Lemma 23.8.3. Let $R$ be a regular ring. Let $\mathfrak p \subset R$ be a prime. Let $I \subset \mathfrak p$ be an ideal. Set $A = (R/I)_\mathfrak p = R_\mathfrak p/I_\mathfrak p$. The following are equivalent

1. the completion of $A$ is a complete intersection in the sense above,

2. $I_\mathfrak p \subset R_\mathfrak p$ is generated by a regular sequence,

3. the module $(I/I^2)_\mathfrak p$ can be generated by $\dim (R_\mathfrak p) - \dim (A)$ elements,

Proof. We may and do replace $R$ by its localization at $\mathfrak p$. Then $\mathfrak p = \mathfrak m$ is the maximal ideal of $R$ and $A = R/I$. Let $f_1, \ldots , f_ r \in I$ be a minimal sequence of generators. The completion of $A$ is equal to $A^\wedge = R^\wedge /(f_1, \ldots , f_ r)R^\wedge$ because completion for finite modules over the Noetherian ring $R_\mathfrak p$ is exact (Algebra, Lemma 10.97.1).

If (1) holds, then the image of the sequence $f_1, \ldots , f_ r$ in $R^\wedge$ is a regular sequence by assumption. Hence it is a regular sequence in $R$ by Algebra, Lemmas 10.97.2 and 10.68.5. Thus (1) implies (2).

Assume (3) holds. Set $c = \dim (R) - \dim (A)$ and let $f_1, \ldots , f_ c \in I$ map to generators of $I/I^2$. by Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $I = (f_1, \ldots , f_ c)$. Since $R$ is regular and hence Cohen-Macaulay (Algebra, Proposition 10.103.4) we see that $f_1, \ldots , f_ c$ is a regular sequence by Algebra, Proposition 10.103.4. Thus (3) implies (2). Finally, (2) implies (1) by Lemma 23.8.2. $\square$

There are also:

• 2 comment(s) on Section 23.8: Local complete intersection rings

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).