# Condition such that the fibres of a polynomial map $p :\mathbb{C}^n\rightarrow \mathbb{C}^n$ are finite

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?

• Is this not essentially Noether normalization? – Francesco Polizzi Jul 15 at 13:05

$$\def\CC{\mathbb{C}}$$User "anon" points out to me that this is Proposition 8.28 in Milne's notes; see also Example 8.36 for a quasi-finite 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 "quasi-finite" (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.

• Well, there is EGA II.6.1.7, but this is probably not at the level you want to stay... – abx Jul 15 at 16:55
• Milne's notes 8.28? – anon Jul 16 at 4:05
• @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 quasi-finite map $\mathbb{C}^2 \to \mathbb{C}^2$ which is not finite. – David E Speyer Jul 16 at 8:28
• Thank you for the references. Your notes give me the exact statement I was looking for. – Vishnu Mangalath Jul 17 at 2:10