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Let $K$ be a field. Let $V/K$ be an affine variety in $A^m$. Let f be a polynomial map (and hence a "morphism of finite type") $f:V\to A^n$. A theorem of Chevalley's tells us that im(f) is either a variety or "almost" a variety - that is, im(f) is a variety $W$ with perhaps a few varieties of lower dimension cut out from it.

Question: is the degree of $W$ (= Zariski closure of im(f)) bounded solely in terms of m, n, deg(V) and the degree of the polynomials $f_1,\dots ,f_n$ defining f (i.e. $f(\vec{x}) = (f_1(\vec{x}),...,f_n(\vec{x}))$?).

This seems intuitively obvious, but I do not know where to look for a reference. (It's also non-obvious how to adapt the proof of Chevalley's theorem I'm looking at so as to give this.) Does anybody where to look this up and/or how to prove this?

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2 Answers 2

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Call $r$ the dimension of $W$, $d$ the degree of $V$, and $e$ the largest degree of one of the $f_i$. Then I claim that $\deg W$ is at most $de^r$.

We may assume that $K$ is algebraically closed. We may also assume that the dimensions of $V$ and $W$ are the same (otherwise, cut $V$ with generic hyperplane sections until they become equal). Then the morphism $V \to W$ generically finite and non-empty fibers.

Now, let $L \subseteq \mathbb A^n$ be a generic linear subspace of codimension $r$ of $W$. The cardinality of the intersection $L \cap W$ is $\deg W$, and this is bounded above by the cardinality of $f^{-1}L$, which is finite. But $f^{-1}L$ is the intersection of $V$ with the zero loci of $r$ generic linear combinations of the $f_i$. Now use the following fact: if $X$ is a (not necessarily equidimensional) closed subset of an affine space, call $\deg X$ its degree (that is, the sum of the degrees of its irreducible components), and $H$ is a hypersurface of degree $e$, then $\deg(X\cap H) \leq e\deg X$. This reduces easily to the case that $X$ is irreducible, which is standard. This implies that the degree of $f^{-1}L$, that is, its cardinality, is at most $de^r$, which gives us what we want.

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  • $\begingroup$ I like this argument (and have upvoted it), but why should the morphism $V \to W$ have nonempty fibers, given that $W$ is, by definition, the closure of the image of $V$? (I imagine this can be repaired, but I think it needs to be pointed out.) $\endgroup$ Apr 29, 2011 at 22:12
  • $\begingroup$ Hi Charles: I guess he meant $generically$ non-empty fibers. $\endgroup$
    – pinaki
    Apr 30, 2011 at 0:10
  • $\begingroup$ Yes, instead of "generically finite and non-empty fibers", I should have written "finite and non-empty general fibers". $\endgroup$
    – Angelo
    Apr 30, 2011 at 5:36
  • $\begingroup$ Great - thanks! This feels like it should be a standard argument - does anybody have a reference? $\endgroup$ May 2, 2011 at 18:40
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[This was a bit long for a comment.]

Let $I \subset K[x_1, \dotsc, x_m]$, $J \subset K[y_1, \dotsc, y_n]$ be the ideals defining $V, W$, respectively. Then $J$ is precisely the kernel of the map $$K[y_1, \dotsc, y_n] \to K[x_1, \dotsc, x_m] / I$$ given by sending $$y_i \mapsto f_i.$$ If we have explicit generators for $I$, then there is a Groebner-basis algorithm for finding explicit generators for this kernel (see, e.g., Cox, Little, and O'Shea, Ideals, Varieties, and Algorithms). I would be very surprised if a careful analysis of this algorithm would not provide explicit bounds on the number and degrees of the generators it gives for $J$, and hence on the degree of the variety $V(J) = W$. Of course, these bounds would be in terms of the number and degrees of the generators of $I$ (as well as the $f_i$), so you would need some way of bounding these from the degree of $V$.

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