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
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