TITLE: Show there exists a finite flat morphism
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.