
Lemma 10.47.2. Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent

1. for every field extension $k \subset k'$ the spectrum of $R \otimes _ k k'$ is connected, and

2. for every finite separable field extension $k \subset k'$ the spectrum of $R \otimes _ k k'$ is connected.

Proof. For any extension of fields $k \subset k'$ the connectivity of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ having no nontrivial idempotents, see Lemma 10.20.4. Assume (2). Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Using Lemma 10.42.4 we see that (2) is equivalent to $R \otimes _ k \overline{k}$ having no nontrivial idempotents. For any field extension $k \subset k'$, there exists a field extension $\overline{k} \subset \overline{k}'$ with $k' \subset \overline{k}'$. By Lemma 10.47.1 we see that $R \otimes _ k \overline{k}'$ has no nontrivial idempotents. If $R \otimes _ k k'$ has a nontrivial idempotent, then also $R \otimes _ k \overline{k}'$, contradiction. $\square$

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