5
$\begingroup$

Suppose $f: \widetilde{X} \to X$ is a finite dominant morphism between connected, normal, Noetherian schemes, and that this morphism induces a dominant morphism $f_W: \widetilde{W} \to W$ between connected normal subschemes of $\widetilde{X}$ and $X$, respectively.

My question is: can we bound $\deg(f_W)$ in terms of $\deg(f)$?

I know that if $f$ is assumed to be flat, and $\widetilde{W} = \widetilde{X} \times_X W$, then the degrees are equal (see Elencwajg's answer to this question, which cites Q. Liu's "Algebraic Geometry and Arithmetic Curves", ex. 1.25 on pg 176) but I don't see how to obtain a general bound, nor what a likely counterexample would be.

$\endgroup$
3
  • $\begingroup$ It seems to me that this might be false in general, because (scheme-theoretic) fibre length is upper semi-continuous (so it can be really large on some closed subscheme). But it's pretty hard to make counterexamples, for example because of miracle flatness. $\endgroup$ Jun 9, 2017 at 19:29
  • $\begingroup$ To make examples where $\tilde W$ is not reduced, you can consider maps like $k[x^n, x^{n-1}y, \ldots, y^n] \subseteq k[x,y]$. This is a finite morphism of degree $n$ of normal domains, but the scheme-theoretic fibre above $(0,\ldots,0)$ has length $\binom{n+1}{2}$. $\endgroup$ Jun 9, 2017 at 19:47
  • $\begingroup$ I see. You raise a good point, though, about upper semicontinuity: compactness implies that fiber length admits a maximum, and so using the general bound on degree by fiber length, the degree on any subscheme is bounded by some (potentially huge) multiple of the degree of $f$. $\endgroup$ Jun 9, 2017 at 19:49

2 Answers 2

5
$\begingroup$

In your case, it is bounded. Notice that, it suffices to show that for any point $p\in X$, the cardinality of $f^{-1}(p)$ is bounded. You can replace $f$ by a separable map, since purely inseparable maps are bijection on points. Then, you can replace $\widetilde{X}$ by the Galois closure, say $Z$. Thus you are reduced to the case $g:Z\to X$, finite and Galois. Then, cardinality of $g^{-1}(p)$ is bounded by the order of the Galois group.

This fails if you want to look at scheme theoretic inverse images. In other words for a $W\subset X$, the induced map $f^{-1}(W)\to W$ (scheme theoretic inverse image) can have unbounded degree. For an example, consider the affine variety defined by $k[x^iy^j|i+j=n]$, $n$ odd and let $\mathbb{Z}/2\mathbb{Z}$ act on it by $x\mapsto -x, y\mapsto -y$. Consider the quotient map, which is of degree 2 (everything normal), but the fiber over the origin has length at least $n+1$.

$\endgroup$
2
  • 1
    $\begingroup$ And in order to bound the size of the Galois group: it's at most $(\deg f)!$. $\endgroup$ Jun 9, 2017 at 19:55
  • $\begingroup$ @R.vanDobbendeBruyn Thank you, forgot to mention that. $\endgroup$
    – Mohan
    Jun 9, 2017 at 20:01
2
$\begingroup$

If you have a degree $d$ finite dominant morphism $f : X \to Y$ between normal integral schemes, then every geometric fibre $X_y$ of $f$ has at most $d$ points. More generally, this is true if we replace the assumption that $X$ is integral by "every irreducible component of $X$ dominates $Y$" (but we keep all the other assumptions). In this form, the statement can be reduced to the case where $Y$ is strictly henselian and $y$ is the closed point. This case is obvious because $X$ splits into one piece for each $x$ lying over $y$. Now apply this to a geometric point of $Y$ lying over the generic point of an integral closed subscheme $W \subset Y$ and see what you get. Enjoy!

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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