Definition 10.49.1. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is *geometrically integral over $k$* if for every field extension $k'/k$ the ring of $S \otimes _ k k'$ is a domain.

## 10.49 Geometrically integral algebras

Here is the definition.

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

Lemma 10.49.2. 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 10.49.3. Let $k$ be a field. Let $S$ be a $k$-algebra. The following are equivalent

$S$ is geometrically integral over $k$,

for every finite extension $k'/k$ of fields the ring $S \otimes _ k k'$ is a domain,

$S \otimes _ k \overline{k}$ is a domain where $\overline{k}$ is the algebraic closure of $k$.

**Proof.**
Follows from Lemmas 10.49.2, 10.44.3, and 10.47.3.
$\square$

Lemma 10.49.4. 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 10.43.5 the ring $R \otimes _ k S$ is reduced and by Lemma 10.47.7 the ring $R \otimes _ k S$ is irreducible (the spectrum has just one irreducible component), so $R \otimes _ k S$ is an integral domain.
$\square$

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