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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: