Lemma 15.108.6. Let $(A, \mathfrak m)$ be a Noetherian local ring. The punctured spectrum of $A^\wedge$ is disconnected if and only if the punctured spectrum of $A^ h$ is disconnected.

Proof. Since the completion of $A$ and $A^ h$ are the same, we may assume that $A$ is henselian (Lemma 15.45.3).

Since $A \to A^\wedge$ is faithfully flat (see reference just given) the map from the punctured spectrum of $A^\wedge$ to the punctured spectrum of $A$ is surjective (see Algebra, Lemma 10.39.16). Hence if the punctured spectrum of $A$ is disconnected, then the same is true for $A^\wedge$.

Assume the punctured spectrum of $A^\wedge$ is disconnected. This means that

$\mathop{\mathrm{Spec}}(A^\wedge ) \setminus \{ \mathfrak m^\wedge \} = Z \amalg Z'$

with $Z$ and $Z'$ closed. Let $\overline{Z}, \overline{Z}' \subset \mathop{\mathrm{Spec}}(A^\wedge )$ be the closures. Say $\overline{Z} = V(J)$, $\overline{Z}' = V(J')$ for some ideals $J, J' \subset A^\wedge$. Then $V(J + J') = \{ \mathfrak m^\wedge \}$ and $V(JJ') = \mathop{\mathrm{Spec}}(A^\wedge )$. The first equality means that $\mathfrak m^\wedge = \sqrt{J + J'}$ which implies $(\mathfrak m^\wedge )^ e \subset J + J'$ for some $e \geq 1$. The second equality implies every element of $JJ'$ is nilpotent hence $(JJ')^ n = 0$ for some $n \geq 1$. Combined this means that $J^ n/J^{2n}$ is annihilated by $J^ n$ and $(J')^ n$ and hence by $(\mathfrak m^\wedge )^{2en}$. Thus we may apply Lemma 15.108.4 to see that there is a finite type $A$-algebra $C$ and an isomorphism $A^\wedge /J^ n = C^\wedge$.

The rest of the proof is exactly the same as the second part of the proof of Lemma 15.108.5; of course that lemma is a special case of this one! We have $C/\mathfrak m C = A/\mathfrak m$ because $A^\wedge /J^ n$ has the same property. Hence $\mathfrak m_ C = \mathfrak m C$ is a maximal ideal and $A \to C$ is unramified at $\mathfrak m_ C$ (Algebra, Lemma 10.151.7). After replacing $C$ by a principal localization we may assume that $C$ is a quotient of an étale $A$-algebra $B$, see Algebra, Proposition 10.152.1. However, since the residue field extension of $A \to C_{\mathfrak m_ C}$ is trivial and $A$ is henselian, we conclude that $B = A$ again after a localization. Thus $C = A/I$ for some ideal $I \subset A$ and it follows that $J^ n = IA^\wedge$ (because completion is exact in our situation by Algebra, Lemma 10.97.2) and $I = J^ n \cap A$ (by flatness of $A \to A^\wedge$). By symmetry $I' = (J')^ n \cap A$ satisfies $(J')^ n = I'A^\wedge$. Then $\mathfrak m^ e \subset I + I'$ and $II' = 0$ and we conclude that $V(I)$ and $V(I')$ are closed subschemes which give the desired disjoint union decomposition of the punctured spectrum of $A$. $\square$

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