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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.

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