Let $(R, \mathfrak m, k)$ be a regular local ring with unique maximal ideal $\mathfrak m$ and residue field $k = R / \mathfrak m.$ Let $\dim(R) = d.$ Let $I$ be a proper ideal of $R.$ Proposition 2.2.4 of Bruns and Herzog asserts the following.
(*) If $R/I$ is a regular local ring, then $I$ is generated by a subset of a regular system of parameters of $R.$
By hypothesis that $R/I$ is a regular local ring, it follows that $\mu(\mathfrak m/I) = \dim(R/I).$ Call this invariant $d'.$ By definition, we have that $\mu(\mathfrak m/I)$ is the $k$-vector space dimension of $(\mathfrak m/I)/ \mathfrak m(\mathfrak m/I)) = \mathfrak m/(\mathfrak m^2 + I).$ Bruns and Herzog claim that one can appeal to Nakayama's Lemma to conclude that $I$ contains some elements $x_1, \dots, x_{d – d'}$ that belong to a minimal system of generators of $\mathfrak m$; however, this is unclear to me. On the other hand, by Nakayama's Lemma and the previous sentence, we can find some elements $y_1, \dots, y_d'$ of $\mathfrak m$ that belong to a minimal system of generators of $\mathfrak m.$
Given that such elements $x_1, \dots, x_{d – d'}$ of $I$ can be found (i.e., once I understand why they can be found), the authors claim that $R / (x_1, \dots, x_{d – d'})$ has dimension $d – (d – d') = d'.$ But does this not a priori require that $R$ is a Cohen-Macaulay local ring so that the generators of $\mathfrak m$ form a regular sequence? (Of course, a regular local ring is a Cohen-Macaulay local ring, but in my line of argument, the statement (*) that we seek to prove is part of the proof of this fact.)
Thank you very much for your time and consideration.
Best Answer
By Nakayama's lemma, we have $\dim R=\mu_R(\mathfrak{m})=\dim_k(\mathfrak{m}/\mathfrak{m}^2)$ and $\dim R/I=\mu_R(\mathfrak{m}/I)=\dim_k \mathfrak{m}/(\mathfrak{m}^2+I).$ There is an exact sequence $$0 \to I/\mathfrak{m}^2 \cap I \to \mathfrak{m}/\mathfrak{m}^2 \to \mathfrak{m}/(\mathfrak{m}^2+I) \to 0.$$ By additivity of dimension, we have $\dim_k I/\mathfrak{m}^2 \cap I=\dim R-\dim R/I$. Letting $x_1,\dots, x_{\dim R-\dim R/I}$ be preimages of chosen basis elements provides elements of $I$ which are part of a minimal generating set of $\mathfrak{m}$.
One does not need to know a priori, that $R$ is Cohen-Macaulay to calculate that the dimension of $R/(x_1,\dots,x_{\dim R-\dim R/I})$ is $\dim R/I$. This follows as $x_1,\dots,x_{\dim R-\dim R/I}$ are part of a system of parameters of $R$.