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*}

10.48 Geometrically integral algebras

Definition 10.48.1. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is geometrically integral over $k$ if for every field extension $k \subset k'$ the ring of $S \otimes _ k k'$ is a domain.

Any question about geometrically integral algebras can be translated in a question about geometrically reduced and irreducible algebras.

Lemma 10.48.2. Let $k$ be a field. Let $S$ be a $k$-algebra. In this case $S$ is geometrically integral over $k$ if and only if $S$ is geometrically irreducible as well as geometrically reduced over $k$.

Proof. Omitted. $\square$

Lemma 10.48.3. Let $k$ be a field. Let $S$ be a geometrically integral $k$-algebra. Let $R$ be a $k$-algebra and an integral domain. Then $R \otimes _ k S$ is an integral domain.

Proof. By Lemma 10.42.5 the ring $R \otimes _ k S$ is reduced and by Lemma 10.46.7 the ring $R \otimes _ k S$ is irreducible (the spectrum has just one irreducible component), so $R \otimes _ k S$ is an integral domain. $\square$


Comments (3)

Comment #306 by UT on

I think it should be instead of in the definition.

Comment #307 by UT on

Perhaps it is useful to add the following lemma (similar to 034N and 037O): Let be a field. Let be a geometrically integral -algebra. Let be any integral -algebra. Then is integral.

Proof: By 034N is reduced and by 037O is irreducible (it has just one irreducible component), so is integral.


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