
Lemma 36.34.5. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point with image $s \in S$. Assume

1. $f$ is smooth at $x$, and

2. $x$ is a closed point of $X_ s$ and $\kappa (s) \subset \kappa (x)$ is separable.

Then there exists an immersion $Z \to X$ containing $x$ such that

1. $Z \to S$ is étale, and

2. $Z_ s = \{ x\}$ set theoretically.

Proof. We may and do replace $S$ by an affine open neighbourhood of $s$. We may and do replace $X$ by an affine open neighbourhood of $x$ such that $X \to S$ is smooth. We will prove the lemma for smooth morphisms of affines by induction on $d = \dim _ x(X_ s)$.

The case $d = 0$. In this case we show that we may take $Z$ to be an open neighbourhood of $x$. Namely, if $d = 0$, then $X \to S$ is quasi-finite at $x$, see Morphisms, Lemma 28.28.5. Hence there exists an affine open neighbourhood $U \subset X$ such that $U \to S$ is quasi-finite, see Morphisms, Lemma 28.53.2. Thus after replacing $X$ by $U$ we see that $X$ is quasi-finite and smooth over $S$, hence smooth of relative dimension $0$ over $S$, hence étale over $S$. Moreover, the fibre $X_ s$ is a finite discrete set. Hence after replacing $X$ by a further affine open neighbourhood of $X$ we see that $f^{-1}(\{ s\} ) = \{ x\}$ (because the topology on $X_ s$ is induced from the topology on $X$, see Schemes, Lemma 25.18.5). This proves the lemma in this case.

Next, assume $d > 0$. Note that because $x$ is a closed point of its fibre the extension $\kappa (s) \subset \kappa (x)$ is finite (by the Hilbert Nullstellensatz, see Morphisms, Lemma 28.19.3). Thus we see $\Omega _{\kappa (x)/\kappa (s)} = 0$ as this holds for algebraic separable field extensions. Thus we may apply Lemma 36.34.2 to find a diagram

$\xymatrix{ D \ar[r] \ar[rrd] & U \ar[r] \ar[rd] & X \ar[d] \\ & & S }$

with $x \in D$. Note that $\dim _ x(D_ s) = \dim _ x(X_ s) - 1$ for example because $\dim (\mathcal{O}_{D_ s, x}) = \dim (\mathcal{O}_{X_ s, x}) - 1$ by Algebra, Lemma 10.59.12 (also $D_ s \subset X_ s$ is effective Cartier, see Divisors, Lemma 30.18.1) and then using Morphisms, Lemma 28.27.1. Thus the morphism $D \to S$ is smooth with $\dim _ x(D_ s) = \dim _ x(X_ s) - 1 = d - 1$. By induction hypothesis we can find an immersion $Z \to D$ as desired, which finishes the proof. $\square$

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