**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 10.96.1).

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 10.96.2 and 10.67.5. 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 10.19.1) we see that $I = (f_1, \ldots , f_ c)$. Since $R$ is regular and hence Cohen-Macaulay (Algebra, Proposition 10.102.4) we see that $f_1, \ldots , f_ c$ is a regular sequence by Algebra, Proposition 10.102.4. Thus (3) implies (2). Finally, (2) implies (1) by Lemma 23.8.2.
$\square$

## Comments (0)