Suppose $f\colon X \to Y $ is a morphism of schemes. We can define a function on the topological space $Y$ by sending $y\in Y$ to the dimension of the fiber of $f$ over $y$.

When is this function upper semi-continuous?

I have the following "concrete" application in mind. If an algebraic group $G$ acts on a scheme $X$, I'm pretty sure the stabilizer dimension is an upper semi-continuous function on $X$ (i.e. it can jump up on closed sub-schemes), but I don't know a proof. The stabilizers of points are the fibers of the map $\text{Stab}\to X$ in the following Cartesian square: \begin{equation} \require{AMScd} \begin{CD} \text{Stab} @>>> G \times X \\ @VVV @VV{\alpha}V \\ X @>{\Delta}>> X \times X. \end{CD} \end{equation} where $\alpha\colon G\times X\to X\times X $ is given by $(g,x) \mapsto (g\cdot x,x)$, and $\Delta\colon X\to X\times X $ is the diagonal map $x\mapsto (x,x)$. It would be nice to have a condition satisfied by $\alpha\colon G\times X \to X\times X$ that would guarantee the upper semi-continuity of fiber dimension.

  • 3
    $\begingroup$ @Transcendental: Thanks for the suggested edit. I've updated the formatting. This question was asked before MathOverflow had any way to display math! $\endgroup$ Commented Oct 21, 2014 at 21:46
  • 2
    $\begingroup$ You’re most welcome, and thank you for your understanding. This question is very old indeed, and I thought that the best form of respect that any member of the Math Overflow community could show toward it was simply to update its formatting so that everybody can enjoy reading it. It therefore seems rather amusing that some people have chosen to make my initial edit the basis for some kind of editing tug-of-war, with nostalgic claims in support of archaic mathematical typesetting, which, I’m pretty sure you would agree, is the reason why we have MathJax today! $\endgroup$ Commented Oct 21, 2014 at 23:59

2 Answers 2


Theorem (EGA IV 13.1.3): Let $f \colon X \to Y$ be a morphism of schemes, locally of finite type. Then $$x \mapsto \dim_x(X_{f(x)})$$ is upper semi-continuous.

Corollary (Chevalley's upper semi-continuous theorem, EGA IV 13.1.5): Let $f \colon X \to Y$ be proper, then: $$y \mapsto \dim(X_y)$$ is upper semi-continuous.

Corollary (SGA3, ??): Let $X/Y$ be a group scheme, locally of finite type. Then $$y \mapsto \dim(X_y)$$ is upper semi-continuous.

Proof: The dimension of a group scheme over a field is the same as the dimension at the identity. Thus the function $$y \mapsto \dim(X_y)$$ is the composition of the continuous function $y \to e(y)$ and the upper semi-continuous function $x \mapsto \dim_x(X_{f(x)})$.

Concerning your application: The fiber dimensions of the stabilizer group scheme Stab/X is upper semi-continuous, but the "diagonal" $G \times X \to X \times X$ does not always have this property (unless it is proper, i.e., "$G$ acts properly").

  • 2
    $\begingroup$ Oh, I see; you get semi-continuity on X much more generally than semi-continuity on Y. I got a little confused in the last Corollary, so I'll add a comment to clarify to my future self. X/S is a group scheme with an S-morphism f:X-->Y. The map y-->e(y) is the composition of the structure morphism Y-->S and the identity S-->X. To get the application from this Corollary, take X=Stab and Y=X. $\endgroup$ Commented Oct 8, 2009 at 17:53
  • $\begingroup$ Well, there was a typo: It should be X/Y and not X/S in the last Corollary (now corrected). As stated before, it was wrong. The reason is that the dimension of the fiber of X->Y at the identity cannot be related to the dimension at other points unless X->Y is equivariant (which is the case if Y=S) so that the fiber acts transitively on itself. $\endgroup$
    – David Rydh
    Commented Oct 9, 2009 at 4:57
  • 5
    $\begingroup$ Stupid example: Let $Y = {\mathbb A}^1$, and $X$ the disjoint union of $(Y\setminus 0) \times {\mathbb P}^1$ and $0$. Then the evident map $X\to Y$ violates Chevalley's theorem, even though it has projective fibers (but is not actually projective). $\endgroup$ Commented Mar 14, 2013 at 0:35
  • $\begingroup$ if x and y are analytic irreducible varieties and the map is proper, Does the Chevalley theorem still holds? $\endgroup$
    – user42804
    Commented Sep 27, 2015 at 14:08

I just discovered that Shavarevich (second edition) has a wrong answer to this question. In Section I.6.3, after Theorem 7 (which is correct), he gives the following Corollary. This quotation combines the Theorem and the Corollary.

Let $f: X \to Y$ be a regular map between irreducible varieties. Suppose that $f$ is surjective ... The sets $Y_k := \{ y \in Y: \dim f^{-1}(y) \geq k \}$ are closed in $Y$.

Note that this differs from the true EGA IV 13.1.5 by replacing "closed" with "surjective". I figured I'd record a counterexample here, which is slightly more public than just creating a handout for my class.

Our map is a composition $X \subset \mathbb{A}^3 \to \mathbb{A}^3 \to Y \subset \mathbb{A}^4$. We'll call the two $\mathbb{A}^3$'s $A$ and $B$ respectively.

$X$ is the quasi-affine variety $A \setminus \{ (0,\ast,0) \}$. We map $A \to B$ by $(x,y,z) \mapsto (x, xy, z)$. We map the $B$ to $\mathbb{A}^4$ by $(p,q,r) \mapsto (p(p-1), p^2(p-1), q,r)$. $Y$ is the affine variety $\{ (a,b,c,d) : a^3 = b(b-a) \}$. In other words, $Y$ is the product of a nodal cubic with $\mathbb{A}^2$.

To see surjectivity, note that $X$ hits every point of $B$ where $p$ is nonzero. The points of $B$ where $p \neq 0$ map to the points of $Y$ where $(a,b) \neq (0,0)$. The points $(0,0,c,d)$ in $Y$ are the images of $(1,c,d) \in B$, which are in turn the images of $(1,c,d)$ in $X$.

Now, let's look at $\dim f^{-1}(0,0,0,r)$. When $r \neq 0$, this is the union of $(1,0,r)$ and $(0, \ast, r)$, so one dimensional. When $r = 0$, the line $(0, \ast, 0)$ is deleted, so the preimage is only a point.

This suggests that something nicer may happen if we insist that fibers are irreducible, or that $Y$ is normal (perhaps Zariski's Main Theorem gets involved?) but I don't have a proposed statement to make.

  • $\begingroup$ Here's another similar example. Let $Y=k[x,y,z]$ and let $Y'=k[x,y,z,w]/w^2-z$. Then let $W$ be the blow-up of $Y'$ along $x=0,w=1$. Finally let $X=W\setminus \{x=0,y=0,w=1\}$. Then $X$ and $Y$ are regular varieties and the morphism $X\to Y$ has projective non-empty fibers. The inverse image of $x=0,z=1$ is the disjoint union of $\mathbb{A}^1$ ($w=-1$) and $(\mathbb{A}^1\setminus \{y=0\})\times \mathbb{P}^1$. The fiber dimension is $0$ outside $\{x=0,z=1\}$ and $1$ over $\{x=0,z=1\}$ except at the point $\{x=0,y=0,z=1\}$ where it is $0$ again. $\endgroup$
    – David Rydh
    Commented Sep 28, 2015 at 17:28
  • 1
    $\begingroup$ Theorem. Let $f\colon X\to Y$ be a separated morphism of finite type between noetherian schemes such that (1) $X$ is irreducible, (2) $Y$ is irreducible and normal, (3) $f$ has proper non-empty fibers, and (4) the generic fiber of $f$ is geometrically connected. Then $f$ is proper (and so satisfies upper-semicontinuity of fiber dimension). $\endgroup$
    – David Rydh
    Commented Sep 28, 2015 at 17:42
  • 1
    $\begingroup$ Proof. There exists a generalized Stein factorization $X\to X'\to Y$ (Thm 2.1, Hall-Rydh, The Hilbert stack, Adv. Math. 253, 2014) where $X\to X'$ is proper, surjective, with geom. connected fibers and $X'\to Y$ is quasi-finite, separated and birational (or K(X')/K(Y) purely inseparable in positive characteristic). The latter is thus an open immersion by ZMT, hence an isomorphism (or finite purely inseparable in pos. char.). QED. $\endgroup$
    – David Rydh
    Commented Sep 28, 2015 at 17:46
  • 2
    $\begingroup$ Another example: Let $Y=\mathrm{Spec}\; k[x,y,z]$, $Y\times \mathbb{A}^1=\mathrm{Spec}\; k[x,y,z,w]$. Let $W$ be the blow-up of $Y\times \mathbb{A}^1$ in $\{x=y=w=0\}$ and let $X=W\setminus \{x=y=z=w=0\}$. Then $f\colon X\to Y$ is a morphism between irreducible regular varieties, which has geometrically connected non-empty non-proper fibers but does not satisfy upper semi-continuity of fiber dimension (general fiber has dimension $1$, fibers over $x=y=0$ has dimension $2$ except $x=y=z=0$ which has $1$ again). $\endgroup$
    – David Rydh
    Commented Sep 28, 2015 at 18:16
  • 2
    $\begingroup$ Dear David, I have taken the liberty to replace your word "projective" by the word "closed", which is the one actually used in the true EGA IV 13.1.5. $\endgroup$ Commented Dec 27, 2020 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.