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 10.160.6. Let $k \subset k'$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically regular over $k$ if and only if it is geometrically regular over $k'$.

Proof. Let $k \subset L$ be a finite purely inseparable field extension. Then $L' = k' \otimes _ k L$ is a field (see material in Fields, Section 9.28) and $A \otimes _ k L = A \otimes _{k'} L'$. Hence if $A$ is geometrically regular over $k'$, then $A$ is geometrically regular over $k$.

Assume $A$ is geometrically regular over $k$. Since $k'$ is the filtered colimit of finite extensions of $k$ we may assume by Lemma 10.160.5 that $k'/k$ is finite separable. Consider the ring maps

\[ k' \to A \otimes _ k k' \to A. \]

Note that $A \otimes _ k k'$ is geometrically regular over $k'$ as a base change of $A$ to $k'$. Note that $A \otimes _ k k' \to A$ is the base change of $k' \otimes _ k k' \to k'$ by the map $k' \to A$. Since $k'/k$ is an ├ętale extension of rings, we see that $k' \otimes _ k k' \to k'$ is ├ętale (Lemma 10.141.3). Hence $A$ is geometrically regular over $k'$ by Lemma 10.160.4. $\square$


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