A)$. I think something like what I'm talking about is covered in chapter 8 of that book, but I really don't know the specifics. $\endgroup$ Commented Apr 4, 2011 at 8:21

0

You must log in to answer this question.

Browse other questions tagged .