The Stacks project

Proof. By induction on $\dim (R)$. The case $\dim (R) = 0$ is trivial, because then $R$ is a field. Assume $\dim (R) > 0$. By (3) this implies that $\dim (S) > 0$. Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the minimal primes of $S$. Note that $\mathfrak q_ i \not\supset \mathfrak m_ R S$ since

the first equality by Lemma 10.104.3 and the inequality by (3). Thus $\mathfrak p_ i = R \cap \mathfrak q_ i$ is not equal to $\mathfrak m_ R$. Pick $x \in \mathfrak m_ R$, $x \not\in \mathfrak m_ R^2$, and $x \not\in \mathfrak p_ i$, see Lemma 10.15.2. Hence we see that $x$ is not contained in any of the minimal primes of $S$. Hence $x$ is a nonzerodivisor on $S$ by (2), see Lemma 10.104.2 and $S/xS$ is Cohen-Macaulay with $\dim (S/xS) = \dim (S) - 1$. By (1) and Lemma 10.106.3 the ring $R/xR$ is regular with $\dim (R/xR) = \dim (R) - 1$. By induction we see that $R/xR \to S/xS$ is flat. Hence we conclude by Lemma 10.99.10 and the remark following it. $\square$