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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.

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  • $\begingroup$ 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. $\endgroup$
    – R. van Dobben de Bruyn
    Commented 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 $\binom{n+1}{2}$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented 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 $f$. $\endgroup$
    – Eric Canton
    Commented Jun 9, 2017 at 19:49

2 Answers 2

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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 $n+1$.

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

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