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?
Best Answer
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.