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 10.160.1. Let $k$ be a field. Let $A$ be a $k$-algebra. Assume $A$ is Noetherian. The following properties of $A$ are equivalent:

  1. $k' \otimes _ k A$ is regular for every finitely generated field extension $k \subset k'$, and

  2. $k' \otimes _ k A$ is regular for every finite purely inseparable extension $k \subset k'$.

Here regular ring is as in Definition 10.109.7.

Proof. The lemma makes sense by the remarks preceding the lemma. It is clear that (1) $\Rightarrow $ (2).

Assume (2) and let $k \subset K$ be a finitely generated field extension. By Lemma 10.44.3 we can find a diagram

\[ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } \]

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is separable. By Lemma 10.152.10 there exists a smooth $k'$-algebra $B$ such that $K'$ is the fraction field of $B$. Now we can argue as follows: Step 1: $k' \otimes _ k A$ is a regular ring because we assumed (2). Step 2: $B \otimes _{k'} k' \otimes _ k A$ is a regular ring as $k' \otimes _ k A \to B \otimes _{k'} k' \otimes _ k A$ is smooth (Lemma 10.135.4) and ascent of regularity along smooth maps (Lemma 10.157.10). Step 3. $K' \otimes _{k'} k' \otimes _ k A = K' \otimes _ k A$ is a regular ring as it is a localization of a regular ring (immediate from the definition). Step 4. Finally $K \otimes _ k A$ is a regular ring by descent of regularity along the faithfully flat ring map $K \otimes _ k A \to K' \otimes _ k A$ (Lemma 10.158.4). This proves the lemma. $\square$


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