I was told that if $A$ is the subring of $\mathbb{C}[x_1,\ldots, x_n]$ generated by the polynomials $p_1(x_1,\ldots, x_n),\ldots, p_1(x_1,\ldots, x_n)$, then the preimage $p^{1}(c)$ via the map $p = (p_1,\ldots, p_n):\mathbb{C}^n\rightarrow \mathbb{C}^n$ is finite for all $c\in \mathbb{C}^n$ if the ring $\mathbb{C}[x_1,\ldots, x_n]$ is integral and flat over the subring $A$. Does anyone know where I can find a reference (preferably a textbook) that has this or an equivalent statement?
1 Answer
$\def\CC{\mathbb{C}}$User "anon" points out to me that this is Proposition 8.28 in Milne's notes; see also Example 8.36 for a quasifinite map $\CC^2 \to \CC^2$ which is not finite. The rest of my answer is probably not as useful now that there is a good reference, but I'll leave it.
Here is the right statement:
Theorem Let $X$ and $Y$ be affine varieties over an algebraically closed field $k$, with corresponding coordinate rings $A$ and $B$. Let $\pi : Y \to X$ be a map and let $\pi^{\ast} : A \to B$ be the corresponding map of rings. If $B$ is a finitely generated as an $A$module, then $\pi^{1}(x)$ is finite for all $x \in X$.
In your case, $X = Y = \CC^n$ and $A=B = \CC[x_1, \ldots, x_n]$. You phrase your hypothesis as $B$ is integral over $\pi^{\ast} A$ but, since $B$ is a finitely generated $\CC$algebra, that is the same as asking that $B$ is finitely generated as an $A$module. Also, the adjective "flat" isn't needed, and actually follows from finiteness in your case by the Miracle Flatness Theorem.
Proof Let $x \in X$ and let $\mathfrak{m}_x$ be the corresponding maximal ideal of $A$. Then $\pi^{1}(x)$ corresponds to the radical ideal $\sqrt{B \pi(\mathfrak{m}_x)}$ so we want to show that $B/\sqrt{B \pi(\mathfrak{m}_x)}$ is a finite dimensional $k$algebra. It is enough to show that $\dim_k B/B\pi(\mathfrak{m}_x)$ is finite, since $B/\sqrt{B \pi(\mathfrak{m}_x)}$ is a quotient of $B/B\pi(\mathfrak{m}_x)$. But $B/B\pi(\mathfrak{m}_x) \cong B \otimes_A A/\mathfrak{m}_x$. Since $B$ is a finite $A$module, $B \otimes_A A/\mathfrak{m}_x$ must be a finite $A/\mathfrak{m}_x$ module, and $A/\mathfrak{m}_x$ is just $k$. $\square$.
Here is where this can be found in some other books: Shaverevich introduces finite maps in Section I.5.3, but doesn't show that they have finite fibers until Section II.6.3 (Theorem 3) and then only under the hypothesis that $X$ is normal. Milne introduces the words "finite" (meaning $B$ is finitely generated as an $A$module) and "quasifinite" (meaning the fibers of $\pi$ are finite) in Definition 2.39, and proves the result we want as Proposition 8.28; as noted above. In Hartshorne, this is Exercise 3.5.(a) in Section II.3. Vakil makes this Important Exercise 7.3.K. When I taught Algebraic Geometry, I got to this one month in, see the notes for October 8.

$\begingroup$ Well, there is EGA II.6.1.7, but this is probably not at the level you want to stay... $\endgroup$– abxCommented Jul 15, 2020 at 16:55

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$\begingroup$ @anon Thanks! That's probably the best choice. I should have known to look harder when it appeared that Milne omitted a basic fact. Also nice to note that Milne's Example 8.36 gives a quasifinite map $\mathbb{C}^2 \to \mathbb{C}^2$ which is not finite. $\endgroup$ Commented Jul 16, 2020 at 8:28

$\begingroup$ Thank you for the references. Your notes give me the exact statement I was looking for. $\endgroup$ Commented Jul 17, 2020 at 2:10