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\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 10.102.13. Let $R$ be a Noetherian ring. Let $M$ be a Cohen-Macaulay module over $R$. Then $M \otimes _ R R[x_1, \ldots , x_ n]$ is a Cohen-Macaulay module over $R[x_1, \ldots , x_ n]$.

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 10.67.5, 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

\[ Q = M[x]_{\mathfrak m}/(f_1, \ldots , f_ d)M[x]_{\mathfrak m} = M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p} \otimes _{R_\mathfrak p} R[x]_{\mathfrak m} \]

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 10.39.4 and 10.39.5). 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 $\kappa (\mathfrak p) \subset \kappa (\mathfrak m)$ is finite (Theorem 10.33.1). 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

\[ M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p} \otimes _ R R[x] = \bigoplus \nolimits _{n \geq 0} M_{\mathfrak p}/(f_1, \ldots , f_ d)M_{\mathfrak p} \cdot x^ n \]

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


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