Lemma 37.28.6 (055I)—The Stacks project

Lemma 37.28.6. Let $f : X \to Y$ be a morphism of schemes. Let $n_{X/Y}$ be the function on $Y$ counting the numbers of geometric connected components of fibres of $f$ introduced in Lemma 37.28.3. Assume $f$ of finite presentation. Then the level sets

\[ E_ n = \{ y \in Y \mid n_{X/Y}(y) = n\} \]

of $n_{X/Y}$ are locally constructible in $Y$.

Proof. Fix $n$. Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E_ n \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation whose base change to $Y$ recovers $f$. By Lemma 37.28.3 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E_ n$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E_ n \cap Z$ either contains a nonempty open subset or is not dense in $Z$. Let $\xi \in Z$ be the generic point. Then Lemma 37.28.5 shows that $n_{X/Y}$ is constant in a neighbourhood of $\xi $ in $Z$. This clearly implies what we want. $\square$

Comments (2)

Comment #784 by Kestutis Cesnavicius on July 02, 2014 at 20:25

I think it would be more accurate to say "geometric connected components" instead of "geometrically connected components" (which could be interpreted to mean "those connected components of fibers that happen to stay connected after base change to the algebraic closure"). Also, I wonder if one could prove this also for algebraic spaces (at least assuming that $Y$ is a scheme)? I think this would be helpful, e.g., for spreading out the property "geometric fibers are connected" for algebraic spaces.

Comment #800 by Johan on July 09, 2014 at 13:38

Yes, you are right and I changed it accordingly. For the moment I am not going to think about your question for algebraic spaces. Why? Just because on the one hand, there is clearly going to be some statement that is true. But one of the goals of the Stacks project is to find and prove the statement in the correct generality, which is going to be kind of awful as you can see if you read Section 68.18 on points of fibres of morphisms between algebraic spaces (which took quite a bit of work to write).

PS: Of course you are right that some condition on the base algebraic space will be needed, maybe decent will be enough, but I am not sure if it is.

There are also:

2 comment(s) on Section 37.28: Connected components of fibres

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