## 58.20 Purity in local case, I

Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$ be the punctured spectrum. We say purity holds for $(A, \mathfrak m)$ if the restriction functor

$\textit{FÉt}_ X \longrightarrow \textit{FÉt}_ U$

is essentially surjective. In this section we try to understand how the question changes when one passes from $X$ to a hypersurface $X_0$ in $X$, in other words, we study a kind of local Lefschetz property for the fundamental groups of punctured spectra. These results will be useful to proceed by induction on dimension in the proofs of our main results on local purity, namely, Lemma 58.21.3, Proposition 58.25.3, and Proposition 58.26.4.

Lemma 58.20.1. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$. Let $\pi : Y \to X$ be a finite morphism such that $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$ for all closed points $y \in Y$. Then $Y$ is the spectrum of $B = \mathcal{O}_ Y(\pi ^{-1}(U))$.

Proof. Set $V = \pi ^{-1}(U)$ and denote $\pi ' : V \to U$ the restriction of $\pi$. Consider the $\mathcal{O}_ X$-module map

$\pi _*\mathcal{O}_ Y \longrightarrow j_*\pi '_*\mathcal{O}_ V$

where $j : U \to X$ is the inclusion morphism. We claim Divisors, Lemma 31.5.11 applies to this map. If so, then $B = \Gamma (Y, \mathcal{O}_ Y)$ and we see that the lemma holds. Let $x \in X$ be the closed point. It suffices to show that $\text{depth}((\pi _*\mathcal{O}_ Y)_ x) \geq 2$. Let $y_1, \ldots , y_ n \in Y$ be the points mapping to $x$. By Algebra, Lemma 10.72.11 it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_ i}) \geq 2$ for $i = 1, \ldots , n$. Since this is the assumption of the lemma the proof is complete. $\square$

Lemma 58.20.2. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$. Let $V$ be finite étale over $U$. Assume $A$ has depth $\geq 2$. The following are equivalent

1. $V = Y \times _ X U$ for some $Y \to X$ finite étale,

2. $B = \Gamma (V, \mathcal{O}_ V)$ is finite étale over $A$.

Proof. Denote $\pi : V \to U$ the given finite étale morphism. Assume $Y$ as in (1) exists. Let $x \in X$ be the point corresponding to $\mathfrak m$. Let $y \in Y$ be a point mapping to $x$. We claim that $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$. This is true because $Y \to X$ is étale and hence $A = \mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ have the same depth (Algebra, Lemma 10.163.2). Hence Lemma 58.20.1 applies and $Y = \mathop{\mathrm{Spec}}(B)$.

The implication (2) $\Rightarrow$ (1) is easier and the details are omitted. $\square$

Lemma 58.20.3. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$. Assume $A$ is normal of dimension $\geq 2$. The functor

$\textit{FÉt}_ U \longrightarrow \left\{ \begin{matrix} \text{finite normal }A\text{-algebras }B\text{ such} \\ \text{that }\mathop{\mathrm{Spec}}(B) \to X\text{ is étale over }U \end{matrix} \right\} , \quad V \longmapsto \Gamma (V, \mathcal{O}_ V)$

is an equivalence. Moreover, $V = Y \times _ X U$ for some $Y \to X$ finite étale if and only if $B = \Gamma (V, \mathcal{O}_ V)$ is finite étale over $A$.

Proof. Observe that $\text{depth}(A) \geq 2$ because $A$ is normal (Serre's criterion for normality, Algebra, Lemma 10.157.4). Thus the final statement follows from Lemma 58.20.2. Given $\pi : V \to U$ finite étale, set $B = \Gamma (V, \mathcal{O}_ V)$. If we can show that $B$ is normal and finite over $A$, then we obtain the displayed functor. Since there is an obvious quasi-inverse functor, this is also all that we have to show.

Since $A$ is normal, the scheme $V$ is normal (Descent, Lemma 35.17.2). Hence $V$ is a finite disjoint union of integral schemes (Properties, Lemma 28.7.6). Thus we may assume $V$ is integral. In this case the function field $L$ of $V$ (Morphisms, Section 29.49) is a finite separable extension of the fraction field of $A$ (because we get it by looking at the generic fibre of $V \to U$ and using Morphisms, Lemma 29.36.7). By Algebra, Lemma 10.161.8 the integral closure $B' \subset L$ of $A$ in $L$ is finite over $A$. By More on Algebra, Lemma 15.23.20 we see that $B'$ is a reflexive $A$-module, which in turn implies that $\text{depth}_ A(B') \geq 2$ by More on Algebra, Lemma 15.23.18.

Let $f \in \mathfrak m$. Then $B_ f = \Gamma (V \times _ U D(f), \mathcal{O}_ V)$ (Properties, Lemma 28.17.1). Hence $B'_ f = B_ f$ because $B_ f$ is normal (see above), finite over $A_ f$ with fraction field $L$. It follows that $V = \mathop{\mathrm{Spec}}(B') \times _ X U$. Then we conclude that $B = B'$ from Lemma 58.20.1 applied to $\mathop{\mathrm{Spec}}(B') \to X$. This lemma applies because the localizations $B'_{\mathfrak m'}$ of $B'$ at maximal ideals $\mathfrak m' \subset B'$ lying over $\mathfrak m$ have depth $\geq 2$ by Algebra, Lemma 10.72.11 and the remark on depth in the preceding paragraph. $\square$

Lemma 58.20.4. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\}$. Let $V$ be finite étale over $U$. Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$, let $X' = \mathop{\mathrm{Spec}}(A^\wedge )$ and let $U'$ and $V'$ be the base changes of $U$ and $V$ to $X'$. The following are equivalent

1. $V = Y \times _ X U$ for some $Y \to X$ finite étale, and

2. $V' = Y' \times _{X'} U'$ for some $Y' \to X'$ finite étale.

Proof. The implication (1) $\Rightarrow$ (2) follows from taking the base change of a solution $Y \to X$. Let $Y' \to X'$ be as in (2). By Lemma 58.19.3 we can find $Y \to X$ finite such that $V = Y \times _ X U$ and $Y' = Y \times _ X X'$. By descent we see that $Y \to X$ is finite étale (Algebra, Lemmas 10.83.2 and 10.143.3). This finishes the proof. $\square$

The point of the following two lemmas is that the assumptions do not force $A$ to have depth $\geq 3$. For example if $A$ is a complete normal local domain of dimension $\geq 3$ and $f \in \mathfrak m$ is nonzero, then the assumptions are satisfied.

Lemma 58.20.5. In Situation 58.19.1. Let $V$ be finite étale over $U$. Assume

1. $A$ has a dualizing complex,

2. the pair $(A, (f))$ is henselian,

3. one of the following is true

1. $A_ f$ is $(S_2)$ and every irreducible component of $X$ not contained in $X_0$ has dimension $\geq 3$, or

2. for every prime $\mathfrak p \subset A$, $f \not\in \mathfrak p$ we have $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 2$.

4. $V_0 = V \times _ U U_0$ is equal to $Y_0 \times _{X_0} U_0$ for some $Y_0 \to X_0$ finite étale.

Then $V = Y \times _ X U$ for some $Y \to X$ finite étale.

Proof. We reduce to the complete case using Lemma 58.20.4. (The assumptions carry over; see proof of Lemma 58.19.5.)

In the complete case we can lift $Y_0 \to X_0$ to a finite étale morphism $Y \to X$ by More on Algebra, Lemma 15.13.2; observe that $(A, fA)$ is a henselian pair by More on Algebra, Lemma 15.11.4. Then we can use Lemma 58.19.5 to see that $V$ is isomorphic to $Y \times _ X U$ and the proof is complete. $\square$

Lemma 58.20.6. In Situation 58.19.1. Let $V$ be finite étale over $U$. Assume

1. $H^1_\mathfrak m(A)$ and $H^2_\mathfrak m(A)$ are annihilated by a power of $f$,

2. $V_0 = V \times _ U U_0$ is equal to $Y_0 \times _{X_0} U_0$ for some $Y_0 \to X_0$ finite étale.

Then $V = Y \times _ X U$ for some $Y \to X$ finite étale.

Proof. We reduce to the complete case using Lemma 58.20.4. (The assumptions carry over; use Dualizing Complexes, Lemma 47.9.3.)

In the complete case we can lift $Y_0 \to X_0$ to a finite étale morphism $Y \to X$ by More on Algebra, Lemma 15.13.2; observe that $(A, fA)$ is a henselian pair by More on Algebra, Lemma 15.11.4. Then we can use Lemma 58.19.6 to see that $V$ is isomorphic to $Y \times _ X U$ and the proof is complete. $\square$

Lemma 58.20.7. In Situation 58.19.1. Let $V$ be finite étale over $U$. Assume

1. $A$ has depth $\geq 3$,

2. $V_0 = V \times _ U U_0$ is equal to $Y_0 \times _{X_0} U_0$ for some $Y_0 \to X_0$ finite étale.

Then $V = Y \times _ X U$ for some $Y \to X$ finite étale.

Proof. The assumption of depth forces $H^1_\mathfrak m(A) = H^2_\mathfrak m(A) = 0$, see Dualizing Complexes, Lemma 47.11.1. Hence Lemma 58.20.6 applies. $\square$

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