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If $f \in k[x_1,...,x_n]$ is irreducible then show there is a finite flat morphism $k[x_1,...,x_{n-1}] \to k[x_1,...,x_n]/(f)$ (i.e. $k[x_1,...,x_{n}]/(f)$ is finitely generated and flat as a module over the image of this morphism).

I know that there exists a finite morphism by Noether normalization, but why is it flat?

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Actually, you have to show that $k[x_1, \ldots, x_n] / (f)$ is finitely-generated as a module over the image. – Zhen Lin Dec 13 '12 at 13:55
Thank you, I corrected it. – Bernard Dec 13 '12 at 15:02
up vote 4 down vote accepted

The trick (which I failed to spot when I had to do this question in my exams last year) is to recall that free modules are flat.

Let $A = k [x_1, \ldots x_{n-1}]$ and $B = k [x_1, \ldots, x_n] / (f (x_1, \ldots, x_n))$. By the preparation lemma and a change of variables, we may assume $f (x_1, \ldots, x_n)$ is of the form $$f(x_1, \ldots, x_n) = a_d {x_n}^{d} + \sum_{i=0}^{d-1} a_i(x_1, \ldots, x_{n-1}) {x_n}^i$$ where $a_d$ is in $k$ and $a_i(x_1, \ldots, x_{n-1})$ is in $A$. It is then clear that an element of $B$ can always be written as $$\sum_{i=0}^{d-1} c_i (x_1, \ldots, x_{n-1}) {x_n}^i \pmod{f (x_1, \ldots, x_n)}$$ for some $c_i (x_1, \ldots x_{n-1})$ in $A$; thus, there is a surjective $A$-module homomorphism $p : A^{\oplus d} \to B$, and in particular $B$ is a finite $A$-module. Consider $\ker p$: suppose for $0 \le i < d$ we have elements $c_i (x_1, \ldots x_{n-1})$ of $A$, such that $$\sum_{i=0}^{d-1} c_i (x_1, \ldots, x_{n-1}) {x_n}^i = g(x_1, \ldots, x_n) f(x_1, \ldots, x_n)$$ for some $g(x_1, \ldots, x_n)$ in $k [x_1, \ldots, x_n]$. By considering the coefficient of the highest power of $x_n$ in $g (x_1, \ldots, x_n)$, we can show that $g = 0$. So in fact $\ker p = 0$, so $A^{\oplus d} \cong B$; in particular $B$ is a flat $A$-module.

Note that we didn't need to assume that $f (x_1, \ldots, x_n)$ is irreducible.

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Ok, that makes perfect sense. Thanks! – Bernard Dec 13 '12 at 15:03
@ZhenLin This is a step in the proof that any hypersurface in $\mathbb A_K^n$ is Cohen-Macaulay. Your argument can be shortened at some moment by using the following elementary remark: if $R$ is a commutative ring and $f\in R[X]$ is a monic polynomial, then $R[X]/(f)$ is a free $R$-module of rank $=\deg f$. – user26857 Dec 14 '12 at 20:08

I have another solution which uses the following characterization of flatness via Cohen-Macaulayness.

Proposition. Let $R$ be a finitely generated algebra over a field and $S\subset R$ a Noether normalization. Then the following are equivalent:

(i) $R$ is $S$-flat,

(ii) $R$ is a free $S$-module,

(iii) $R$ is Cohen-Macaulay.

Proof. (i) $\Rightarrow$ (ii) Assume that $R$ is $S$-flat. Since over an integral domain every finitely generated flat module is projective, then $R$ is a projective $S$-module and by Quillen-Suslin theorem $R$ is a free $S$-module.

(ii) $\Rightarrow$ (iii) We want to prove that $R_P$ is a local Cohen-Macaulay ring for every $P\subset R$ prime ideal. $p=P\cap S$ is a prime ideal of $S$ and $\dim R_P=\dim S_p$. Since $S$ is Cohen-Macaulay we have $\dim S_p=\operatorname{depth}S_p$. As $R_p$ is a free $S_p$-module we get $\operatorname{depth}S_p=\operatorname{depth}_{S_p}R_p$. On the other side, $R_P$ is a localization of $R_p$, and this shows that $\operatorname{depth}_{S_p}R_p\le\operatorname{depth}R_P$ and we are done.

(iii) $\Rightarrow$ (i) Let $M\subset R$ be a maximal ideal and $m=M\cap S$. We want to prove that $R_M$ is $S_m$-flat. Since $S\subset R$ is a finite extension, then $m\subset S$ is also maximal. Furthermore, $S_m$ is regular and let $\underline{x}\subset S$ be a sequence that forms a (regular) system of parameters in $S_m$. Since $S/\underline xS\to R/\underline xR$ is a finite morphism, and $(S/\underline xS)_m$ is artinian, it follows that $(R/\underline xR)_m$ is artinian, too. But $R_M/\underline x R_M$ is a localization of the latter ring, so $R_M/\underline x R_M$ is also artinian. This shows that $\underline x$ is a system of parameters in $R_M$. Since $R_M$ is Cohen-Macaulay it follows that the sequence $\underline x$ is $R_M$-regular and thus we get $\operatorname{depth}R_M\ge\dim S_m$. As $R$ is Cohen-Macaulay we have $\operatorname{depth}R_M=\dim R_M$ and the finiteness of the extension $S\subset R$ gives $\dim R_M\le\dim S_m$. Finally we obtain $\dim R_M=\dim S_m$ and this is enough in order to prove that $R_M$ is $S_m$-flat.

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Have someone a reference for the criterion proved in my answer? (I knew such a result for finitely generated graded algebras over fields, but the proof is slightly different.) I'm curious to see (if there exist) other arguments. – user26857 Dec 14 '12 at 11:50
There is a local version of this in Eisenbud section 18.4. – Andrew Dec 14 '12 at 17:31
@Andrew Thanks. I was aware of this, but my proof is inspired by Bruns and Herzog, Proposition 2.2.11. However, for the last claim (in the last sentence) I've used Eisenbud, Theorem 18.16. – user26857 Dec 14 '12 at 17:59

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