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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.

Dylan C. Beck
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    In my opinion a simpler proof is by induction on $\dim R$. The main point is to notice that if $R/I$ is regular, then $I\not\subseteq\mathfrak m^2$. Then choose an element $x\in I$, $x\notin\mathfrak m^2$ and replace $R$ by $R/(x)$. – user26857 Oct 30 '21 at 08:13

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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 $R$.

user26857
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metalspringpro
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  • Beautiful. It makes sense! On the last paragraph, perhaps I have misunderstood what is meant by a system of parameters. I believe that a system of parameters of a $d$-dimensional Noetherian local ring $(R, \mathfrak m)$ (or finitely generated module $M$ over such a ring) is a collection of elements $x_1, \dots, x_d \in \mathfrak m$ such that $\dim(R / (x_1, \dots, x_d)R) = 0$ (or $\dim(M / (x_1, \dots, x_d)M) = 0$ in the parenthetical case). I see in the literature now that every $M$-regular sequence is part of a system of parameters for $M,$ so I should convince myself of this. – Dylan C. Beck Oct 30 '21 at 05:58
  • On second thought, I think I should try to see that every subset of cardinality $i$ of a system of parameters satisfies $\dim(R/I) = \dim(R) - i,$ where $I$ is the ideal generated by those elements. – Dylan C. Beck Oct 30 '21 at 06:02
  • Yes, that's what you want. – metalspringpro Oct 30 '21 at 07:03
  • Do you have any suggestions toward this end? I believe the proof should proceed by induction on $i,$ but even in the case that $x_1 \in \mathfrak m$ belongs to a system of parameters, I am not sure why $\dim(R / x_1 R) \leq \dim(R) - 1.$ The other inequality holds by Krull's Height Theorem. – Dylan C. Beck Oct 30 '21 at 18:01
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    If $x_1,\dots,x_d$ is a s.o.p. of $R$, you want to claim that $x_2,\dots,x_d$ is an s.o.p. in $R/x_1R$. But this is immediate from the definition you are taking. Indeed, killing $x_2,\dots,x_d$ in $R/x_1R$ is the same as killing $x_1,\dots,x_d$ in $R$. Alternatively, you can show $x_1$ is part of an s.o.p. of $R$ if and only if $x_1$ avoids every minimal prime of $R$. – metalspringpro Oct 30 '21 at 22:00
  • For notational convenience, let's call $I = (x_1, \dots, x_d),$ $I' = (x_2, \dots, x_d),$ and $R' = R / x_1 R.$ It is clear to me that $R/I \cong R'/I'$ so that $\dim(R'/I') = 0$; however, I'm still not sure what that says about $\dim(R / x_1 R).$ I apologize for the confusion. I assume it's just a matter of clearly writing down what it means for $\dim(R'/I') = 0,$ but even that isn't getting me anywhere. – Dylan C. Beck Oct 31 '21 at 06:25
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    As $\dim R'/I'=0$, $I'$ is $\mathfrak{m}$-primary in $R'$. One cannot generate an $\mathfrak{m}$-primary ideal by less elements than the dimension (this follows from Krull's height theorem, and an s.o.p. is defined by achieving this minimum), and so we must have $\dim(R/x_1R) \le d-1$. – metalspringpro Oct 31 '21 at 08:21
  • I see! Very nice. Thank you so much for sticking with me and helping me piece that together. – Dylan C. Beck Oct 31 '21 at 20:14