Dimension theory (algebra)

Local rings

Fundamental theorem

Let ${\displaystyle (R,\mathfrak{m})}$ be a Noetherian local ring and ${\displaystyle I}$ a ${\displaystyle \mathfrak{m}}$-primary ideal (i.e., it sits between some power of ${\displaystyle \mathfrak{m}}$ and ${\displaystyle \mathfrak{m}}$). Let ${\displaystyle F(t)}$ be the Poincaré series of the associated graded ring ${\mathrm{gr}}_{I}R=\underset{0}{\overset{\mathrm{\infty }}{⨁}}{I}^n/I^{n+1}$. That is, $${\displaystyle F(t)=\sum _0^{\mathrm{\infty }}\ell (I^n/I^{n+1})t^n}$$ where ${\displaystyle \ell }$ refers to the length of a module (over an Artinian ring ${\displaystyle ({\mathrm{gr}}_{I}R{)}_0=R/I}$). If ${\displaystyle x_1,\dots ,x_s}$ generate ${\displaystyle I}$, then their image in ${\displaystyle I/I^2}$ have degree 1 and generate ${\displaystyle {\mathrm{gr}}_{I}R}$ as ${\displaystyle R/I}$-algebra. By the Hilbert–Serre theorem, ${\displaystyle F}$ is a rational function with exactly one pole at ${\displaystyle t=1}$ of order ${\displaystyle d\le s}$. Since $${\displaystyle (1-t{)}^{-d}=\sum _0^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{d-1+j}{d-1})t^j,}$$ we find that the coefficient of ${\displaystyle t^n}$ in ${\displaystyle F(t)=(1-t{)}^{d}F(t)(1-t{)}^{-d}}$ is of the form $${\displaystyle \sum _0^N{a}_k(\genfrac{}{}{0ex}{}{d-1+n-k}{d-1})=(1-t{)}^{d}F(t){|}_{t=1}\frac{n^{d-1}}{d-1!}+O(n^{d-2}).}$$ That is to say, ${\displaystyle \ell (I^n/I^{n+1})}$ is a polynomial ${\displaystyle P}$ in ${\displaystyle n}$ of degree ${\displaystyle d-1}$. ${\displaystyle P}$ is called the Hilbert polynomial of ${\displaystyle {\mathrm{gr}}_{I}R}$.

We set ${\displaystyle d(R)=d}$. We also set ${\displaystyle \delta (R)}$ to be the minimum number of elements of ${\displaystyle R}$ that can generate an ${\displaystyle \mathfrak{m}}$-primary ideal of ${\displaystyle R}$. Our ambition is to prove the fundamental theorem: $${\displaystyle \delta (R)=d(R)=\dim R.}$$ Since we can take ${\displaystyle s}$ to be ${\displaystyle \delta (R)}$, we already have ${\displaystyle \delta (R)\ge d(R)}$ from the above. Next we prove ${\displaystyle d(R)\ge \dim R}$ by induction on ${\displaystyle d(R)}$. Let ${\displaystyle {\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_m}$ be a chain of prime ideals in ${\displaystyle R}$. Let ${\displaystyle D=R/{\mathfrak{p}}_0}$ and ${\displaystyle x}$ a nonzero nonunit element in ${\displaystyle D}$. Since ${\displaystyle x}$ is not a zero-divisor, we have the exact sequence $${\displaystyle 0\to D\stackrel{x}{\to }D\to D/xD\to 0.}$$ The degree bound of the Hilbert-Samuel polynomial now implies that ${\displaystyle d(D)>d(D/xD)\ge d(R/{\mathfrak{p}}_1)}$. (This essentially follows from the Artin–Rees lemma; see Hilbert–Samuel function for the statement and the proof.) In ${\displaystyle R/{\mathfrak{p}}_1}$, the chain ${\displaystyle {\mathfrak{p}}_i}$ becomes a chain of length ${\displaystyle m-1}$ and so, by inductive hypothesis and again by the degree estimate, $${\displaystyle m-1\le \dim (R/{\mathfrak{p}}_1)\le d(R/{\mathfrak{p}}_1)\le d(D)-1\le d(R)-1.}$$ The claim follows. It now remains to show ${\displaystyle \dim R\ge \delta (R).}$ More precisely, we shall show:

Lemma—The maximal ideal ${\displaystyle \mathfrak{m}}$ contains elements ${\displaystyle x_1,\dots ,x_d}$, ${\displaystyle d}$ = Krull dimension of ${\displaystyle R}$, such that, for any ${\displaystyle i}$, any prime ideal containing ${\displaystyle (x_1,\dots ,x_i)}$ has height ${\displaystyle \ge i}$.

(Notice: ${\displaystyle (x_1,\dots ,x_d)}$ is then ${\displaystyle \mathfrak{m}}$-primary.) The proof is omitted. It appears, for example, in Atiyah–MacDonald. But it can also be supplied privately; the idea is to use prime avoidance.

Consequences of the fundamental theorem

Let ${\displaystyle (R,\mathfrak{m})}$ be a noetherian local ring and put ${\displaystyle k=R/\mathfrak{m}}$. Then

• ${\displaystyle \dim R\le {\dim }_k\mathfrak{m}/{\mathfrak{m}}^2}$, since a basis of ${\displaystyle \mathfrak{m}/{\mathfrak{m}}^2}$ lifts to a generating set of ${\displaystyle \mathfrak{m}}$ by Nakayama. If the equality holds, then ${\displaystyle R}$ is called a regular local ring.
• ${\displaystyle \dim \hat{R}=\dim R}$, since ${\displaystyle \mathrm{gr}R=\mathrm{gr}\hat{R}}$.
• (Krull's principal ideal theorem) The height of the ideal generated by elements ${\displaystyle x_1,\dots ,x_s}$ in a Noetherian ring is at most ${\displaystyle s}$. Conversely, a prime ideal of height ${\displaystyle s}$ is minimal over an ideal generated by ${\displaystyle s}$ elements. (Proof: Let ${\displaystyle \mathfrak{p}}$ be a prime ideal minimal over such an ideal. Then ${\displaystyle s\ge \dim R_{\mathfrak{p}}=\mathrm{ht}\mathfrak{p}}$. The converse was shown in the course of the proof of the fundamental theorem.)

Theorem—If ${\displaystyle A\to B}$ is a morphism of noetherian local rings, then $${\displaystyle \dim B/{\mathfrak{m}}_{A}B\ge \dim B-\dim A.}$$ The equality holds if ${\displaystyle A\to B}$ is flat or more generally if it has the going-down property.

Proof: Let ${\displaystyle x_1,\dots ,x_n}$ generate a ${\displaystyle {\mathfrak{m}}_A}$-primary ideal and ${\displaystyle y_1,\dots ,y_m}$ be such that their images generate a ${\displaystyle {\mathfrak{m}}_B/{\mathfrak{m}}_{A}B}$-primary ideal. Then ${\displaystyle {{\mathfrak{m}}_B}^s\subset (y_1,\dots ,y_m)+{\mathfrak{m}}_{A}B}$ for some ${\displaystyle s}$. Raising both sides to higher powers, we see some power of ${\displaystyle {\mathfrak{m}}_B}$ is contained in ${\displaystyle (y_1,\dots ,y_m,x_1,\dots ,x_n)}$; i.e., the latter ideal is ${\displaystyle {\mathfrak{m}}_B}$-primary; thus, ${\displaystyle m+n\ge \dim B}$. The equality is a straightforward application of the going-down property. Q.E.D.

Proposition—If ${\displaystyle R}$ is a noetherian ring, then $${\displaystyle \dim R+1=\dim R[x]=\dim R[[x]].}$$

Proof: If ${\displaystyle {\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊\cdots ⊊{\mathfrak{p}}_n}$ are a chain of prime ideals in ${\displaystyle R}$, then ${\displaystyle {\mathfrak{p}}_{i}R[x]}$ are a chain of prime ideals in ${\displaystyle R[x]}$ while ${\displaystyle {\mathfrak{p}}_{n}R[x]}$ is not a maximal ideal. Thus, ${\displaystyle \dim R+1\le \dim R[x]}$. For the reverse inequality, let ${\displaystyle \mathfrak{m}}$ be a maximal ideal of ${\displaystyle R[x]}$ and ${\displaystyle \mathfrak{p}=R\cap \mathfrak{m}}$. Clearly, ${\displaystyle R[x{]}_{\mathfrak{m}}=R_{\mathfrak{p}}[x{]}_{\mathfrak{m}}}$. Since ${\displaystyle R[x{]}_{\mathfrak{m}}/\mathfrak{p}{R}_{\mathfrak{p}}R[x{]}_{\mathfrak{m}}=(R_{\mathfrak{p}}/\mathfrak{p}{R}_{\mathfrak{p}})[x{]}_{\mathfrak{m}}}$ is then a localization of a principal ideal domain and has dimension at most one, we get ${\displaystyle 1+\dim R\ge 1+\dim R_{\mathfrak{p}}\ge \dim R[x{]}_{\mathfrak{m}}}$ by the previous inequality. Since ${\displaystyle \mathfrak{m}}$ is arbitrary, it follows ${\displaystyle 1+\dim R\ge \dim R[x]}$. Q.E.D.

Nagata's altitude formula

Theorem—Let ${\displaystyle R\subset R^{\prime }}$ be integral domains, ${\displaystyle {\mathfrak{p}}^{\prime }\subset R^{\prime }}$ be a prime ideal and ${\displaystyle \mathfrak{p}=R\cap {\mathfrak{p}}^{\prime }}$. If ${\displaystyle R}$ is a Noetherian ring, then $${\displaystyle \dim R_{{\mathfrak{p}}^{\prime }}^{\prime }+{\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_{R/\mathfrak{p}}{R}^{\prime }/{\mathfrak{p}}^{\prime }\le \dim R_{\mathfrak{p}}+{\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_R{R}^{\prime }}$$ where the equality holds if either (a) ${\displaystyle R}$ is universally catenary and ${\displaystyle R^{\prime }}$ is finitely generated ${\displaystyle R}$-algebra or (b) ${\displaystyle R^{\prime }}$ is a polynomial ring over ${\displaystyle R}$.

Proof: First suppose ${\displaystyle R^{\prime }}$ is a polynomial ring. By induction on the number of variables, it is enough to consider the case ${\displaystyle R^{\prime }=R[x]}$. Since ${\displaystyle R^{\prime }}$ is flat over ${\displaystyle R}$, $${\displaystyle \dim R_{{\mathfrak{p}}^{\prime }}^{\prime }=\dim R_{\mathfrak{p}}+\dim \kappa (\mathfrak{p}){\otimes }_R{R^{\prime }}_{{\mathfrak{p}}^{\prime }}.}$$ By Noether's normalization lemma, the second term on the right side is: $${\displaystyle \dim \kappa (\mathfrak{p}){\otimes }_R{R}^{\prime }-\dim \kappa (\mathfrak{p}){\otimes }_R{R}^{\prime }/{\mathfrak{p}}^{\prime }=1-{\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_{\kappa (\mathfrak{p})}\kappa ({\mathfrak{p}}^{\prime })={\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_R{R}^{\prime }-\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}\kappa ({\mathfrak{p}}^{\prime }).}$$ Next, suppose ${\displaystyle R^{\prime }}$ is generated by a single element; thus, ${\displaystyle R^{\prime }=R[x]/I}$. If ${\displaystyle I}$ = 0, then we are already done. Suppose not. Then ${\displaystyle R^{\prime }}$ is algebraic over ${\displaystyle R}$ and so ${\displaystyle {\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_R{R}^{\prime }=0}$. Since ${\displaystyle R}$ is a subring of ${\displaystyle R^{\prime }}$, ${\displaystyle I\cap R=0}$ and so ${\displaystyle \mathrm{ht}I=\dim R[x{]}_I=\dim Q(R)[x{]}_I=1-{\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_{Q(R)}\kappa (I)=1}$ since ${\displaystyle \kappa (I)=Q(R^{\prime })}$ is algebraic over ${\displaystyle Q(R)}$. Let ${\displaystyle {\mathfrak{p}}^{\mathrm{\prime }c}}$ denote the pre-image in ${\displaystyle R[x]}$ of ${\displaystyle {\mathfrak{p}}^{\prime }}$. Then, as ${\displaystyle \kappa ({\mathfrak{p}}^{\mathrm{\prime }c})=\kappa (\mathfrak{p})}$, by the polynomial case, $${\displaystyle \mathrm{ht}{\mathfrak{p}}^{\prime }=\mathrm{ht}{\mathfrak{p}}^{\mathrm{\prime }c}/I\le \mathrm{ht}{\mathfrak{p}}^{\mathrm{\prime }c}-\mathrm{ht}I=\dim R_{\mathfrak{p}}-{\mathrm{t}\mathrm{r}.\mathrm{d}\mathrm{e}\mathrm{g}}_{\kappa (\mathfrak{p})}\kappa ({\mathfrak{p}}^{\prime }).}$$ Here, note that the inequality is the equality if ${\displaystyle R^{\prime }}$ is catenary. Finally, working with a chain of prime ideals, it is straightforward to reduce the general case to the above case. Q.E.D.

Homological methods

Regular rings

Let ${\displaystyle R}$ be a noetherian ring. The projective dimension of a finite ${\displaystyle R}$-module ${\displaystyle M}$ is the shortest length of any projective resolution of ${\displaystyle M}$ (possibly infinite) and is denoted by ${\displaystyle {\mathrm{pd}}_{R}M}$. We set ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R=sup\{{\mathrm{pd}}_{R}M\mid M\text{ is a finite module}\}}$; it is called the global dimension of ${\displaystyle R}$.

Assume ${\displaystyle R}$ is local with residue field ${\displaystyle k}$.

Lemma—${\displaystyle {\mathrm{pd}}_{R}k=\mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R}$ (possibly infinite).

Proof: We claim: for any finite ${\displaystyle R}$-module ${\displaystyle M}$, $${\displaystyle {\mathrm{pd}}_{R}M\le n\iff {\mathrm{Tor}}_{n+1}^R(M,k)=0.}$$ By dimension shifting (cf. the proof of Theorem of Serre below), it is enough to prove this for ${\displaystyle n=0}$. But then, by the local criterion for flatness, ${\displaystyle {\mathrm{Tor}}_1^R(M,k)=0\Rightarrow M\text{ flat }\Rightarrow M\text{ free }\Rightarrow {\mathrm{pd}}_R(M)\le 0.}$ Now, $${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R\le n\Rightarrow {\mathrm{pd}}_{R}k\le n\Rightarrow {\mathrm{Tor}}_{n+1}^R(-,k)=0\Rightarrow {\mathrm{pd}}_R-\le n\Rightarrow \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R\le n,}$$ completing the proof. Q.E.D.

Remark: The proof also shows that ${\displaystyle {\mathrm{pd}}_{R}K={\mathrm{pd}}_{R}M-1}$ if ${\displaystyle M}$ is not free and ${\displaystyle K}$ is the kernel of some surjection from a free module to ${\displaystyle M}$.

Lemma—Let ${\displaystyle R_1=R/fR}$, ${\displaystyle f}$ a non-zerodivisor of ${\displaystyle R}$. If ${\displaystyle f}$ is a non-zerodivisor on ${\displaystyle M}$, then $${\displaystyle {\mathrm{pd}}_{R}M\ge {\mathrm{pd}}_{R_1}(M\otimes R_1).}$$

Proof: If ${\displaystyle {\mathrm{pd}}_{R}M=0}$, then ${\displaystyle M}$ is ${\displaystyle R}$-free and thus ${\displaystyle M\otimes R_1}$ is ${\displaystyle R_1}$-free. Next suppose ${\displaystyle {\mathrm{pd}}_{R}M>0}$. Then we have: ${\displaystyle {\mathrm{pd}}_{R}K={\mathrm{pd}}_{R}M-1}$ as in the remark above. Thus, by induction, it is enough to consider the case ${\displaystyle {\mathrm{pd}}_{R}M=1}$. Then there is a projective resolution: ${\displaystyle 0\to P_1\to P_0\to M\to 0}$, which gives: $${\displaystyle {\mathrm{Tor}}_1^R(M,R_1)\to P_1\otimes R_1\to P_0\otimes R_1\to M\otimes R_1\to 0.}$$ But ${\displaystyle {\mathrm{Tor}}_1^R(M,R_1)={}_{f}M=\{m\in M\mid fm=0\}=0.}$ Hence, ${\displaystyle {\mathrm{pd}}_R(M\otimes R_1)}$ is at most 1. Q.E.D.

Theorem of Serre—${\displaystyle R}$ regular ${\displaystyle \iff \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R<\mathrm{\infty }\iff \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R=\dim R.}$

Proof: If ${\displaystyle R}$ is regular, we can write ${\displaystyle k=R/(f_1,\dots ,f_n)}$, ${\displaystyle f_i}$ a regular system of parameters. An exact sequence ${\displaystyle 0\to M\stackrel{f}{\to }M\to M_1\to 0}$, some ${\displaystyle f}$ in the maximal ideal, of finite modules, ${\displaystyle {\mathrm{pd}}_{R}M<\mathrm{\infty }}$, gives us: $${\displaystyle 0={\mathrm{Tor}}_{i+1}^R(M,k)\to {\mathrm{Tor}}_{i+1}^R(M_1,k)\to {\mathrm{Tor}}_i^R(M,k)\stackrel{f}{\to }{\mathrm{Tor}}_i^R(M,k),i\ge {\mathrm{pd}}_{R}M.}$$ But ${\displaystyle f}$ here is zero since it kills ${\displaystyle k}$. Thus, ${\displaystyle {\mathrm{Tor}}_{i+1}^R(M_1,k)\simeq {\mathrm{Tor}}_i^R(M,k)}$ and consequently ${\displaystyle {\mathrm{pd}}_R{M}_1=1+{\mathrm{pd}}_{R}M}$. Using this, we get: $${\displaystyle {\mathrm{pd}}_{R}k=1+{\mathrm{pd}}_R(R/(f_1,\dots ,f_{n-1}))=\cdots =n.}$$ The proof of the converse is by induction on ${\displaystyle \dim R}$. We begin with the inductive step. Set ${\displaystyle R_1=R/f_{1}R}$, ${\displaystyle f_1}$ among a system of parameters. To show ${\displaystyle R}$ is regular, it is enough to show ${\displaystyle R_1}$ is regular. But, since ${\displaystyle \dim R_1<\dim R}$, by inductive hypothesis and the preceding lemma with ${\displaystyle M=\mathfrak{m}}$, $${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R<\mathrm{\infty }\Rightarrow \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}{R}_1={\mathrm{pd}}_{R_1}k\le {\mathrm{pd}}_{R_1}\mathfrak{m}/f_1\mathfrak{m}<\mathrm{\infty }\Rightarrow R_1\text{ regular}.}$$

The basic step remains. Suppose ${\displaystyle \dim R=0}$. We claim ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R=0}$ if it is finite. (This would imply that ${\displaystyle R}$ is a semisimple local ring; i.e., a field.) If that is not the case, then there is some finite module ${\displaystyle M}$ with ${\displaystyle 0<{\mathrm{pd}}_{R}M<\mathrm{\infty }}$ and thus in fact we can find ${\displaystyle M}$ with ${\displaystyle {\mathrm{pd}}_{R}M=1}$. By Nakayama's lemma, there is a surjection ${\displaystyle F\to M}$ from a free module ${\displaystyle F}$ to ${\displaystyle M}$ whose kernel ${\displaystyle K}$ is contained in ${\displaystyle \mathfrak{m}F}$. Since ${\displaystyle \dim R=0}$, the maximal ideal ${\displaystyle \mathfrak{m}}$ is an associated prime of ${\displaystyle R}$; i.e., ${\displaystyle \mathfrak{m}=\mathrm{ann}(s)}$ for some nonzero ${\displaystyle s}$ in ${\displaystyle R}$. Since ${\displaystyle K\subset \mathfrak{m}F}$, ${\displaystyle sK=0}$. Since ${\displaystyle K}$ is not zero and is free, this implies ${\displaystyle s=0}$, which is absurd. Q.E.D.

Corollary—A regular local ring is a unique factorization domain.

Proof: Let ${\displaystyle R}$ be a regular local ring. Then ${\displaystyle \mathrm{gr}R\simeq k[x_1,\dots ,x_d]}$, which is an integrally closed domain. It is a standard algebra exercise to show this implies that ${\displaystyle R}$ is an integrally closed domain. Now, we need to show every divisorial ideal is principal; i.e., the divisor class group of ${\displaystyle R}$ vanishes. But, according to Bourbaki, Algèbre commutative, chapitre 7, §. 4. Corollary 2 to Proposition 16, a divisorial ideal is principal if it admits a finite free resolution, which is indeed the case by the theorem. Q.E.D.

Theorem—Let ${\displaystyle R}$ be a ring. Then $${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R[x_1,\dots ,x_n]=\mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R+n.}$$

Depth

Let ${\displaystyle R}$ be a ring and ${\displaystyle M}$ a module over it. A sequence of elements ${\displaystyle x_1,\dots ,x_n}$ in ${\displaystyle R}$ is called an ${\displaystyle M}$-regular sequence if ${\displaystyle x_1}$ is not a zero-divisor on ${\displaystyle M}$ and ${\displaystyle x_i}$ is not a zero divisor on ${\displaystyle M/(x_1,\dots ,x_{i-1})M}$ for each ${\displaystyle i=2,\dots ,n}$. A priori, it is not obvious whether any permutation of a regular sequence is still regular (see the section below for some positive answer).

Let ${\displaystyle R}$ be a local Noetherian ring with maximal ideal ${\displaystyle \mathfrak{m}}$ and put ${\displaystyle k=R/\mathfrak{m}}$. Then, by definition, the depth of a finite ${\displaystyle R}$-module ${\displaystyle M}$ is the supremum of the lengths of all ${\displaystyle M}$-regular sequences in ${\displaystyle \mathfrak{m}}$. For example, we have ${\displaystyle \mathrm{depth}M=0\iff \mathfrak{m}}$ consists of zerodivisors on ${\displaystyle M}$ ${\displaystyle \iff \mathfrak{m}}$ is associated with ${\displaystyle M}$. By induction, we find $${\displaystyle \mathrm{depth}M\le \dim R/\mathfrak{p}}$$ for any associated primes ${\displaystyle \mathfrak{p}}$ of ${\displaystyle M}$. In particular, ${\displaystyle \mathrm{depth}M\le \dim M}$. If the equality holds for ${\displaystyle M}$ = ${\displaystyle R}$, ${\displaystyle R}$ is called a Cohen–Macaulay ring.

Example: A regular Noetherian local ring is Cohen–Macaulay (since a regular system of parameters is an ${\displaystyle R}$-regular sequence).

In general, a Noetherian ring is called a Cohen–Macaulay ring if the localizations at all maximal ideals are Cohen–Macaulay. We note that a Cohen–Macaulay ring is universally catenary. This implies for example that a polynomial ring ${\displaystyle k[x_1,\dots ,x_d]}$ is universally catenary since it is regular and thus Cohen–Macaulay.

Proposition (Rees)—Let ${\displaystyle M}$ be a finite ${\displaystyle R}$-module. Then the Ext functor satisfies ${\displaystyle \mathrm{depth}M=sup\{n\mid {\mathrm{Ext}}_R^i(k,M)=0,i<n\}}$.

More generally, for any finite ${\displaystyle R}$-module ${\displaystyle N}$ whose support is exactly ${\displaystyle \{\mathfrak{m}\}}$, $${\displaystyle \mathrm{depth}M=sup\{n\mid {\mathrm{Ext}}_R^i(N,M)=0,i<n\}.}$$

Proof: We first prove by induction on ${\displaystyle n}$ the following statement: for every ${\displaystyle R}$-module ${\displaystyle M}$ and every ${\displaystyle M}$-regular sequence ${\displaystyle x_1,\dots ,x_n}$ in ${\displaystyle \mathfrak{m}}$,

The basic step ${\displaystyle n}$ = 0 is trivial. Next, by inductive hypothesis, ${\displaystyle {\mathrm{Ext}}_R^{n-1}(N,M)\simeq {\mathrm{Hom}}_R(N,M/(x_1,\dots ,x_{n-1})M)}$. But the latter is zero since the annihilator of ${\displaystyle N}$ contains some power of ${\displaystyle x_n}$. Thus, from the exact sequence ${\displaystyle 0\to M\stackrel{x_1}{\to }M\to M_1\to 0}$ and the fact that ${\displaystyle x_1}$ kills ${\displaystyle N}$, using the inductive hypothesis again, we get $${\displaystyle {\mathrm{Ext}}_R^n(N,M)\simeq {\mathrm{Ext}}_R^{n-1}(N,M/x_{1}M)\simeq {\mathrm{Hom}}_R(N,M/(x_1,\dots ,x_n)M),}$$ proving (⁎). Now, if ${\displaystyle n<\mathrm{depth}M}$, then we can find an ${\displaystyle M}$-regular sequence of length more than ${\displaystyle n}$ and so by (⁎) we see ${\displaystyle {\mathrm{Ext}}_R^n(N,M)=0}$. It remains to show ${\displaystyle {\mathrm{Ext}}_R^n(N,M)\ne 0}$ if ${\displaystyle n=\mathrm{depth}M}$. By (⁎) we can assume ${\displaystyle n}$ = 0. Then ${\displaystyle \mathfrak{m}}$ is associated with ${\displaystyle M}$; thus is in the support of ${\displaystyle M}$. On the other hand, ${\displaystyle \mathfrak{m}\in \mathrm{Supp}(N).}$ It follows by linear algebra that there is a nonzero homomorphism from ${\displaystyle N}$ to ${\displaystyle M}$ modulo ${\displaystyle \mathfrak{m}}$; hence, one from ${\displaystyle N}$ to ${\displaystyle M}$ by Nakayama's lemma. Q.E.D.

The Auslander–Buchsbaum formula relates depth and projective dimension.

Theorem—Let ${\displaystyle M}$ be a finite module over a noetherian local ring ${\displaystyle R}$. If ${\displaystyle {\mathrm{pd}}_{R}M<\mathrm{\infty }}$, then $${\displaystyle {\mathrm{pd}}_{R}M+\mathrm{depth}M=\mathrm{depth}R.}$$

Proof: We argue by induction on ${\displaystyle {\mathrm{pd}}_{R}M}$, the basic case (i.e., ${\displaystyle M}$ free) being trivial. By Nakayama's lemma, we have the exact sequence ${\displaystyle 0\to K\stackrel{f}{\to }F\to M\to 0}$ where ${\displaystyle F}$ is free and the image of ${\displaystyle f}$ is contained in ${\displaystyle \mathfrak{m}F}$. Since ${\displaystyle {\mathrm{pd}}_{R}K={\mathrm{pd}}_{R}M-1,}$ what we need to show is ${\displaystyle \mathrm{depth}K=\mathrm{depth}M+1}$. Since ${\displaystyle f}$ kills ${\displaystyle k}$, the exact sequence yields: for any ${\displaystyle i}$, $${\displaystyle {\mathrm{Ext}}_R^i(k,F)\to {\mathrm{Ext}}_R^i(k,M)\to {\mathrm{Ext}}_R^{i+1}(k,K)\to 0.}$$ Note the left-most term is zero if ${\displaystyle i<\mathrm{depth}R}$. If ${\displaystyle i<\mathrm{depth}K-1}$, then since ${\displaystyle \mathrm{depth}K\le \mathrm{depth}R}$ by inductive hypothesis, we see ${\displaystyle {\mathrm{Ext}}_R^i(k,M)=0.}$ If ${\displaystyle i=\mathrm{depth}K-1}$, then ${\displaystyle {\mathrm{Ext}}_R^{i+1}(k,K)\ne 0}$ and it must be ${\displaystyle {\mathrm{Ext}}_R^i(k,M)\ne 0.}$ Q.E.D.

As a matter of notation, for any ${\displaystyle R}$-module ${\displaystyle M}$, we let $${\displaystyle {\mathrm{\Gamma }}_{\mathfrak{m}}(M)=\{s\in M\mid \mathrm{supp}(s)\subset \{\mathfrak{m}\}\}=\{s\in M\mid {\mathfrak{m}}^{j}s=0\text{ for some }j\}.}$$ One sees without difficulty that ${\displaystyle {\mathrm{\Gamma }}_{\mathfrak{m}}}$ is a left-exact functor and then let ${\displaystyle H_{\mathfrak{m}}^j=R^j{\mathrm{\Gamma }}_{\mathfrak{m}}}$ be its ${\displaystyle j}$-th right derived functor, called the local cohomology of ${\displaystyle R}$. Since ${\displaystyle {\mathrm{\Gamma }}_{\mathfrak{m}}(M)=\underrightarrow{\lim }{\mathrm{Hom}}_R(R/{\mathfrak{m}}^j,M)}$, via abstract nonsense, $${\displaystyle H_{\mathfrak{m}}^i(M)=\underrightarrow{\lim }{\mathrm{Ext}}_R^i(R/{\mathfrak{m}}^j,M).}$$ This observation proves the first part of the theorem below.

Theorem (Grothendieck)—Let ${\displaystyle M}$ be a finite ${\displaystyle R}$-module. Then

1. ${\displaystyle \mathrm{depth}\mathrm{M}=sup\{n\mid H_{\mathfrak{m}}^i(M)=0,i<n\}}$.
2. ${\displaystyle H_{\mathfrak{m}}^i(M)=0,i>\dim M}$ and ${\displaystyle \ne 0}$ if ${\displaystyle i=\dim M.}$
3. If ${\displaystyle R}$ is complete and ${\displaystyle d}$ its Krull dimension and if ${\displaystyle E}$ is the injective hull of ${\displaystyle k}$, then $${\displaystyle {\mathrm{Hom}}_R(H_{\mathfrak{m}}^d(-),E)}$$ is representable (the representing object is sometimes called the canonical module especially if ${\displaystyle R}$ is Cohen–Macaulay).

Proof: 1. is already noted (except to show the nonvanishing at the degree equal to the depth of ${\displaystyle M}$; use induction to see this) and 3. is a general fact by abstract nonsense. 2. is a consequence of an explicit computation of a local cohomology by means of Koszul complexes (see below). ${\displaystyle ◻}$

Koszul complex

Let ${\displaystyle R}$ be a ring and ${\displaystyle x}$ an element in it. We form the chain complex ${\displaystyle K}$(${\displaystyle x}$) given by ${\displaystyle K(x{)}_i=R}$ for ${\displaystyle i}$ = 0, 1 and ${\displaystyle K(x{)}_i=0}$ for any other ${\displaystyle i}$ with the differential $${\displaystyle d:K_1(R)\to K_0(R),r\mapsto xr.}$$ For any ${\displaystyle R}$-module ${\displaystyle M}$, we then get the complex ${\displaystyle K(x,M)=K(x){\otimes }_{R}M}$ with the differential ${\displaystyle d\otimes 1}$ and let ${\displaystyle {\mathrm{H}}_{\ast }(x,M)={\mathrm{H}}_{\ast }(K(x,M))}$ be its homology. Note: $${\displaystyle {\mathrm{H}}_0(x,M)=M/xM,}$$ $${\displaystyle {\mathrm{H}}_1(x,M)={}_{x}M=\{m\in M\mid xm=0\}.}$$

More generally, given a finite sequence ${\displaystyle x_1,\dots ,x_n}$ of elements in a ring ${\displaystyle R}$, we form the tensor product of complexes: $${\displaystyle K(x_1,\dots ,x_n)=K(x_1)\otimes \cdots \otimes K(x_n)}$$ and let ${\displaystyle {\mathrm{H}}_{\ast }(x_1,\dots ,x_n,M)={\mathrm{H}}_{\ast }(K(x_1,\dots ,x_n,M))}$ its homology. As before, $${\displaystyle {\mathrm{H}}_0(\underset{\_}{x},M)=M/(x_1,\dots ,x_n)M,}$$ $${\displaystyle {\mathrm{H}}_n(\underset{\_}{x},M)={\mathrm{Ann}}_M((x_1,\dots ,x_n)).}$$

We now have the homological characterization of a regular sequence.

Theorem—Suppose ${\displaystyle R}$ is Noetherian, ${\displaystyle M}$ is a finite module over ${\displaystyle R}$ and ${\displaystyle x_i}$ are in the Jacobson radical of ${\displaystyle R}$. Then the following are equivalent

1. ${\displaystyle \underset{\_}{x}}$ is an ${\displaystyle M}$-regular sequence.
2. ${\displaystyle {\mathrm{H}}_i(\underset{\_}{x},M)=0,i\ge 1}$.
3. ${\displaystyle {\mathrm{H}}_1(\underset{\_}{x},M)=0}$.

Corollary—The sequence ${\displaystyle x_i}$ is ${\displaystyle M}$-regular if and only if any of its permutations is so.

Corollary—If ${\displaystyle x_1,\dots ,x_n}$ is an ${\displaystyle M}$-regular sequence, then ${\displaystyle x_1^j,\dots ,x_n^j}$ is also an ${\displaystyle M}$-regular sequence for each positive integer ${\displaystyle j}$.

A Koszul complex is a powerful computational tool. For instance, it follows from the theorem and the corollary $${\displaystyle {\mathrm{H}}_{\mathfrak{m}}^i(M)\simeq \underrightarrow{\lim }{\mathrm{H}}^i(K(x_1^j,\dots ,x_n^j;M))}$$ (Here, one uses the self-duality of a Koszul complex; see Proposition 17.15. of Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.)

Another instance would be

Theorem—Assume ${\displaystyle R}$ is local. Then let $${\displaystyle s={\dim }_k\mathfrak{m}/{\mathfrak{m}}^2,}$$ the dimension of the Zariski tangent space (often called the embedding dimension of ${\displaystyle R}$). Then $${\displaystyle (\genfrac{}{}{0ex}{}{s}{i})\le {\dim }_k{\mathrm{Tor}}_i^R(k,k).}$$

Remark: The theorem can be used to give a second quick proof of Serre's theorem, that ${\displaystyle R}$ is regular if and only if it has finite global dimension. Indeed, by the above theorem, ${\displaystyle {\mathrm{Tor}}_s^R(k,k)\ne 0}$ and thus ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R\ge s}$. On the other hand, as ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R={\mathrm{pd}}_{R}k}$, the Auslander–Buchsbaum formula gives ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R=\dim R}$. Hence, ${\displaystyle \dim R\le s\le \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R=\dim R}$.

We next use a Koszul homology to define and study complete intersection rings. Let ${\displaystyle R}$ be a Noetherian local ring. By definition, the first deviation of ${\displaystyle R}$ is the vector space dimension $${\displaystyle {ϵ}_1(R)={\dim }_k{\mathrm{H}}_1(\underset{\_}{x})}$$ where ${\displaystyle \underset{\_}{x}=(x_1,\dots ,x_d)}$ is a system of parameters. By definition, ${\displaystyle R}$ is a complete intersection ring if ${\displaystyle \dim R+{ϵ}_1(R)}$ is the dimension of the tangent space. (See Hartshorne for a geometric meaning.)

Theorem—${\displaystyle R}$ is a complete intersection ring if and only if its Koszul algebra is an exterior algebra.

Injective dimension and Tor dimensions

Let ${\displaystyle R}$ be a ring. The injective dimension of an ${\displaystyle R}$-module ${\displaystyle M}$ denoted by ${\displaystyle {\mathrm{id}}_{R}M}$ is defined just like a projective dimension: it is the minimal length of an injective resolution of ${\displaystyle M}$. Let ${\displaystyle {\mathrm{Mod}}_R}$ be the category of ${\displaystyle R}$-modules.

Theorem—For any ring ${\displaystyle R}$, $${\displaystyle \begin{array}{cc}\mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R& =\sup \{{\mathrm{id}}_{R}M\mid M\in {\mathrm{Mod}}_R\}\\ & =inf\{n\mid {\mathrm{Ext}}_R^i(M,N)=0,i>n,M,N\in {\mathrm{Mod}}_R\}\end{array}}$$

Proof: Suppose ${\displaystyle \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R\le n}$. Let ${\displaystyle M}$ be an ${\displaystyle R}$-module and consider a resolution $${\displaystyle 0\to M\to I_0\stackrel{{\varphi }_0}{\to }{I}_1\to \cdots \to I_{n-1}\stackrel{{\varphi }_{n-1}}{\to }N\to 0}$$ where ${\displaystyle I_i}$ are injective modules. For any ideal ${\displaystyle I}$, $${\displaystyle {\mathrm{Ext}}_R^1(R/I,N)\simeq {\mathrm{Ext}}_R^2(R/I,\ker ({\varphi }_{n-1}))\simeq \cdots \simeq {\mathrm{Ext}}_R^{n+1}(R/I,M),}$$ which is zero since ${\displaystyle {\mathrm{Ext}}_R^{n+1}(R/I,-)}$ is computed via a projective resolution of ${\displaystyle R/I}$. Thus, by Baer's criterion, ${\displaystyle N}$ is injective. We conclude that ${\displaystyle sup\{{\mathrm{id}}_{R}M|M\}\le n}$. Essentially by reversing the arrows, one can also prove the implication in the other way. Q.E.D.

The theorem suggests that we consider a sort of a dual of a global dimension: $${\displaystyle \mathrm{w}.\mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}=inf\{n\mid {\mathrm{Tor}}_i^R(M,N)=0,i>n,M,N\in {\mathrm{Mod}}_R\}.}$$ It was originally called the weak global dimension of ${\displaystyle R}$ but today it is more commonly called the Tor dimension of ${\displaystyle R}$.

Remark: for any ring ${\displaystyle R}$, ${\displaystyle \mathrm{w}.\mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R\le \mathrm{g}\mathrm{l}.\mathrm{d}\mathrm{i}\mathrm{m}R}$.

Proposition—A ring has weak global dimension zero if and only if it is von Neumann regular.

Dimensions of non-commutative rings

Let ${\displaystyle A}$ be a graded algebra over a field ${\displaystyle k}$. If ${\displaystyle V}$ is a finite-dimensional generating subspace of ${\displaystyle A}$, then we let ${\displaystyle f(n)={\dim }_k{V}^n}$ and then put $${\displaystyle \mathrm{gk}(A)=\underset{n\to \mathrm{\infty }}{lim sup}\frac{\log f(n)}{\log n}.}$$ It is called the Gelfand–Kirillov dimension of ${\displaystyle A}$. It is easy to show ${\displaystyle \mathrm{gk}(A)}$ is independent of a choice of ${\displaystyle V}$. Given a graded right (or left) module ${\displaystyle M}$ over ${\displaystyle A}$ one may similarly define the Gelfand-Kirillov dimension ${\displaystyle gk(M)}$ of ${\displaystyle M}$.

Example: If ${\displaystyle A}$ is finite-dimensional, then gk(${\displaystyle A}$) = 0. If ${\displaystyle A}$ is an affine ring, then gk(${\displaystyle A}$) = Krull dimension of ${\displaystyle A}$.

Example: If ${\displaystyle A_n=k[x_1,...,x_n,{\mathrm{\partial }}_1,...,{\mathrm{\partial }}_n]}$ is the n-th Weyl algebra then ${\displaystyle \mathrm{gk}(A_n)=2n.}$