**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)

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