The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.127.1. Let $R \to S$ be a local homomorphism of Noetherian local rings. Assume

  1. $R$ is regular,

  2. $S$ Cohen-Macaulay,

  3. $\dim (S) = \dim (R) + \dim (S/\mathfrak m_ R S)$.

Then $R \to S$ is flat.

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

\[ \dim (S/\mathfrak q_ i) = \dim (S) > \dim (S/\mathfrak m_ R S) \]

the first equality by Lemma 10.103.3 and the inequality by (3). Thus $\mathfrak p_ i = R \cap \mathfrak q_ i$ is not equal to $\mathfrak m_ R$. Pick $x \in \mathfrak m$, $x \not\in \mathfrak m^2$, and $x \not\in \mathfrak p_ i$, see Lemma 10.14.2. 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 10.103.2 and $S/xS$ is Cohen-Macaulay with $\dim (S/xS) = \dim (S) - 1$. By (1) and Lemma 10.105.3 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 10.98.10 and the remark following it. $\square$


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