Lemma 58.28.2. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a finite type morphism. Let $x \in X$ be a point. Assume

1. $A$ is an excellent regular local ring,

2. $\mathcal{O}_{X, x}$ is normal of dimension $2$,

3. $f$ is étale outside of $\overline{\{ x\} }$.

Then $f$ is étale at $x$.

Proof. We first replace $X$ by an affine open neighbourhood of $x$. Observe that $\mathcal{O}_{X, x}$ is an excellent local ring (More on Algebra, Lemma 15.52.2). Thus we can choose a minimal resolution of singularities $W \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$, see Resolution of Surfaces, Theorem 54.14.5. After possibly replacing $X$ by an affine open neighbourhood of $x$ we can find a proper morphism $b : X' \to X$ such that $X' \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) = W$, see Limits, Lemma 32.19.1. After shrinking $X$ further, we may assume $X'$ is regular. Namely, we know $W$ is regular and $X'$ is excellent and the regular locus of the spectrum of an excellent ring is open. Since $W \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is projective (as a sequence of normalized blowing ups), we may assume after shrinking $X$ that $b$ is projective (details omitted). Let $U = X \setminus \overline{\{ x\} }$. Since $W \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is an isomorphism over the punctured spectrum, we may assume $b : X' \to X$ is an isomorphism over $U$. Thus we may and will think of $U$ as an open subscheme of $X'$ as well. Set $f' = f \circ b : X' \to \mathop{\mathrm{Spec}}(A)$.

Since $A$ is regular we see that $\mathcal{O}_ Y$ is a dualizing complex for $Y$. Hence $f^!\mathcal{O}_ Y$ is a dualzing complex on $X$ (Duality for Schemes, Lemma 48.17.7). The Cohen-Macaulay locus of $X$ is open by Duality for Schemes, Lemma 48.23.1 (this can also be proven using excellency). Since $\mathcal{O}_{X, x}$ is Cohen-Macaulay, after shrinking $X$ we may assume $X$ is Cohen-Macaulay. Observe that an étale morphism is a local complete intersection. Thus Duality for Schemes, Lemma 48.29.3 applies with $r = 0$ and we get a map

$\mathcal{O}_ X \longrightarrow \omega _{X/Y} = H^0(f^!\mathcal{O}_ Y)$

which is an isomorphism over $X \setminus \overline{\{ x\} }$. Since $\omega _{X/Y}$ is $(S_2)$ by Duality for Schemes, Lemma 48.21.5 we find this map is an isomorphism by Divisors, Lemma 31.2.11. This already shows that $X$ and in particular $\mathcal{O}_{X, x}$ is Gorenstein.

Set $\omega _{X'/Y} = H^0((f')^!\mathcal{O}_ Y)$. Arguing in exactly the same manner as above we find that $(f')^!\mathcal{O}_ Y = \omega _{X'/Y}[0]$ is a dualizing complex for $X'$. Since $X'$ is regular the morphism $X' \to Y$ is a local complete intersection morphism, see More on Morphisms, Lemma 37.59.11. By Duality for Schemes, Lemma 48.29.2 there exists a map

$\mathcal{O}_{X'} \longrightarrow \omega _{X'/Y}$

which is an isomorphism over $U$. We conclude $\omega _{X'/Y} = \mathcal{O}_{X'}(E)$ for some effective Cartier divisor $E \subset X'$ disjoint from $U$.

Since $\omega _{X/Y} = \mathcal{O}_ Y$ we see that $\omega _{X'/Y} = b^! f^!\mathcal{O}_ Y = b^!\mathcal{O}_ X$. Returning to $W \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ we see that $\omega _ W = \mathcal{O}_ W(E|_ W)$. By Lemma 58.28.1 we find $E|_ W = 0$. This means that $f' : X' \to Y$ is étale by (the already used) Duality for Schemes, Lemma 48.29.2. This immediately finishes the proof, as étaleness of $f'$ forces $b$ to be an isomorphism. $\square$

Comment #3228 by Fei Hu on

a typo, "Then $f$ is \'etale at $X$." should be "Then $f$ is \'etale at $x$."

Comment #3330 by on

Thanks, fixed here. Let me know if your name is as you wish it to be in the contributors file.

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