**Proof.**
Parts (3) and (4) follow from parts (1) and (2) via Properties, Lemma 28.15.6.

Proof of (1). Let $\mathop{\mathrm{Spec}}(A)$ be an affine open neighbourhood of $x$ and let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. We may replace $X$ by $\mathop{\mathrm{Spec}}(A)$ and it suffices to consider affine elementary étale neighbourhoods $(U, u)$ in the supremum as they form a cofinal subsystem. Recall that the henselization $A_\mathfrak p^ h$ is the colimit of the rings $B_\mathfrak q$ over the category of pairs $(B, \mathfrak q)$ where $B$ is an étale $A$-algebra and $\mathfrak q$ is a prime lying over $\mathfrak p$ with $\kappa (\mathfrak q) = \kappa (\mathfrak p)$, see Algebra, Lemma 10.155.7. These pairs $(B, \mathfrak q)$ correspond exactly to the affine elementary étale neighbourhoods $(U, u)$ by the correspondence between rings and affine schemes. Observe that irreducible components of $\mathop{\mathrm{Spec}}(B)$ passing through $\mathfrak q$ are exactly the minimal prime ideals of $B_\mathfrak q$. The number of minimal primes of $A_\mathfrak p^ h$ is the number of branches of $X$ at $x$ by Properties, Definition 28.15.4. Observe that the transition maps $B_\mathfrak q \to B'_{\mathfrak q'}$ in the system are all flat. Since a flat local ring map is faithfully flat (Algebra, Lemma 10.39.17) we see that the lemma follows from Lemma 37.35.1.

Proof of (2). The proof is the same as the proof of (1), except that we use Algebra, Lemma 10.155.11. There is a tiny difference: given a separable algebraic closure $\kappa ^{sep}$ of $\kappa (x)$ for every étale neighbourhood $(U, u)$ we can choose a $\kappa (x)$-embedding $\phi : \kappa (u) \to \kappa ^{sep}$ because $\kappa (u)/\kappa (x)$ is finite separable (Morphisms, Lemma 29.36.15). Hence we can look at the supremum over all triples $(U, u, \phi )$ where $(U, u) \to (X, x)$ is an affine étale neighbourhood and $\phi : \kappa (u) \to \kappa ^{sep}$ is a $\kappa (x)$-embedding. These triples correspond exactly to the triples in Algebra, Lemma 10.155.11 and the rest of the proof is exactly the same.
$\square$

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