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 10.96.3). 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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