The Stacks project

Proof. There is a trace map $Rf_*\omega _ X \to \omega _ A$, see Duality for Schemes, Section 48.7. By Grauert-Riemenschneider (Resolution of Surfaces, Proposition 54.7.8) we have $R^1f_*\omega _ X = 0$. Thus the trace map is a map $f_*\omega _ X \to \omega _ A$. Then we can consider

where the first map comes from the map $\mathcal{O}_ X \to \mathcal{O}_ X(E) = \omega _ X$ which is assumed to exist in the statement of the lemma. The composition is an isomorphism by Divisors, Lemma 31.2.11 as it is an isomorphism over the punctured spectrum of $A$ (by the assumption in the lemma and the fact that $f$ is an isomorphism over the punctured spectrum) and $A$ and $\omega _ A$ are $A$-modules of depth $2$ (by Algebra, Lemma 10.157.4 and Dualizing Complexes, Lemma 47.17.5). Hence $f_*\omega _ X \to \omega _ A$ is surjective whence an isomorphism. Thus $Rf_*\omega _ X = \omega _ A$ which by duality implies $Rf_*\mathcal{O}_ X = \mathcal{O}_{\mathop{\mathrm{Spec}}(A)}$. Whence $H^1(X, \mathcal{O}_ X) = 0$ which implies that $A$ defines a rational singularity (see discussion in Resolution of Surfaces, Section 54.8 in particular Lemmas 54.8.7 and 54.8.1). If $f$ is minimal, then $E = 0$ because the map $f^*\omega _ A \to \omega _ X$ is surjective by a repeated application of Resolution of Surfaces, Lemma 54.9.7 and $\omega _ A \cong A$ as we've seen above. $\square$

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 32.20.1. 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

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 37.62.11. By Duality for Schemes, Lemma 48.29.2 there exists a map

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. $\square$

Proof. We will prove the lemma by induction on $d = \dim (\mathcal{O}_{X, x})$.

An uninteresting case is $d = 1$ since in that case the morphism $f$ is étale at $x$ by assumption. Assume $d \geq 2$.

We can base change by $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to Y$ without affecting the conclusion of the lemma, see Morphisms, Lemma 29.36.17. Thus we may assume $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a regular local ring and $y$ corresponds to the maximal ideal $\mathfrak m$ of $A$.

Let $x' \leadsto x$ be a specialization with $x' \not= x$. Then $\mathcal{O}_{X, x'}$ is normal as a localization of $\mathcal{O}_{X, x}$. If $x'$ is not a generic point of $X$, then $1 \leq \dim (\mathcal{O}_{X, x'}) < d$ and we conclude that $f$ is étale at $x'$ by induction hypothesis. Thus we may assume that $f$ is étale at all points specializing to $x$. Since the set of points where $f$ is étale is open in $X$ (by definition) we may after replacing $X$ by an open neighbourhood of $x$ assume that $f$ is étale away from $\overline{\{ x\} }$. In particular, we see that $f$ is étale except at points lying over the closed point $y \in Y = \mathop{\mathrm{Spec}}(A)$.

Let $X' = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$. Let $x' \in X'$ be the unique point lying over $x$. By the above we see that $X'$ is étale over $\mathop{\mathrm{Spec}}(A^\wedge )$ away from the closed fibre and hence $X'$ is normal away from the closed fibre. Since $X$ is normal we conclude that $X'$ is normal by Resolution of Surfaces, Lemma 54.11.6. Then if we can show $X' \to \mathop{\mathrm{Spec}}(A^\wedge )$ is étale at $x'$, then $f$ is étale at $x$ (by the aforementioned Morphisms, Lemma 29.36.17). Thus we may and do assume $A$ is a regular complete local ring.

The case $d = 2$ now follows from Lemma 58.28.2.

Assume $d > 2$. Let $t \in \mathfrak m$, $t \not\in \mathfrak m^2$. Set $Y_0 = \mathop{\mathrm{Spec}}(A/tA)$ and $X_0 = X \times _ Y Y_0$. Then $X_0 \to Y_0$ is étale away from the fibre over the closed point. Since $d > 2$ we have $\dim (\mathcal{O}_{X_0, x}) = d - 1$ is $\geq 2$. The normalization $X_0' \to X_0$ is surjective and finite (as we're working over a complete local ring and such rings are Nagata). Let $x' \in X_0'$ be a point mapping to $x$. By induction hypothesis the morphism $X'_0 \to Y$ is étale at $x'$. From the inclusions $\kappa (y) \subset \kappa (x) \subset \kappa (x')$ we conclude that $\kappa (x)$ is finite over $\kappa (y)$. Hence $x$ is a closed point of the fibre of $X \to Y$ over $y$. But since $x$ is also a generic point of this fibre, we conclude that $f$ is quasi-finite at $x$ and we reduce to the case of purity of branch locus, see Lemma 58.21.4. $\square$