
## 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 $S \otimes_k k'$ instead of $R \otimes_k k'$ in the definition.

Comment #307 by UT on

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

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

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