# Irreducible “family” of relative effective divisors of a smooth morphism

Let $\pi:X\rightarrow Y$ be a smooth proper (assume projective if needed) morphism of schemes with $Y$ locally noetherian, and let $Z\subset X$ be an irreducible integral closed subscheme containing no fiber of $\pi$.

• Is the locus $Pic_\pi(Z)=\{y\in Y:Z_y \text{ is Cartier in }X_y\}$ closed in $Y$?
• If not, what extra hypotheses would make it closed?

As $\pi$ is smooth, Cartier and Weil divisors on fibers are the same, and as it is proper, the dimension of fibers is semicontinuous, so the issue is actually about components of smaller dimension in the fibers. Thus I'd drop the hypothesis on not containing fibers and replace it by the hypotheses that $Z$ dominates $Y$; then the question would be:

• Is the locus of $y$ such that $Z_y$ has a (possibly embedded) component of codimension $\ge 2$ in $X_y$ open in $Y$?

I have the feeling that this is related to Zariski's main theorem, although in the first formulation it seems closer to asking whether the subscheme of relative effective divisors is closed in the Hilbert scheme when $\pi$ is smooth. But I can't pin it down.

(In my situation, $Z$ is actually of codimension 2, and everything is over the complex field, but I don't think this is necessary).

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Is a Picard divisor just another name for a Cartier divisor? – Jack Huizenga Jul 11 '12 at 7:02
Haha! Yes, it is its name before the morning coffee. Thanks for the correction. – quim Jul 11 '12 at 7:08

In fact both sets are constructible in $Y$.

Suppose for simplicity that $X$ is connected. Then the dimension of the (non-empty) fibers of $X\to Y$ is constant (EGA IV.12.1.1(i), and flatness is enough), denote it by $d$. Let $Y'$ be the (integral) image of $Z$ in $Y$. We can replace $X\to Y$ by $X\times_Y Y'\to Y'$ and suppose that $Z\to Y$ is surjective. Let $e$ be the dimension of the generic fiber of $Z\to Y$.

Special case: $e=d-1$. By Chevalley's theorem (EGA IV.13.1.1), for any $y\in Y$, the irreducible components of $Z_y$ all have dimension $\ge d-1$. By hypothesis, $Z_y\ne X_y$, hence $\dim Z_y\le d-1$. Therefore $\mathrm{Pic}_\pi(Z) = Y$ (and it is closed in $Y$ if $Z$ doesn't dominate $Y$).

General case. The set of $x\in Z$ such that $\dim_x Z_{\pi(x)}\le d-2$ is open in $Z$ (EGA IV.13.1.3). The complementary of the image by $\pi$ of this open subset is constructible and is your ${Pic}_{\pi}(Z)$.

In case $Z\to Y$ is flat, then $\mathrm{Pic}_\pi(Z)$ is closed by openess of $\pi|_Z$.

For the second question, if $F$ denotes the set of $x\in Z$ such that all associated components of $Z_{\pi(x)}$ passing through $x$ have dimension $\ge d-1$, then you are considering $\pi(X\setminus F)$. By EGA, IV.9.9.2(iii), $F$ is constructible, so your set is constructible.

If $Z\to Y$ is moreover flat, then $F$ is open by EGA, IV.12.1.1(i) (I learn recently this reference from an anonymous referee.), hence your set $\pi(X\setminus F)$ is in fact closed.

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Thanks a lot for taking the time to write this answer. The references were really useful, and mentioning flatness pointed me to the right direction. – quim Nov 8 '12 at 11:58
@Qing Liu: Apparently, Chevalley's semicontinuity does not say anything about the dimension of embedded components of $Z_y$. Consider the typical example of a family of twisted cubics degenerating to a nodal plane cubic with an embedded point (Hartshorne A.G. Example 9.8.4). So I have doubts now about the "Special case". Am I missing something? – quim Nov 19 '12 at 13:43
@quim: I will think about it. I probably didn't take care of embededd components. – Qing Liu Nov 20 '12 at 14:01
@Qing Liu: It seems (by the last reference in your answer or, rather equivalently, reading Kleiman's exposition on relative effective divisors in his "FGA Explained" chapter) that if $Z\rightarrow Y$ is flat, then $Pic_\pi$ is open. So using the flattening stratification would yield constructibility. I also see that your proof attempts to be more elementary and avoid flattening stratification. OTOH, these results do not require $\pi$ smooth, and I am still wondering whether $Pic_\pi$ is closed in the case that $X\rightarrow Y$ is smooth, and $Z\rightarrow Y$ is flat? – quim Nov 29 '12 at 9:57
Finally found a reference that settles the flat "special case". In Grothendieck's FGA, last éxposé (VI), Theorème 2.1 (i). A maybe simpler explanation (which avoids part of the algebra but focuses on the surface case) appears in Mumford's Lectures on Curves..., in lecture 15, VIII. But I didn't understand why embedded components of fibers are impossible until I read Grothendieck's proof. – quim Dec 11 '12 at 17:52

Take $Y=\mathbb{A}^1$ (with coordinate $t$), and $X=\mathbb{P}^2_Y$ with homogeneous coordinates $u$, $v$, $w$. Now let $Z$ be the zero scheme of $(tu, u^2, uv)$. Over any point $y$ where $t\neq0$, $X_y$ is the line $u=0$ in $\mathbb{P}^2$, while $X_0$ is defined by $u^2=uv=0$, hence has an embedded component.

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Thanks! But Z itself has an embedded component, which I don't want. That's what I ment by irreducible, but I'll clarify the question. – quim Jul 11 '12 at 8:58

This should be a counterexample.

Let $n$ be sufficiently large, and let $Y = \mathbb{G}(2,n)$ be the Grassmannian of $\mathbb{P}^2$'s in $\mathbb{P}^n$. Take $X = \{ (p,\Lambda):p\in\Lambda \}$ to be the universal $2$-plane over $Y$. Let $S\subset \mathbb{P}^n$ be a surface which contains a line $L$ (perhaps a rational surface scroll, but there are lots of things to try), and let $Z = \{(p,\Lambda): p\in S \}\subset X.$ Then $Z$ is irreducible. I'd expect that the general $2$-plane $\Lambda_0$ which contains $L$ will not intersect $S$ in any other points (say $n\geq 5$), i.e. that the corresponding fiber $Z_{\Lambda_0} \subset X_{\Lambda_0} \cong \mathbb{P}^2$ is a Cartier divisor. But for more special choices of $2$-plane containing $L$, the intersection will be $L$ together with a finite collection of points, and so the fiber will not be a divisor.

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Fascinating (but frustrating). I won't accept the answer yet, just in case someone can shed light on what extra hypotheses would make it closed. (Btw, in this example everything takes place in YxW, where W is projective space; X has codimension 3 there, and Z is the intersection of X with YxS, ie something of codimension 3 as well. My situation is analogous, but both X and the thing I intersect it with to construct Z have codimension 2). – quim Jul 11 '12 at 17:34
If we restrict this family to the set of $2$-planes which meet $S$, don't we obtain a codimension 2 counterexample? (Although with $X$ still of codimension 3 in terms of your setup) – Jack Huizenga Jul 11 '12 at 18:08
Yes, Z is codimension 2 in X, so it meets all requirements of the question. My remark is that (even restricting to the set Y of 2-planes which meet S) X has codimension 3 in YxW, YxS has codimension 3 in YxW, and Z is what YxS cuts in X, but it is only codimension 5 (or 2 inside X). There may be an irreducible hypersurface in YxW containing both X and YxS, where these meet properly, but right now I don't see it. – quim Jul 12 '12 at 5:40
Actually such a hypersurface must exist, I think. I'd like to see a "natural" one. It is not at all a criticism on the example. – quim Jul 12 '12 at 5:57
I wish one could accept two answers! Yours was perfect. Thanks again. But Qing Liu's extra information ended up being exactly what I needed. – quim Nov 8 '12 at 11:55