**Proof.**
It is clear that (1) implies (2).

Assume (2) and let $\overline{k}$ is the separable algebraic closure of $k$. Suppose $\mathfrak q_ i \subset R \otimes _ k \overline{k}$, $i = 1, 2$ are two minimal prime ideals. For every finite subextension $\overline{k}/k'/k$ the extension $k'/k$ is separable and the ring map $R \otimes _ k k' \to R \otimes _ k \overline{k}$ is flat. Hence $\mathfrak p_ i = (R \otimes _ k k') \cap \mathfrak q_ i$ are minimal prime ideals (as we have going down for flat ring maps by Lemma 10.39.19). Thus we see that $\mathfrak p_1 = \mathfrak p_2$ by assumption (2). Since $\overline{k} = \bigcup k'$ we conclude $\mathfrak q_1 = \mathfrak q_2$. Hence $\mathop{\mathrm{Spec}}(R \otimes _ k \overline{k})$ is irreducible.

Assume (3) and let $\overline{k}$ be the algebraic closure of $k$. Let $\overline{k}/\overline{k}'/k$ be the corresponding separable algebraic closure of $k$. Then $\overline{k}/\overline{k}'$ is purely inseparable (in positive characteristic) or trivial. Hence $R \otimes _ k \overline{k}' \to R \otimes _ k \overline{k}$ induces a homeomorphism on spectra, for example by Lemma 10.46.7. Thus we have (4).

Assume (4). Let $k'/k$ be an arbitrary field extension and let $\overline{k}$ be the algebraic closure of $k$. We may choose a field $F$ such that both $k'$ and $\overline{k}$ are isomorphic to subfields of $F$. Then

\[ R \otimes _ k F = (R \otimes _ k \overline{k}) \otimes _{\overline{k}} F \]

and hence we see from Lemma 10.47.2 that $R \otimes _ k F$ has a unique minimal prime. Finally, the ring map $R \otimes _ k k' \to R \otimes _ k F$ is flat and injective and hence any minimal prime of $R \otimes _ k k'$ is the image of a minimal prime of $R \otimes _ k F$ (by Lemma 10.30.5 and going down). We conclude that there is only one such minimal prime and the proof is complete.
$\square$

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