Lemma 15.108.5. Let (A, \mathfrak m) be a Noetherian local ring with henselization A^ h. Let \mathfrak q \subset A^\wedge be a minimal prime with \dim (A^\wedge /\mathfrak q) = 1. Then there exists a minimal prime \mathfrak q^ h of A^ h such that \mathfrak q = \sqrt{\mathfrak q^ hA^\wedge }.
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
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