The Stacks project

Proof. By induction on the number of variables it suffices to prove this for $M[x] = M \otimes _ R R[x]$ over $R[x]$. Let $\mathfrak m \subset R[x]$ be a maximal ideal, and let $\mathfrak p = R \cap \mathfrak m$. Let $f_1, \ldots , f_ d$ be a $M_\mathfrak p$-regular sequence in the maximal ideal of $R_{\mathfrak p}$ of length $d = \dim (\text{Supp}(M_{\mathfrak p}))$. Note that since $R[x]$ is flat over $R$ the localization $R[x]_{\mathfrak m}$ is flat over $R_{\mathfrak p}$. Hence, by Lemma 10.68.5, the sequence $f_1, \ldots , f_ d$ is a $M[x]_{\mathfrak m}$-regular sequence of length $d$ in $R[x]_{\mathfrak m}$. The quotient

has support equal to the primes lying over $\mathfrak p$ because $R_\mathfrak p \to R[x]_\mathfrak m$ is flat and the support of $M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p}$ is equal to $\{ \mathfrak p\}$ (details omitted; hint: follows from Lemmas 10.40.4 and 10.40.5). Hence the dimension is $1$. To finish the proof it suffices to find an $f \in \mathfrak m$ which is a nonzerodivisor on $Q$. Since $\mathfrak m$ is a maximal ideal, the field extension $\kappa (\mathfrak m)/\kappa (\mathfrak p)$ is finite (Theorem 10.34.1). Hence we can find $f \in \mathfrak m$ which viewed as a polynomial in $x$ has leading coefficient not in $\mathfrak p$. Such an $f$ acts as a nonzerodivisor on

and hence acts as a nonzerodivisor on $Q$. $\square$