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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: