The Stacks project

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

  1. $S$ is geometrically integral over $k$,

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

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

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

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.

Comment #4656 by Hao on

A possibly helpful equivalent definition: If is a algebra, then is geometrically integral iff is a domain.

Comment #4658 by on

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 to be irreducible with equal to either the algebraic closure or the separable algebraic closure. I will add this the next time I go through the comments.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05DW. Beware of the difference between the letter 'O' and the digit '0'.



View Section 10.49 as pdf