Throughout, let $R$ be a commutative Noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k=R/\mathfrak{m}$.
The notion of Cohen-Macaulay ring is sufficiently general to a wealth of examples in algebraic geometry, invariance theory and combinatorics; meanwhile it is sufficiently strict to allow a rich theory. The notion of Cohen-Macaulay is a workhorse of commutative algebra. In this post, we discover an important subclass of Cohen-Macaulay ring - regular local rings (one would be thinking about $k[[x_1,\dots,x_n]]$). See also “Why Cohen-Macaulay rings have become important in commutative algebra?” on MathOverflow.
It is recommended to be familiar with basic commutative algebra tools such as Nakayama’s lemma and minimal prime ideals.
The content can be generalised to modules to a good extent, but we are not doing it for sake of quick accessibility.
Definition 1. The Krull dimension of $R$, written as $\dim{R}$, is the supremum taking over the length of prime ideal chains
This definition was introduced to define dimension of affine varieties, in a global sense. Locally, we have the following definition.
Definition 2. The embedding dimension of $R$ is the dimension of a vector space
The right hand side is the dimension of a $k$-vector space $\mathfrak{m}/\mathfrak{m}^2$.
Let $R$ be the local ring of a complex variety $X$ at a point $P$, in other words we write $R=\mathcal{O}_{P,X}$. Then $(\mathfrak{m}/\mathfrak{m}^2)^\ast$ is the Zariski tangent space of $X$ at $P$, whose dimension equals $\dim_k(\mathfrak{m}/\mathfrak{m}^2)=emb.\dim{R}$. Here for a vector space $V$ over $k$, $V^\ast$ is the dual space of $V$. The embedding dimension of $R$ is the smallest integer $n$ such that some analytic neighbourhood of $P$ in $X$ embeds into $\mathbb{C}^n$. If this dimension equals the dimension of $X$, then $X$ is “smooth” at $P$. For this reason we define regular local ring.
Definition 3. The ring $R$ is called regular if $\dim{R}=emb.\dim{R}$.
The most immediate intuitive example of regular local ring has to be rings of the form
where $K$ is a field. These kind of rings are regular local rings of Krull dimension $n$. As one would imagine, this ring contains much more information than $K[x_1,\dots,x_n]$. Power series in complex analysis is much more powerful than polynomials.
But by working on regular local rings, we are not essentially restricting ourselves into the ring of power series over a field. For example, the ring $\mathbb{Z}[X]_{(2,X)}$ is also a regular local ring, but it does not even contain a field.
Nevertheless, our primary psychological model of regular local rings is still a ring of the form $A=K[[x_1,\dots,x_n]]$, which has a maximal ideal $\mathfrak{m}=(x_1,\dots,x_n)$. To study local rings in the flavour of $A$, we develop an analogy of elements $\{x_1,\dots,x_n\}$.
Definition 4. A regular sequence of $R$, also written as $R$-sequence, is a sequence $[x_1,\dots,x_n]$ of elements in $\mathfrak{m}$ such that $x_1$ is a non-zero-divisor in $R$, and such that given $i>1$, each $x_i$ is a non-zero-divisor in $R/(x_1,\dots,x_i)$.
The grade of $R$, $G(R)$, is the longest length of regular sequences. If $G(R)=\dim{R}$, then $R$ is called Cohen-Macaulay.
It is quite intuitive that, for $A=K[[x_1,\dots,x_n]]$, the longest $R$-sequence has to be $[x_1,\dots,x_n]$, and therefore $A$ is Cohen-Macaulay. But such an argument does not bring us to the conclusion that quick. We will show later, anyway, every regular local ring is a Cohen-Macaulay ring.
Amongst many sequences, we are in particular interested in the sequence that are mapped onto a basis of the $k$-vector space $\mathfrak{m}/\mathfrak{m}^2$. We will show later that this “regular” sequence is indeed the regular sequence.
Proposition 1. Let $x_1,\dots,x_n$ be elements in $\mathfrak{m} \subset R$ whose images form a basis of $\mathfrak{m}/\mathfrak{m}^2$, then $x_1,\dots,x_n$ generate the maximal ideal $\mathfrak{m}$.
Proof. Nakayama’s lemma (8). Notice that as $R$ is local, the Jacobson radical is $\mathfrak{m}$. Besides, we take $I=M=\mathfrak{m}$. $\square$
Proposition 2. If $R$ is a regular local ring of dimension $n$ and $x_1, \dots,x_n \in \mathfrak{m}$ map to a basis of $\mathfrak{m}/\mathfrak{m}^2$, then $R/(x_1,\dots,x_i)$ is a regular local ring of dimension $n-i$.
Proof. By proposition 1, we have $\mathfrak{m}=(x_1,\dots,x_i,x_{i+1},\dots,x_n)$. The dimension of $R/(x_1,\dots,x_i)$ is determined by the chain in $R$:
which has length $n-i$. That is, $\dim R/(x_1,\dots,x_i)=n-i$. On the other hand, the maximal ideal $\mathfrak{M}$ in $R/(x_1,\dots,x_i)$ is isomorphic to $(x_{i+1},\dots,x_n)$, and $x_{i+1},\dots,x_n$ map to a basis of $\mathfrak{M}/\mathfrak{M}^2$, which consequently has dimension $n-i$. $\square$
It looks quite promising now that the sequence of basis can get everything down to earth, and we will show that in the following section.
Proposition 3. If $R$ is regular, then $R$ is an integral domain.
Proof. We use induction on $\dim R$. When $\dim{R}=0$ and $R$ is regular, $R$ has to be a field, hence an integral domain by definition. Next we assume that $\dim{R}>0$ and the argument has been proved for $\dim{R}-1$.
Pick $x \in \mathfrak{m} \setminus \mathfrak{m}^2$. Then this element map to a nonzero element in $\mathfrak{m}/\mathfrak{m}^2$. There exists a basis of $\mathfrak{m}/\mathfrak{m}^2$ that contains $\overline{x}$. Therefore by proposition 2, $R/(x)$ is a regular local ring of dimension $\dim{R}-1$, which is an integral domain by assumption. It follows that $(x)$ is prime.
We claim that there exists $x \in \mathfrak{m}/\mathfrak{m}^2$ such that $(x)$ has height $1$. If not, then for all $x \in \mathfrak{m}/\mathfrak{m}^2$, $(x)$ is a minimal. It follows that there exists finitely many minimal prime ideals $\mathfrak{p}_1,\dots,\mathfrak{p}_r$ such that
and consequently $\mathfrak{m} \subset \mathfrak{p}_j$ for some $1 \le j \le r$. It follows that $\dim{R}=0$, contradicting our assumption that $\dim{R}>0$. [Note: the prime avoidance allows at most two ideals to be non-prime. See P. 90 of Eisenbud’s Commutative Algebra, with a View Toward Algebraic Geometry.]
Thus, as our claim is true, we can write $\mathfrak{p} \subsetneq (x)$ with $\mathfrak{p}$ prime and $x \in \mathfrak{m} \setminus \mathfrak{m}^2$. We see $\mathfrak{p} \in (x^n)$ for all $n$ because if $p=rx^n \in \mathfrak{p}$, then $r \in \mathfrak{q} \subset (x)$ and therefore we write $r=sx$ or equivalently $p = sx^{n+1} \in (x^{n+1})$. When this is the case, we have $\mathfrak{p} \subset \bigcap_{n=1}^{\infty}(x^n)=0$. Therefore $R/\mathfrak{p}=R/0=R$ is an integral domain.
We now reach our conclusion of this post.
Proposition 4. If $R$ is regular and of Krull dimension $n$, any $x_1,\dots,x_n \in \mathfrak{m}$ mapping to a basis of $\mathfrak{m}/\mathfrak{m}^2$ gives rise to a regular sequence ($R$-sequence). Hence $G(R)=\dim{R}$ and therefore $R$ is Cohen-Macaulay.
Proof. As $G(R) \le \dim{R}$, once we have shown that $[x_1,\dots,x_n]$ is a regular sequence, we have $G(R) \ge \dim{R}$. To show it being a regular sequence, first of all notice that $x_1$ is non-zero-divisor (because $R$ is an integral domain). For any $i>1$, we see $R/(x_1,\dots,x_i)$ is a regular local ring of dimension $d-i$, hence again an integral domain. Therefore $x_{i+1},\dots,x_i$ are non-zero-divisors. $\square$
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