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Geometric irreducibility can be tested over a separable algebraic closure of the base field.

Lemma 33.8.8. Let k be a field. Let X be a scheme over k. The following are equivalent:

  1. X is geometrically irreducible over k,

  2. for every finite separable field extension k'/k the scheme X_{k'} is irreducible, and

  3. X_{\overline{k}} is irreducible, where k \subset \overline{k} is a separable algebraic closure of k.

Proof. Assume X_{\overline{k}} is irreducible, i.e., assume (3). Let k'/k be a field extension. There exists a field extension \overline{k}'/\overline{k} such that k' embeds into \overline{k}' as an extension of k. By Lemma 33.8.7 we see that X_{\overline{k}'} is irreducible. Since X_{\overline{k}'} \to X_{k'} is surjective we conclude that X_{k'} is irreducible. Hence (1) holds.

Let k \subset \overline{k} be a separable algebraic closure of k. Assume not (3), i.e., assume X_{\overline{k}} is reducible. Our goal is to show that also X_{k'} is reducible for some finite subextension \overline{k}/k'/k. Let X = \bigcup _{i \in I} U_ i be an affine open covering with U_ i not empty. If for some i the scheme U_ i is reducible, or if for some pair i \not= j the intersection U_ i \cap U_ j is empty, then X is reducible (Properties, Lemma 28.3.3) and we are done. In particular we may assume that U_{i, \overline{k}} \cap U_{j, \overline{k}} for all i, j \in I is nonempty and we conclude that U_{i, \overline{k}} has to be reducible for some i. According to Algebra, Lemma 10.47.3 this means that U_{i, k'} is reducible for some finite separable field extension k'/k. Hence also X_{k'} is reducible. Thus we see that (2) implies (3).

The implication (1) \Rightarrow (2) is immediate. This proves the lemma. \square


Comments (3)

Comment #776 by Keenan Kidwell on

Quasi-compactness is unnecessary and not used in the proof.

Comment #794 by on

Wow, that is weird (but correct)! Thanks!

Comment #1282 by on

Suggested slogan: Geometric irreducibility can be tested over a separable algebraic closure of the base field.

There are also:

  • 2 comment(s) on Section 33.8: Geometrically irreducible schemes

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