Dimension theory (algebra)

Let ${\displaystyle (R,{\mathfrak {m}})}$ be a noetherian local ring and 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 ${\textstyle \operatorname {gr} _{I}R=\bigoplus _{0}^{\infty }I^{n}/I^{n+1}}$. That is,

$${\displaystyle F(t)=\sum _{0}^{\infty }\ell (I^{n}/I^{n+1})t^{n}}$$

where ${\displaystyle \ell }$ refers to the length of a module (over an artinian ring ${\displaystyle (\operatorname {gr} _{I}R)_{0}=R/I}$). If ${\displaystyle x_{1},\dots ,x_{s}}$ generate I, then their image in ${\displaystyle I/I^{2}}$ have degree 1 and generate ${\displaystyle \operatorname {gr} _{I}R}$ as ${\displaystyle R/I}$-algebra. By the Hilbert–Serre theorem, F is a rational function with exactly one pole at ${\displaystyle t=1}$ of order ${\displaystyle d\leq s}$. Since

$${\displaystyle (1-t)^{-d}=\sum _{0}^{\infty }{\binom {d-1+j}{d-1}}t^{j},}$$

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

$${\displaystyle \sum _{0}^{N}a_{k}{\binom {d-1+n-k}{d-1}}=(1-t)^{d}F(t){\big |}_{t=1}{n^{d-1} \over {d-1}!}+O(n^{d-2}).}$$

That is to say, ${\displaystyle \ell (I^{n}/I^{n+1})}$ is a polynomial ${\displaystyle P}$ in n of degree ${\displaystyle d-1}$. P is called the Hilbert polynomial of ${\displaystyle \operatorname {gr} _{I}R}$.

We set ${\displaystyle d(R)=d}$. We also set ${\displaystyle \delta (R)}$ to be the minimum number of elements of R that can generate an ${\displaystyle {\mathfrak {m}}}$-primary ideal of R. Our ambition is to prove the fundamental theorem:

$${\displaystyle \delta (R)=d(R)=\dim R.}$$

Since we can take s to be ${\displaystyle \delta (R)}$, we already have ${\displaystyle \delta (R)\geq d(R)}$ from the above. Next we prove ${\displaystyle d(R)\geq \dim R}$ by induction on ${\displaystyle d(R)}$. Let ${\displaystyle {\mathfrak {p}}_{0}\subsetneq \cdots \subsetneq {\mathfrak {p}}_{m}}$ be a chain of prime ideals in R. Let ${\displaystyle D=R/{\mathfrak {p}}_{0}}$ and x a nonzero nonunit element in D. Since x is not a zero-divisor, we have the exact sequence

$${\displaystyle 0\to D{\overset {x}{\to }}D\to D/xD\to 0.}$$

The degree bound of the Hilbert-Samuel polynomial now implies that ${\displaystyle d(D)>d(D/xD)\geq d(R/{\mathfrak {p}}_{1})}$. (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,

$${\displaystyle m-1\leq \dim(R/{\mathfrak {p}}_{1})\leq d(R/{\mathfrak {p}}_{1})\leq d(D)-1\leq d(R)-1.}$$

The claim follows. It now remains to show ${\displaystyle \dim R\geq \delta (R).}$ More precisely, we shall show:

(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.

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

• ${\displaystyle \dim R\leq \dim _{k}{\mathfrak {m}}/{\mathfrak {m}}^{2}}$, 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 R is called a regular local ring.
• ${\displaystyle \dim {\widehat {R}}=\dim R}$, since ${\displaystyle \operatorname {gr} R=\operatorname {gr} {\widehat {R}}}$.
• (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 s. Conversely, a prime ideal of height s is minimal over an ideal generated by s elements. (Proof: Let ${\displaystyle {\mathfrak {p}}}$ be a prime ideal minimal over such an ideal. Then ${\displaystyle s\geq \dim R_{\mathfrak {p}}=\operatorname {ht} {\mathfrak {p}}}$. 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[1]

$${\displaystyle \dim B/{\mathfrak {m}}_{A}B\geq \dim B-\dim A.}$$

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 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, ${\displaystyle m+n\geq \dim B}$. The equality is a straightforward application of the going-down property. Q.E.D.

Proposition — If R is a noetherian ring, then

$${\displaystyle \dim R+1=\dim R[x]=\dim R[\![x]\!].}$$

Proof: If ${\displaystyle {\mathfrak {p}}_{0}\subsetneq {\mathfrak {p}}_{1}\subsetneq \cdots \subsetneq {\mathfrak {p}}_{n}}$ are a chain of prime ideals in 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, ${\displaystyle \dim R+1\leq \dim R[x]}$. 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 ${\displaystyle 1+\dim R\geq 1+\dim R_{\mathfrak {p}}\geq \dim R[x]_{\mathfrak {m}}}$ by the previous inequality. Since ${\displaystyle {\mathfrak {m}}}$ is arbitrary, it follows ${\displaystyle 1+\dim R\geq \dim R[x]}$. Q.E.D.

Theorem — Let ${\displaystyle R\subset R'}$ be integral domains, ${\displaystyle {\mathfrak {p}}'\subset R'}$ be a prime ideal and ${\displaystyle {\mathfrak {p}}=R\cap {\mathfrak {p}}'}$. If R is a Noetherian ring, then

$${\displaystyle \dim R'_{{\mathfrak {p}}'}+\operatorname {tr.deg} _{R/{\mathfrak {p}}}{R'/{\mathfrak {p}}'}\leq \dim R_{\mathfrak {p}}+\operatorname {tr.deg} _{R}{R'}}$$

where the equality holds if either (a) R is universally catenary and R' is finitely generated R-algebra or (b) R' is a polynomial ring over R.

Proof:[2] First suppose ${\displaystyle R'}$ is a polynomial ring. By induction on the number of variables, it is enough to consider the case ${\displaystyle R'=R[x]}$. Since R' is flat over R,

$${\displaystyle \dim R'_{\mathfrak {p'}}=\dim R_{\mathfrak {p}}+\dim \kappa ({\mathfrak {p}})\otimes _{R}{R'}_{{\mathfrak {p}}'}.}$$

By Noether's normalization lemma, the second term on the right side is:

$${\displaystyle \dim \kappa ({\mathfrak {p}})\otimes _{R}R'-\dim \kappa ({\mathfrak {p}})\otimes _{R}R'/{\mathfrak {p}}'=1-\operatorname {tr.deg} _{\kappa ({\mathfrak {p}})}\kappa ({\mathfrak {p}}')=\operatorname {tr.deg} _{R}R'-\operatorname {tr.deg} \kappa ({\mathfrak {p}}').}$$

Next, suppose ${\displaystyle R'}$ is generated by a single element; thus, ${\displaystyle R'=R[x]/I}$. If I = 0, then we are already done. Suppose not. Then ${\displaystyle R'}$ is algebraic over R and so ${\displaystyle \operatorname {tr.deg} _{R}R'=0}$. Since R is a subring of R', ${\displaystyle I\cap R=0}$ and so ${\displaystyle \operatorname {ht} I=\dim R[x]_{I}=\dim Q(R)[x]_{I}=1-\operatorname {tr.deg} _{Q(R)}\kappa (I)=1}$ since ${\displaystyle \kappa (I)=Q(R')}$ is algebraic over ${\displaystyle Q(R)}$. Let ${\displaystyle {\mathfrak {p}}^{\prime c}}$ denote the pre-image in ${\displaystyle R[x]}$ of ${\displaystyle {\mathfrak {p}}'}$. Then, as ${\displaystyle \kappa ({\mathfrak {p}}^{\prime c})=\kappa ({\mathfrak {p}})}$, by the polynomial case,

$${\displaystyle \operatorname {ht} {{\mathfrak {p}}'}=\operatorname {ht} {{\mathfrak {p}}^{\prime c}/I}\leq \operatorname {ht} {{\mathfrak {p}}^{\prime c}}-\operatorname {ht} {I}=\dim R_{\mathfrak {p}}-\operatorname {tr.deg} _{\kappa ({\mathfrak {p}})}\kappa ({\mathfrak {p}}').}$$

Here, note that the inequality is the equality if R' 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.

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

Assume R is local with residue field k.

Proof: We claim: for any finite R-module M,

$${\displaystyle \operatorname {pd} _{R}M\leq n\Leftrightarrow \operatorname {Tor} _{n+1}^{R}(M,k)=0.}$$

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, ${\displaystyle \operatorname {Tor} _{1}^{R}(M,k)=0\Rightarrow M{\text{ flat }}\Rightarrow M{\text{ free }}\Rightarrow \operatorname {pd} _{R}(M)\leq 0.}$ Now,

$${\displaystyle \operatorname {gl.dim} R\leq n\Rightarrow \operatorname {pd} _{R}k\leq n\Rightarrow \operatorname {Tor} _{n+1}^{R}(-,k)=0\Rightarrow \operatorname {pd} _{R}-\leq n\Rightarrow \operatorname {gl.dim} R\leq n,}$$

completing the proof. Q.E.D.

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

Lemma — Let ${\displaystyle R_{1}=R/fR}$, f a non-zerodivisor of R. If f is a non-zerodivisor on M, then

$${\displaystyle \operatorname {pd} _{R}M\geq \operatorname {pd} _{R_{1}}(M\otimes R_{1}).}$$

Proof: If ${\displaystyle \operatorname {pd} _{R}M=0}$, then M is R-free and thus ${\displaystyle M\otimes R_{1}}$ is ${\displaystyle R_{1}}$-free. Next suppose ${\displaystyle \operatorname {pd} _{R}M>0}$. Then we have: ${\displaystyle \operatorname {pd} _{R}K=\operatorname {pd} _{R}M-1}$ as in the remark above. Thus, by induction, it is enough to consider the case ${\displaystyle \operatorname {pd} _{R}M=1}$. Then there is a projective resolution: ${\displaystyle 0\to P_{1}\to P_{0}\to M\to 0}$, which gives:

$${\displaystyle \operatorname {Tor} _{1}^{R}(M,R_{1})\to P_{1}\otimes R_{1}\to P_{0}\otimes R_{1}\to M\otimes R_{1}\to 0.}$$

But ${\displaystyle \operatorname {Tor} _{1}^{R}(M,R_{1})={}_{f}M=\{m\in M\mid fm=0\}=0.}$ Hence, ${\displaystyle \operatorname {pd} _{R}(M\otimes R_{1})}$ is at most 1. Q.E.D.

Proof:[3] If 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 ${\displaystyle 0\to M{\overset {f}{\to }}M\to M_{1}\to 0}$, some f in the maximal ideal, of finite modules, ${\displaystyle \operatorname {pd} _{R}M<\infty }$, gives us:

$${\displaystyle 0=\operatorname {Tor} _{i+1}^{R}(M,k)\to \operatorname {Tor} _{i+1}^{R}(M_{1},k)\to \operatorname {Tor} _{i}^{R}(M,k){\overset {f}{\to }}\operatorname {Tor} _{i}^{R}(M,k),\quad i\geq \operatorname {pd} _{R}M.}$$

But f here is zero since it kills k. Thus, ${\displaystyle \operatorname {Tor} _{i+1}^{R}(M_{1},k)\simeq \operatorname {Tor} _{i}^{R}(M,k)}$ and consequently ${\displaystyle \operatorname {pd} _{R}M_{1}=1+\operatorname {pd} _{R}M}$. Using this, we get:

$${\displaystyle \operatorname {pd} _{R}k=1+\operatorname {pd} _{R}(R/(f_{1},\dots ,f_{n-1}))=\cdots =n.}$$

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 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 ${\displaystyle M={\mathfrak {m}}}$,

$${\displaystyle \operatorname {gl.dim} R<\infty \Rightarrow \operatorname {gl.dim} R_{1}=\operatorname {pd} _{R_{1}}k\leq \operatorname {pd} _{R_{1}}{\mathfrak {m}}/f_{1}{\mathfrak {m}}<\infty \Rightarrow R_{1}{\text{ regular}}.}$$

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

Proof: Let R be a regular local ring. Then ${\displaystyle \operatorname {gr} R\simeq k[x_{1},\dots ,x_{d}]}$, which is an integrally closed domain. It is a standard algebra exercise to show this implies that R is an integrally closed domain. Now, we need to show every divisorial ideal is principal; i.e., the divisor class group of 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 R be a ring. Then

$${\displaystyle \operatorname {gl.dim} R[x_{1},\dots ,x_{n}]=\operatorname {gl.dim} R+n.}$$

Let R be a ring and M a module over it. A sequence of elements ${\displaystyle x_{1},\dots ,x_{n}}$ in ${\displaystyle R}$ is called an 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 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 R-module M is the supremum of the lengths of all M-regular sequences in ${\displaystyle {\mathfrak {m}}}$. For example, we have ${\displaystyle \operatorname {depth} M=0\Leftrightarrow {\mathfrak {m}}}$ consists of zerodivisors on M ${\displaystyle \Leftrightarrow {\mathfrak {m}}}$ is associated with M. By induction, we find

$${\displaystyle \operatorname {depth} M\leq \dim R/{\mathfrak {p}}}$$

for any associated primes ${\displaystyle {\mathfrak {p}}}$ of M. In particular, ${\displaystyle \operatorname {depth} M\leq \dim M}$. If the equality holds for M = R, R is called a Cohen–Macaulay ring.

Example: A regular Noetherian local ring is Cohen–Macaulay (since a regular system of parameters is an 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 M be a finite R-module. Then ${\displaystyle \operatorname {depth} M=\sup\{n\mid \operatorname {Ext} _{R}^{i}(k,M)=0,i<n\}}$.

More generally, for any finite R-module N whose support is exactly ${\displaystyle \{{\mathfrak {m}}\}}$,

$${\displaystyle \operatorname {depth} M=\sup\{n\mid \operatorname {Ext} _{R}^{i}(N,M)=0,i<n\}.}$$

Proof: We first prove by induction on n the following statement: for every R-module M and every M-regular sequence ${\displaystyle x_{1},\dots ,x_{n}}$ in ${\displaystyle {\mathfrak {m}}}$,

$${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)\simeq \operatorname {Hom} _{R}(N,M/(x_{1},\dots ,x_{n})M).}$$

(⁎)

The basic step n = 0 is trivial. Next, by inductive hypothesis, ${\displaystyle \operatorname {Ext} _{R}^{n-1}(N,M)\simeq \operatorname {Hom} _{R}(N,M/(x_{1},\dots ,x_{n-1})M)}$. But the latter is zero since the annihilator of N contains some power of ${\displaystyle x_{n}}$. Thus, from the exact sequence ${\displaystyle 0\to M{\overset {x_{1}}{\to }}M\to M_{1}\to 0}$ and the fact that ${\displaystyle x_{1}}$ kills N, using the inductive hypothesis again, we get

$${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)\simeq \operatorname {Ext} _{R}^{n-1}(N,M/x_{1}M)\simeq \operatorname {Hom} _{R}(N,M/(x_{1},\dots ,x_{n})M),}$$

proving (⁎). Now, if ${\displaystyle n<\operatorname {depth} M}$, then we can find an M-regular sequence of length more than n and so by (⁎) we see ${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)=0}$. It remains to show ${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)\neq 0}$ if ${\displaystyle n=\operatorname {depth} M}$. By (⁎) we can assume n = 0. Then ${\displaystyle {\mathfrak {m}}}$ is associated with M; thus is in the support of M. On the other hand, ${\displaystyle {\mathfrak {m}}\in \operatorname {Supp} (N).}$ It follows by linear algebra that there is a nonzero homomorphism from N to M modulo ${\displaystyle {\mathfrak {m}}}$; hence, one from N to M by Nakayama's lemma. Q.E.D.

The Auslander–Buchsbaum formula relates depth and projective dimension.

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

$${\displaystyle \operatorname {pd} _{R}M+\operatorname {depth} M=\operatorname {depth} R.}$$

Proof: We argue by induction on ${\displaystyle \operatorname {pd} _{R}M}$, the basic case (i.e., M free) being trivial. By Nakayama's lemma, we have the exact sequence ${\displaystyle 0\to K{\overset {f}{\to }}F\to M\to 0}$ where F is free and the image of f is contained in ${\displaystyle {\mathfrak {m}}F}$. Since ${\displaystyle \operatorname {pd} _{R}K=\operatorname {pd} _{R}M-1,}$ what we need to show is ${\displaystyle \operatorname {depth} K=\operatorname {depth} M+1}$. Since f kills k, the exact sequence yields: for any i,

$${\displaystyle \operatorname {Ext} _{R}^{i}(k,F)\to \operatorname {Ext} _{R}^{i}(k,M)\to \operatorname {Ext} _{R}^{i+1}(k,K)\to 0.}$$

Note the left-most term is zero if ${\displaystyle i<\operatorname {depth} R}$. If ${\displaystyle i<\operatorname {depth} K-1}$, then since ${\displaystyle \operatorname {depth} K\leq \operatorname {depth} R}$ by inductive hypothesis, we see ${\displaystyle \operatorname {Ext} _{R}^{i}(k,M)=0.}$ If ${\displaystyle i=\operatorname {depth} K-1}$, then ${\displaystyle \operatorname {Ext} _{R}^{i+1}(k,K)\neq 0}$ and it must be ${\displaystyle \operatorname {Ext} _{R}^{i}(k,M)\neq 0.}$ Q.E.D.

As a matter of notation, for any R-module M, we let

$${\displaystyle \Gamma _{\mathfrak {m}}(M)=\{s\in M\mid \operatorname {supp} (s)\subset \{{\mathfrak {m}}\}\}=\{s\in M\mid {\mathfrak {m}}^{j}s=0{\text{ for some }}j\}.}$$

One sees without difficulty that ${\displaystyle \Gamma _{\mathfrak {m}}}$ is a left-exact functor and then let ${\displaystyle H_{\mathfrak {m}}^{j}=R^{j}\Gamma _{\mathfrak {m}}}$ be its j-th right derived functor, called the local cohomology of R. Since ${\displaystyle \Gamma _{\mathfrak {m}}(M)=\varinjlim \operatorname {Hom} _{R}(R/{\mathfrak {m}}^{j},M)}$, via abstract nonsense,

$${\displaystyle H_{\mathfrak {m}}^{i}(M)=\varinjlim \operatorname {Ext} _{R}^{i}(R/{\mathfrak {m}}^{j},M).}$$

This observation proves the first part of the theorem below.

Proof: 1. is already noted (except to show the nonvanishing at the degree equal to the depth of 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). ${\displaystyle \square }$

Let R be a ring and x an element in it. We form the chain complex K(x) given by ${\displaystyle K(x)_{i}=R}$ for i = 0, 1 and ${\displaystyle K(x)_{i}=0}$ for any other i with the differential

$${\displaystyle d:K_{1}(R)\to K_{0}(R),\,r\mapsto xr.}$$

For any R-module M, we then get the complex ${\displaystyle K(x,M)=K(x)\otimes _{R}M}$ with the differential ${\displaystyle d\otimes 1}$ and let ${\displaystyle \operatorname {H} _{*}(x,M)=\operatorname {H} _{*}(K(x,M))}$ be its homology. Note:

$${\displaystyle \operatorname {H} _{0}(x,M)=M/xM,}$$

$${\displaystyle \operatorname {H} _{1}(x,M)={}_{x}M=\{m\in M\mid xm=0\}.}$$

More generally, given a finite sequence ${\displaystyle x_{1},\dots ,x_{n}}$ of elements in a ring R, we form the tensor product of complexes:

$${\displaystyle K(x_{1},\dots ,x_{n})=K(x_{1})\otimes \dots \otimes K(x_{n})}$$

and let ${\displaystyle \operatorname {H} _{*}(x_{1},\dots ,x_{n},M)=\operatorname {H} _{*}(K(x_{1},\dots ,x_{n},M))}$ its homology. As before,

$${\displaystyle \operatorname {H} _{0}({\underline {x}},M)=M/(x_{1},\dots ,x_{n})M,}$$

$${\displaystyle \operatorname {H} _{n}({\underline {x}},M)=\operatorname {Ann} _{M}((x_{1},\dots ,x_{n})).}$$

We now have the homological characterization of a regular sequence.

A Koszul complex is a powerful computational tool. For instance, it follows from the theorem and the corollary

$${\displaystyle \operatorname {H} _{\mathfrak {m}}^{i}(M)\simeq \varinjlim \operatorname {H} ^{i}(K(x_{1}^{j},\dots ,x_{n}^{j};M))}$$

(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 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 R). Then

$${\displaystyle {\binom {s}{i}}\leq \dim _{k}\operatorname {Tor} _{i}^{R}(k,k).}$$

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

We next use a Koszul homology to define and study complete intersection rings. Let R be a Noetherian local ring. By definition, the first deviation of R is the vector space dimension

$${\displaystyle \epsilon _{1}(R)=\dim _{k}\operatorname {H} _{1}({\underline {x}})}$$

where ${\displaystyle {\underline {x}}=(x_{1},\dots ,x_{d})}$ is a system of parameters. By definition, R is a complete intersection ring if ${\displaystyle \dim R+\epsilon _{1}(R)}$ is the dimension of the tangent space. (See Hartshorne for a geometric meaning.)

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

Theorem — For any ring R,

$${\displaystyle {\begin{aligned}\operatorname {gl.dim} R\,&=\operatorname {sup} \{\operatorname {id} _{R}M\mid M\in \operatorname {Mod} _{R}\}\\&=\inf\{n\mid \operatorname {Ext} _{R}^{i}(M,N)=0,\,i>n,M,N\in \operatorname {Mod} _{R}\}\end{aligned}}}$$

Proof: Suppose ${\displaystyle \operatorname {gl.dim} R\leq n}$. Let M be an R-module and consider a resolution

$${\displaystyle 0\to M\to I_{0}{\overset {\phi _{0}}{\to }}I_{1}\to \dots \to I_{n-1}{\overset {\phi _{n-1}}{\to }}N\to 0}$$

where ${\displaystyle I_{i}}$ are injective modules. For any ideal I,

$${\displaystyle \operatorname {Ext} _{R}^{1}(R/I,N)\simeq \operatorname {Ext} _{R}^{2}(R/I,\operatorname {ker} (\phi _{n-1}))\simeq \dots \simeq \operatorname {Ext} _{R}^{n+1}(R/I,M),}$$

which is zero since ${\displaystyle \operatorname {Ext} _{R}^{n+1}(R/I,-)}$ is computed via a projective resolution of ${\displaystyle R/I}$. Thus, by Baer's criterion, N is injective. We conclude that ${\displaystyle \sup\{\operatorname {id} _{R}M|M\}\leq n}$. 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:

$${\displaystyle \operatorname {w.gl.dim} =\inf\{n\mid \operatorname {Tor} _{i}^{R}(M,N)=0,\,i>n,M,N\in \operatorname {Mod} _{R}\}.}$$

It was originally called the weak global dimension of R but today it is more commonly called the Tor dimension of R.

Remark: for any ring R, ${\displaystyle \operatorname {w.gl.dim} R\leq \operatorname {gl.dim} R}$.