The Stacks project

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. $\square$

Comments (2)

Comment #3228 by Fei Hu on

a typo, "Then is \'etale at ." should be "Then is \'etale at ."

Comment #3330 by on

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

There are also:

  • 2 comment(s) on Section 58.28: Purity of ramification locus

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EA3. Beware of the difference between the letter 'O' and the digit '0'.