The Stacks project

Proof. Since the completion of $A$ and $A^ h$ are the same, we may assume that $A$ is henselian (Lemma 15.45.3). We will apply Lemma 15.108.4 to $A^\wedge \to A^\wedge /J$ where $J = \mathop{\mathrm{Ker}}(A^\wedge \to (A^\wedge )_{\mathfrak q})$. Since $\dim ((A^\wedge )_\mathfrak q) = 0$ we see that $\mathfrak q^ n \subset J$ for some $n$. Hence $J/J^2$ is annihilated by $\mathfrak q^ n$. On the other hand $(J/J^2)_\mathfrak q = 0$ because $J_\mathfrak q = 0$. Hence $\mathfrak m$ is the only associated prime of $J/J^2$ and we find that a power of $\mathfrak m$ annihilates $J/J^2$. Thus the lemma applies and we find that $A^\wedge /J = C^\wedge $ for some finite type $A$-algebra $C$.

Then $C/\mathfrak m C = A/\mathfrak m$ because $A^\wedge /J$ 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 = IA^\wedge $ (because completion is exact in our situation by Algebra, Lemma 10.97.2) and $I = J \cap A$ (by flatness of $A \to A^\wedge $). Since $\mathfrak q^ n \subset J \subset \mathfrak q$ we see that $\mathfrak p = \mathfrak q \cap A$ satisfies $\mathfrak p^ n \subset I \subset \mathfrak p$. Then $\sqrt{\mathfrak p A^\wedge } = \mathfrak q$ and the proof is complete. $\square$