Lemma 10.106.4. Let $R$ be a regular local ring. Let $I \subset R$ be an ideal such that $R/I$ is a regular local ring as well. Then there exists a minimal set of generators $x_1, \ldots , x_ d$ for the maximal ideal $\mathfrak m$ of $R$ such that $I = (x_1, \ldots , x_ c)$ for some $0 \leq c \leq d$.

Proof. Say $\dim (R) = d$ and $\dim (R/I) = d - c$. Denote $\overline{\mathfrak m} = \mathfrak m/I$ the maximal ideal of $R/I$. Let $\kappa = R/\mathfrak m$. We have

$\dim _\kappa ((I + \mathfrak m^2)/\mathfrak m^2) = \dim _\kappa (\mathfrak m/\mathfrak m^2) - \dim (\overline{\mathfrak m}/\overline{\mathfrak m}^2) = d - (d - c) = c$

by the definition of a regular local ring. Hence we can choose $x_1, \ldots , x_ c \in I$ whose images in $\mathfrak m/\mathfrak m^2$ are linearly independent and supplement with $x_{c + 1}, \ldots , x_ d$ to get a minimal system of generators of $\mathfrak m$. The induced map $R/(x_1, \ldots , x_ c) \to R/I$ is a surjection between regular local rings of the same dimension (Lemma 10.106.3). It follows that the kernel is zero, i.e., $I = (x_1, \ldots , x_ c)$. Namely, if not then we would have $\dim (R/I) < \dim (R/(x_1, \ldots , x_ c))$ by Lemmas 10.106.2 and 10.60.13. $\square$

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