Section 10.49 (05DW): Geometrically integral algebras

10.49 Geometrically integral algebras

Here is the definition.

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.

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, 10.44.4, 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. $\square$

Comments (6)

Comment #306 by UT on September 08, 2013 at 17:54

I think it should be $S \otimes_k k'$ instead of $R \otimes_k k'$ in the definition.

Comment #307 by UT on September 08, 2013 at 18:01

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.

Comment #308 by Johan on September 12, 2013 at 19:22

OK, here is the commit. Thanks!

Comment #4656 by Hao on November 11, 2019 at 09:40

A possibly helpful equivalent definition: If $A$ is a $k$ algebra, then $A$ is geometrically integral iff $A\otimes_k \overline{k}$ is a domain.

Comment #4658 by Johan on November 11, 2019 at 15:41

Although you are right that it is equivalent, it is not completely trivial to see the equivalence. The corresponding result for "geometrically reduced" is Lemma 10.44.4 and the corresponding result for "geometrically irreducible" is Lemma 10.47.3 except that unfortunately it is missing the part where we say it is enough to look for the spectrum of $A \otimes_k \overline{k}$ to be irreducible with $\overline{k}$ equal to either the algebraic closure or the separable algebraic closure. I will add this the next time I go through the comments.

Comment #4797 by Johan on December 10, 2019 at 10:18

Thanks again. I have no added the necessary material in this commit.

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10.49 Geometrically integral algebras

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