Lemma 10.48.2. Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent
for every field extension $k'/k$ the spectrum of $R \otimes _ k k'$ is connected, and
for every finite separable field extension $k'/k$ the spectrum of $R \otimes _ k k'$ is connected.
Proof. For any extension of fields $k'/k$ the connectivity of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ having no nontrivial idempotents, see Lemma 10.21.4. Assume (2). Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Using Lemma 10.43.4 we see that (2) is equivalent to $R \otimes _ k \overline{k}$ having no nontrivial idempotents. For any field extension $k'/k$, there exists a field extension $\overline{k}'/\overline{k}$ with $k' \subset \overline{k}'$. By Lemma 10.48.1 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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