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?



Condition such that the fibres of a polynomial map $p :\mathbb{C}^n\rightarrow \mathbb{C}^n$ are finite
$\def\CC{\mathbb{C}}$User "anon" points out to me that this is Proposition 8.28 in [Milne's notes](https://www.jmilne.org/math/CourseNotes/AG.pdf); 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](https://en.wikipedia.org/wiki/Local_criterion_for_flatness#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](https://www.jmilne.org/math/CourseNotes/AG.pdf) 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](http://math.stanford.edu/%7Evakil/216blog/FOAGnov1817public.pdf) 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](http://www.math.lsa.umich.edu/%7Espeyer/631/Update.pdf).


