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 10.46.3. Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent

  1. for every field extension $k \subset k'$ the spectrum of $R \otimes _ k k'$ is irreducible, and

  2. for every finite separable field extension $k \subset k'$ the spectrum of $R \otimes _ k k'$ is irreducible.

Proof. Let $k \subset k^{perf}$ be a perfect closure of $k$, see Definition 10.44.5. By Lemma 10.45.7 we may replace $R$ by $(R \otimes _ k k^{perf})_{reduction}$ and $k$ by $k^{perf}$ (some details omitted). Hence we may assume that $R$ is geometrically reduced over $k$.

Assume $R$ is geometrically reduced over $k$. For any extension of fields $k \subset k'$ we see irreducibility of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ being a domain. Assume (2). Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Using Lemma 10.42.4 we see that (2) is equivalent to $R \otimes _ k \overline{k}$ being a domain. For any field extension $k \subset k'$, there exists a field extension $\overline{k} \subset \overline{k}'$ with $k' \subset \overline{k}'$. By Lemma 10.46.2 we see that $R \otimes _ k \overline{k}'$ is a domain. If $R \otimes _ k k'$ is not a domain, then also $R \otimes _ k \overline{k}'$ is not a domain, contradiction. $\square$

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