Lemma 23.8.6. Let (A, \mathfrak m) be a Noetherian local ring. Let \mathfrak p \subset A be a prime ideal. If A is a complete intersection, then A_\mathfrak p is a complete intersection too.
Proof. Choose a prime \mathfrak q of A^\wedge lying over \mathfrak p (this is possible as A \to A^\wedge is faithfully flat by Algebra, Lemma 10.97.3). Then A_\mathfrak p \to (A^\wedge )_\mathfrak q is a flat local ring homomorphism. Thus by Proposition 23.8.4 we see that A_\mathfrak p is a complete intersection if and only if (A^\wedge )_\mathfrak q is a complete intersection. Thus it suffices to prove the lemma in case A is complete (this is the key step of the proof).
Assume A is complete. By definition we may write A = R/(f_1, \ldots , f_ r) for some regular sequence f_1, \ldots , f_ r in a regular local ring R. Let \mathfrak q \subset R be the prime corresponding to \mathfrak p. Observe that f_1, \ldots , f_ r \in \mathfrak q and that A_\mathfrak p = R_\mathfrak q/(f_1, \ldots , f_ r)R_\mathfrak q. Hence A_\mathfrak p is a complete intersection by Lemma 23.8.2. \square
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