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 10.129.1. Let $S$ be a finite type algebra over a field $k$. Let $\varphi : k[y_1, \ldots , y_ d] \to S$ be a quasi-finite ring map. As subsets of $\mathop{\mathrm{Spec}}(S)$ we have

\[ \{ \mathfrak q \mid S_{\mathfrak q} \text{ flat over }k[y_1, \ldots , y_ d]\} = \{ \mathfrak q \mid S_{\mathfrak q} \text{ CM and }\dim _{\mathfrak q}(S/k) = d\} \]

For notation see Definition 10.124.1.

Proof. Let $\mathfrak q \subset S$ be a prime. Denote $\mathfrak p = k[y_1, \ldots , y_ d] \cap \mathfrak q$. Note that always $\dim (S_{\mathfrak q}) \leq \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$ by Lemma 10.124.4 for example. Moreover, the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite and hence $\text{trdeg}_ k(\kappa (\mathfrak p)) = \text{trdeg}_ k(\kappa (\mathfrak q))$.

Let $\mathfrak q$ be an element of the left hand side. Then Lemma 10.111.9 applies and we conclude that $S_{\mathfrak q}$ is Cohen-Macaulay and $\dim (S_{\mathfrak q}) = \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$. Combined with the equality of transcendence degrees above and Lemma 10.115.3 this implies that $\dim _{\mathfrak q}(S/k) = d$. Hence $\mathfrak q$ is an element of the right hand side.

Let $\mathfrak q$ be an element of the right hand side. By the equality of transcendence degrees above, the assumption that $\dim _{\mathfrak q}(S/k) = d$ and Lemma 10.115.3 we conclude that $\dim (S_{\mathfrak q}) = \dim (k[y_1, \ldots , y_ d]_{\mathfrak p})$. Hence Lemma 10.127.1 applies and we see that $\mathfrak q$ is an element of the left hand side. $\square$


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