The Stacks project

Proof. By induction on $\dim (A)$. If $\dim (A) = 1$, then $\mathfrak p = \mathfrak m$ works. If $\dim (A) > 1$, then let $Z \subset \mathop{\mathrm{Spec}}(A)$ be an irreducible component of dimension $> 1$. Then $V(f) \cap Z$ has dimension $> 0$ (Algebra, Lemma 10.60.13). Pick a prime $\mathfrak q \in V(f) \cap Z$, $\mathfrak q \not= \mathfrak m$ corresponding to a closed point of the punctured spectrum of $A$; this is possible by Properties, Lemma 28.6.4. Then $\mathfrak q$ is not the generic point of $Z$. Hence $0 < \dim (A_\mathfrak q) < \dim (A)$ and $f \in \mathfrak q A_\mathfrak q$. By induction on the dimension we can find $f \in \mathfrak p \subset A_\mathfrak q$ with $\dim ((A_\mathfrak q)_\mathfrak p) = 1$. Then $\mathfrak p \cap A$ works. $\square$

Proof. Observe that the set of points where $f$ is unramified is the same as the set of points where $f$ is étale and that this set is open. See Morphisms, Definitions 29.35.1 and 29.36.1 and Lemma 29.36.16. To check $f$ is étale at $x$ we may work étale locally on the base and on the target (Descent, Lemmas 35.23.29 and 35.31.1). Thus we can apply More on Morphisms, Lemma 37.41.1 and assume that $f : X \to Y$ is finite and that $x$ is the unique point of $X$ lying over $y = f(x)$. Then it follows that $f$ is finite locally free (Morphisms, Lemma 29.48.2).

Assume $f$ is finite locally free and that $x$ is the unique point of $X$ lying over $y = f(x)$. By Discriminants, Lemma 49.3.1 we find a locally principal closed subscheme $D_\pi \subset Y$ such that $y' \in D_\pi $ if and only if there exists an $x' \in X$ with $f(x') = y'$ and $f$ ramified at $x'$. Thus we have to prove that $y \not\in D_\pi $. Assume $y \in D_\pi $ to get a contradiction.

By condition (3) we have $\dim (\mathcal{O}_{X, x}) \geq 1$. We have $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y})$ by Algebra, Lemma 10.112.7. By Lemma 58.21.1 we can find $y' \in D_\pi $ specializing to $y$ with $\dim (\mathcal{O}_{Y, y'}) = 1$. Choose $x' \in X$ with $f(x') = y'$ where $f$ is ramified. Since $f$ is finite it is closed, and hence $x' \leadsto x$. We have $\dim (\mathcal{O}_{X, x'}) = \dim (\mathcal{O}_{Y, y'}) = 1$ as before. This contradicts property (4). $\square$

Proof. Recall that a regular local ring is normal by Algebra, Lemma 10.157.5. Hence (1) and (2) are equivalent by Lemma 58.20.3. We prove the lemma by induction on $d$.

The case $d = 2$. In this case $A \to B$ is flat. Namely, we have going down for $A \to B$ by Algebra, Proposition 10.38.7. Then $\dim (B_{\mathfrak m'}) = 2$ for all maximal ideals $\mathfrak m' \subset B$ by Algebra, Lemma 10.112.7. Then $B_{\mathfrak m'}$ is Cohen-Macaulay by Algebra, Lemma 10.157.4. Hence and this is the important step Algebra, Lemma 10.128.1 applies to show $A \to B_{\mathfrak m'}$ is flat. Then Algebra, Lemma 10.39.18 shows $A \to B$ is flat. Thus we can apply Lemma 58.21.2 (or you can directly argue using the easier Discriminants, Lemma 49.3.1) to see that $A \to B$ is étale.

The case $d \geq 3$. Let $V \to U$ be finite étale. Let $f \in \mathfrak m_ A$, $f \not\in \mathfrak m_ A^2$. Then $A/fA$ is a regular local ring of dimension $d - 1 \geq 2$, see Algebra, Lemma 10.106.3. Let $U_0$ be the punctured spectrum of $A/fA$ and let $V_0 = V \times _ U U_0$. By Lemma 58.20.7 it suffices to show that $V_0$ is in the essential image of $\textit{FÉt}_{\mathop{\mathrm{Spec}}(A/fA)} \to \textit{FÉt}_{U_0}$. This follows from the induction hypothesis. $\square$

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

An uninteresting case is when $d = 1$. In that case we are assuming that $f$ is unramified at $x$ and that $\mathcal{O}_{Y, y}$ is a discrete valuation ring (Algebra, Lemma 10.119.7). Then $\mathcal{O}_{X, x}$ is flat over $\mathcal{O}_{Y, y}$ (otherwise the map would not be quasi-finite at $x$) and we see that $f$ is flat at $x$. Since flat $+$ unramified is étale we conclude (some details omitted).

The case $d \geq 2$. We will use induction on $d$ to reduce to the case discussed in Lemma 58.21.3. To check $f$ is étale at $x$ we may work étale locally on the base and on the target (Descent, Lemmas 35.23.29 and 35.31.1). Thus we can apply More on Morphisms, Lemma 37.41.1 and assume that $f : X \to Y$ is finite and that $x$ is the unique point of $X$ lying over $y$. Here we use that étale extensions of local rings do not change dimension, normality, and regularity, see More on Algebra, Section 15.44 and Étale Morphisms, Section 41.19.

Next, we can base change by $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ and assume that $Y$ is the spectrum of a regular local ring. It follows that $X = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ as every point of $X$ necessarily specializes to $x$.

The ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is finite and necessarily injective (by equality of dimensions). We conclude we have going down for $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ by Algebra, Proposition 10.38.7 (and the fact that a regular ring is a normal ring by Algebra, Lemma 10.157.5). Pick $x' \in X$, $x' \not= x$ with image $y' = f(x')$. Then $\mathcal{O}_{X, x'}$ is normal as a localization of a normal domain. Similarly, $\mathcal{O}_{Y, y'}$ is regular (see Algebra, Lemma 10.110.6). We have $\dim (\mathcal{O}_{X, x'}) = \dim (\mathcal{O}_{Y, y'})$ by Algebra, Lemma 10.112.7 (we checked going down above). Of course these dimensions are strictly less than $d$ as $x' \not= x$ and by induction on $d$ we conclude that $f$ is étale at $x'$.

Thus we arrive at the following situation: We have a finite local homomorphism $A \to B$ of Noetherian local rings of dimension $d \geq 2$, with $A$ regular, $B$ normal, which induces a finite étale morphism $V \to U$ on punctured spectra. Our goal is to show that $A \to B$ is étale. This follows from Lemma 58.21.3 and the proof is complete. $\square$

Proof. Observe that $\pi _*\mathcal{O}_ V$ is a finite locally free $\mathcal{O}_ U$-module, in particular reflexive. By Divisors, Lemma 31.12.12 the module $j_*\pi _*\mathcal{O}_ V$ is the unique reflexive coherent module on $X$ restricting to $\pi _*\mathcal{O}_ V$ over $U$. This proves (1).

By construction $Y \times _ X U = V$. Since $\mathcal{B}$ is coherent, we see that $Y \to X$ is finite. We have $\text{depth}(\mathcal{B}_ x) \geq 2$ for $x \in X \setminus U$ by Divisors, Lemma 31.12.11. Hence $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$ for $y \in Y \setminus V$ by Algebra, Lemma 10.72.11. Conversely, suppose that $\pi ' : Y' \to X$ is a finite morphism such that $V = Y' \times _ X U$ and $\text{depth}(\mathcal{O}_{Y', y'}) \geq 2$ for $y' \in Y' \setminus V$. Then $\pi '_*\mathcal{O}_{Y'}$ restricts to $\pi _*\mathcal{O}_ V$ over $U$ and satisfies $\text{depth}((\pi '_*\mathcal{O}_{Y'})_ x) \geq 2$ for $x \in X \setminus U$ by Algebra, Lemma 10.72.11. Then $\pi '_*\mathcal{O}_{Y'}$ is canonically isomorphic to $j_*\pi _*\mathcal{O}_ V$ for example by Divisors, Lemma 31.5.11. This proves (2).

If $Y \to X$ is étale at $y$, then $Y \to X$ is flat at $y$. Conversely, suppose that $Y \to X$ is flat at $y$. If $y \in V$, then $Y \to X$ is étale at $y$. If $y \not\in V$, then we check (1), (2), (3), and (4) of Lemma 58.21.2 hold to see that $Y \to X$ is étale at $y$. Parts (1) and (2) are clear and so is (3) since $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$. If $y' \leadsto y$ is a specialization and $\dim (\mathcal{O}_{Y, y'}) = 1$, then $y' \in V$ since otherwise the depth of this local ring would be $2$ a contradiction by Algebra, Lemma 10.72.3. Hence $Y \to X$ is étale at $y'$ and we conclude (4) of Lemma 58.21.2 holds too. This finishes the proof of (3).

Part (4) follows from (3) and the fact that $((Y \to X)_*\mathcal{O}_ Y)_ x$ is a flat $\mathcal{O}_{X, x}$-module if and only if $\mathcal{O}_{Y, y}$ is a flat $\mathcal{O}_{X, x}$-module for all $y \in Y$ mapping to $x$, see Algebra, Lemma 10.39.18. Here we also use that a finite flat module over a Noetherian ring is finite locally free, see Algebra, Lemma 10.78.2 (and Algebra, Lemma 10.31.4).

As to the final assertions of the lemma, part (a) follows from flat base change, see Cohomology of Schemes, Lemma 30.5.2 and part (b) follows from the uniqueness in (2) applied to the restriction $Y \times _ X U'$. $\square$

Proof. Let $V \to U$ be a finite étale morphism. By Noetherian induction it suffices to extend $V \to U$ to a finite étale morphism to a strictly larger open subset of $X$. Let $x \in X \setminus U$ be the generic point of an irreducible component of $X \setminus U$. Then the inverse image $U_ x$ of $U$ in $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is the punctured spectrum of $\mathcal{O}_{X, x}$. By assumption $V_ x = V \times _ U U_ x$ is the restriction of a finite étale morphism $Y_ x \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ to $U_ x$. By Limits, Lemma 32.20.3 we find an open subscheme $U \subset U' \subset X$ containing $x$ and a morphism $V' \to U'$ of finite presentation whose restriction to $U$ recovers $V \to U$ and whose restriction to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ recovering $Y_ x$. Finally, the morphism $V' \to U'$ is finite étale after possible shrinking $U'$ to a smaller open by Limits, Lemma 32.20.4. $\square$