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 10.42.6. Let $k$ be a field. Let $S$ be a reduced $k$-algebra. Let $k \subset K$ be either a separable field extension, or a separably generated field extension. Then $K \otimes _ k S$ is reduced.

Proof. Assume $k \subset K$ is separable. By Lemma 10.42.4 we may assume that $S$ is of finite type over $k$ and $K$ is finitely generated over $k$. Then $S$ embeds into a finite product of fields, namely its total ring of fractions (see Lemmas 10.24.1 and 10.24.4). Hence we may actually assume that $S$ is a domain. We choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.41.3. Let $P \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $x_{r + 1}$. It is a separable polynomial. It is easy to see that $k[x_1, \ldots , x_ r] \otimes _ k S = S[x_1, \ldots , x_ r]$ is a domain. This implies $k(x_1, \ldots , x_ r) \otimes _ k S$ is a domain as it is a localization of $S[x_1, \ldots , x_ r]$. The ring extension $k(x_1, \ldots , x_ r) \otimes _ k S \subset K \otimes _ k S$ is generated by a single element $x_{r + 1}$ with a single equation, namely $P$. Hence $K \otimes _ k S$ embeds into $F[T]/(P)$ where $F$ is the fraction field of $k(x_1, \ldots , x_ r) \otimes _ k S$. Since $P$ is separable this is a finite product of fields and we win.

At this point we do not yet know that a separably generated field extension is separable, so we have to prove the lemma in this case also. To do this suppose that $\{ x_ i\} _{i \in I}$ is a separating transcendence basis for $K$ over $k$. For any finite set of elements $\lambda _ j \in K$ there exists a finite subset $T \subset I$ such that $k(\{ x_ i\} _{i\in T}) \subset k(\{ x_ i\} _{i \in T} \cup \{ \lambda _ j\} )$ is finite separable. Hence we see that $K$ is a directed colimit of finitely generated and separably generated extensions of $k$. Thus the argument of the preceding paragraph applies to this case as well. $\square$


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