**Proof.**
More on Algebra, Lemma 15.106.5 shows that (1) implies that the punctured spectra in (2) are irreducible and in particular connected.

Assume (2). Let $x \in X$. We have to show that $\mathcal{O}_{X, x}$ is geometrically unibranch. By induction on $\dim (\mathcal{O}_{X, x})$ we may assume that the result holds for every nontrivial generalization of $x$. We may replace $X$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. In other words, we may assume that $X = \mathop{\mathrm{Spec}}(A)$ with $A$ local and that $A_\mathfrak p$ is geometrically unibranch for each nonmaximal prime $\mathfrak p \subset A$.

Let $A^{sh}$ be the strict henselization of $A$. If $\mathfrak q \subset A^{sh}$ is a prime lying over $\mathfrak p \subset A$, then $A_\mathfrak p \to A^{sh}_\mathfrak q$ is a filtered colimit of étale algebras. Hence the strict henselizations of $A_\mathfrak p$ and $A^{sh}_\mathfrak q$ are isomorphic. Thus by More on Algebra, Lemma 15.106.5 we conclude that $A^{sh}_\mathfrak q$ has a unique minimal prime ideal for every nonmaximal prime $\mathfrak q$ of $A^{sh}$.

Let $\mathfrak q_1, \ldots , \mathfrak q_ r$ be the minimal primes of $A^{sh}$. We have to show that $r = 1$. By the above we see that $V(\mathfrak q_1) \cap V(\mathfrak q_ j) = \{ \mathfrak m^{sh}\} $ for $j = 2, \ldots , r$. Hence $V(\mathfrak q_1) \setminus \{ \mathfrak m^{sh}\} $ is an open and closed subset of the punctured spectrum of $A^{sh}$ which is a contradiction with the assumption that this punctured spectrum is connected unless $r = 1$.
$\square$

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