Degree of the preimage of a variety

Question

Let $V\subset \mathbb{A}^m$ and $W\subset \mathbb{A}^n$ be affine varieties defined over an arbitrary field. Let $f:V\to \mathbb{A}^n$ be a morphism given by polynomials of degree $\leq D$.

Is it true that $$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim(\overline{f(V)})}?$$

Here, by "degree", we mean "sum of the degrees of the irreducible components".

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Here is what I have:

• a proof of the above when $\dim(W) = 0$ (call this Lemma 1),
• a proof of the simple bound $\deg(\overline{f(V)})\leq \deg(V) D^{\dim(\overline{f(V)})}$ (call this Lemma 2),
• a proof of the weaker bound $\leq \deg(V) \deg(W) D^{\dim V}$ (UPDATE - I do use this (call it Lemma 3), and prove it, in my self-answer. The proof uses Lemma 2.)

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Is there a short, clean actual proof? UPDATE: I claim there is one, and in fact I can prove a stronger, tight bound; see self-answer. I use Lemma 2 and 3 above, but not Lemma 1. Lemma 2 was pointed out to me here (Degree of image of a polynomial map) a long time ago.

PS. Talk of "wonky steps" in the comments refers to a previous attempt at a proof by me (with steps clearly marked as "wonky" by me). The current proof would seem to be correct.

PPS (added a few weeks later). I was recently talking to two algebraic geometers while on semi-vacation. I realize now that there's a potential ambiguity that may make the lack of additional technical conditions in the above rub people the wrong way. When I write $f^{-1}(W)$, I mean the set-theoretic ("geometric"?) preimage, that is, the variety consisting of the points $x$ such that $f(x)$ lies in $W$. It is not the same as the scheme-theoretic or ideal-theoretic preimage, for which the statement would not be true in general without added conditions.

Indeed, Fulton-Macpherson's generalized Bézout theorem (which is the howitzer in the proof) would not be true without extra conditions if the intersection of two varieties $V$ and $W$ were interpreted in the ideal-theoretic sense.

Fortunately the set-theoretic ("geometric"?) preimage (that is, the reduced subscheme of the scheme-theoretic preimage) is what I actually need.

I feel a tad wiser now, and would like to explain the above, but I'm not sure of the format to use; it is not a question. Should I give another self-answer (crediting the two algebraic geometers)?

Answer 1

[I'm editing my original answer slightly so as to work in affine space, for the sake of clarity, since, in the proof of Lemma 3 in the comments, I am really working in affine space. If one needs degree bounds of the kind I'm giving in projective space (not that I do), one can just deduce them from the bounds in affine space, by choosing a copy of $\mathbb{A}^n$ in $\mathbb{P}^n$ such that the complement of $\mathbb{A}^n$ does not contain any of the varieties we are working with. Of course one can do exactly that to prove a projective version of Lemma 3 using the affine version whose proof is in the comments.]

Let $L\subset \mathbb{A}^m$ be a hyperplane of codimension $\dim f^{-1}(W)$ intersecting $f^{-1}(W)$ at $\deg f^{-1}(W)$ points. We restrict to $L$, and have thus reduced matters to the case where $f^{-1}(W)$ is zero-dimensional. (We have exchanged $V$ for $V\cap L$, which of course has degree $\leq \deg(V)$; we haven't touched $W$.)

We can assume now that $V$ (our new $V$) is irreducible, without loss of generality. Then Theorem 2 in section 1.8 of Mumford's Red Book shows that, because $f^{-1}(W)$ is zero-dimensional, we must have $\dim \overline{f(V)} = \dim V$. But then we have our result by Lemma 3 above: $$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim V} = \deg(V) \deg(W) D^{\dim \overline{f(V)}}\;\;\;\;\;\;\;\;\;\;\;\;\textrm{QED}.$$

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In fact, since we can take $L$ to be generic, we can assume that $V\cap L$ (our new $V$) has dimension $\dim V - \textrm{codim}(L) = \dim V - \dim f^{-1}(W)$. By the same result in Mumford, $\dim f^{-1}(W) \geq \dim(\overline{f(V)\cap W}) + \dim V - \dim \overline{f(V)}$, and so we have really proved that $$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim V - \dim f^{-1}(V)} = \deg(V) \deg(W) D^{\dim(\overline{f(V)}) - \dim(\overline{f(V)\cap W})}.$$ (EDIT: I earlier wrote $\overline{f(V)}\cap W$ here. With the argument as written above, one must work with $\overline{f(V)\cap W}$, as the result in Mumford requires dominance.

It seems to me that one can indeed get $\overline{f(V)}\cap W$ by being a little more roundabout: prove a projective version of Lemma 3 using the affine version of Lemma 3 (choosing a copy $U$ of $\mathbb{A}^n$ in $\mathbb{P}^n$ such that $\mathbb{P}^n\setminus U$ does not contain any components of $f^{-1}(W)$, where $W$ is a projective variety) and then work projectively. This is a bit of a detour.)

At least if $\overline{f(V)} = \mathbb{A}^n$, this new bound is tight in general: if $W$ is given as a complete intersection of hypersurfaces $g_i(\vec{y})=0$, $1\leq i\leq r$ of degree $D_i$ with $\prod_i D_i = \deg(W)$, then $f^{-1}(W)$ is the intersection of $V$ with the hypersurfaces $(g_i\circ f)(\vec{x})=0$ of degree $D_i D$, and so the degree of $f^{-1}(W)$ should be $$\deg(V) \prod_{i=1}^r D_i D = \deg(V) \prod_{i=1}^r D_i \cdot D^{\textrm{codim}(W)} = \deg(V) \deg(W)D^{\dim(\overline{f(V)}) - \dim(\overline{f(V)}\cap W)}.$$