In mathematics, dimension theory is the study in terms of commutative algebra of the notion dimension of an algebraic variety (and by extension that of a scheme). The need of a theory for such an apparently simple notion results from the existence of many definitions of dimension that are equivalent only in the most regular cases (see Dimension of an algebraic variety). A large part of dimension theory consists in studying the conditions under which several dimensions are equal, and many important classes of commutative rings may be defined as the rings such that two dimensions are equal; for example, a regular ring is a commutative ring such that the homological dimension is equal to the Krull dimension.

The theory is simpler for commutative rings that are finitely generated algebras over a field, which are also quotient rings of polynomial rings in a finite number of indeterminates over a field. In this case, which is the algebraic counterpart of the case of affine algebraic sets, most of the definitions of the dimension are equivalent. For general commutative rings, the lack of geometric interpretation is an obstacle to the development of the theory; in particular, very little is known for non-noetherian rings. (Kaplansky's Commutative rings gives a good account of the non-noetherian case.)

Throughout the article, ${\displaystyle \dim }$ denotes Krull dimension of a ring and ${\displaystyle \operatorname {ht} }$ the height of a prime ideal (i.e., the Krull dimension of the localization at that prime ideal.) Rings are assumed to be commutative except in the last section on dimensions of non-commutative rings.

Basic results

Let R be a noetherian ring or valuation ring. Then

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

${\displaystyle {\mathfrak {p))}$

$${\displaystyle \operatorname {ht} ({\mathfrak {p))R[x])=\operatorname {ht} ({\mathfrak {p))).}$$

$${\displaystyle \operatorname {ht} ({\mathfrak {q)))=\operatorname {ht} ({\mathfrak {p)))+1}$$

${\displaystyle {\mathfrak {q))\supsetneq {\mathfrak {p))R[x]}$

${\displaystyle R[x]}$

${\displaystyle {\mathfrak {p))}$

${\displaystyle \operatorname {Spec} R[x]\to \operatorname {Spec} R}$

${\displaystyle \geq 2}$

Since an artinian ring (e.g., a field) has dimension zero, by induction one gets a formula: for an artinian ring R,

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

Local rings

Fundamental theorem

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,

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

${\displaystyle \ell }$

${\displaystyle (\operatorname {gr} _{I}R)_{0}=R/I}$

${\displaystyle x_{1},\dots ,x_{s))$

${\displaystyle I/I^{2))$

${\displaystyle \operatorname {gr} _{I}R}$

${\displaystyle R/I}$

${\displaystyle t=1}$

${\displaystyle d\leq s}$

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

${\displaystyle t^{n))$

${\displaystyle F(t)=(1-t)^{d}F(t)(1-t)^{-d))$

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

${\displaystyle \ell (I^{n}/I^{n+1})}$

${\displaystyle P}$

${\displaystyle d-1}$

${\displaystyle \operatorname {gr} _{I}R}$

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:

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

${\displaystyle \delta (R)}$

${\displaystyle \delta (R)\geq d(R)}$

${\displaystyle d(R)\geq \dim R}$

${\displaystyle d(R)}$

${\displaystyle {\mathfrak {p))_{0}\subsetneq \cdots \subsetneq {\mathfrak {p))_{m))$

${\displaystyle D=R/{\mathfrak {p))_{0))$

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

${\displaystyle d(D)>d(D/xD)\geq d(R/{\mathfrak {p))_{1})}$

${\displaystyle R/{\mathfrak {p))_{1))$

${\displaystyle {\mathfrak {p))_{i))$

${\displaystyle m-1}$

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

${\displaystyle \dim R\geq \delta (R).}$

(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

• ${\displaystyle \dim R\leq \dim _{k}{\mathfrak {m))/{\mathfrak {m))^{2))$, 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.

Nagata's altitude formula

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,

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

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

${\displaystyle R'}$

${\displaystyle R'=R[x]/I}$

${\displaystyle R'}$

${\displaystyle \operatorname {tr.deg} _{R}R'=0}$

${\displaystyle I\cap R=0}$

${\displaystyle \operatorname {ht} I=\dim R[x]_{I}=\dim Q(R)[x]_{I}=1-\operatorname {tr.deg} _{Q(R)}\kappa (I)=1}$

${\displaystyle \kappa (I)=Q(R')}$

${\displaystyle Q(R)}$

${\displaystyle {\mathfrak {p))^{\prime c))$

${\displaystyle R[x]}$

${\displaystyle {\mathfrak {p))'}$

${\displaystyle \kappa ({\mathfrak {p))^{\prime c})=\kappa ({\mathfrak {p)))}$

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

Homological methods

Regular rings

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,

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

${\displaystyle n=0}$

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

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

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:

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

${\displaystyle \operatorname {Tor} _{1}^{R}(M,R_{1})={}_{f}M=\{m\in M\mid fm=0\}=0.}$

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

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:

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

${\displaystyle \operatorname {Tor} _{i+1}^{R}(M_{1},k)\simeq \operatorname {Tor} _{i}^{R}(M,k)}$

${\displaystyle \operatorname {pd} _{R}M_{1}=1+\operatorname {pd} _{R}M}$

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

${\displaystyle \dim R}$

${\displaystyle R_{1}=R/f_{1}R}$

${\displaystyle f_{1))$

${\displaystyle R_{1))$

${\displaystyle \dim R_{1}<\dim R}$

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

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

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

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

Depth

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

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

${\displaystyle {\mathfrak {p))}$

${\displaystyle \operatorname {depth} M\leq \dim M}$

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

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

${\displaystyle n<\operatorname {depth} M}$

${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)=0}$

${\displaystyle \operatorname {Ext} _{R}^{n}(N,M)\neq 0}$

${\displaystyle n=\operatorname {depth} M}$

${\displaystyle {\mathfrak {m))}$

${\displaystyle {\mathfrak {m))\in \operatorname {Supp} (N).}$

${\displaystyle {\mathfrak {m))}$

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,

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

${\displaystyle i<\operatorname {depth} R}$

${\displaystyle i<\operatorname {depth} K-1}$

${\displaystyle \operatorname {depth} K\leq \operatorname {depth} R}$

${\displaystyle \operatorname {Ext} _{R}^{i}(k,M)=0.}$

${\displaystyle i=\operatorname {depth} K-1}$

${\displaystyle \operatorname {Ext} _{R}^{i+1}(k,K)\neq 0}$

${\displaystyle \operatorname {Ext} _{R}^{i}(k,M)\neq 0.}$

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\}.}$$

${\displaystyle \Gamma _{\mathfrak {m))}$

${\displaystyle H_{\mathfrak {m))^{j}=R^{j}\Gamma _{\mathfrak {m))}$

${\displaystyle \Gamma _{\mathfrak {m))(M)=\varinjlim \operatorname {Hom} _{R}(R/{\mathfrak {m))^{j},M)}$

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

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). ${\displaystyle \square }$

Koszul complex

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

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

${\displaystyle K(x,M)=K(x)\otimes _{R}M}$

${\displaystyle d\otimes 1}$

${\displaystyle \operatorname {H} _{*}(x,M)=\operatorname {H} _{*}(K(x,M))}$

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

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

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:

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

${\displaystyle \operatorname {H} _{*}(x_{1},\dots ,x_{n},M)=\operatorname {H} _{*}(K(x_{1},\dots ,x_{n},M))}$

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

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

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

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

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

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

${\displaystyle {\underline {x))=(x_{1},\dots ,x_{d})}$

${\displaystyle \dim R+\epsilon _{1}(R)}$

Injective dimension and Tor dimensions

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

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

${\displaystyle I_{i))$

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

${\displaystyle \operatorname {Ext} _{R}^{n+1}(R/I,-)}$

${\displaystyle R/I}$

${\displaystyle \sup\{\operatorname {id} _{R}M|M\}\leq n}$

The theorem suggests that we consider a sort of a dual of a global dimension:

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

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

Dimensions of non-commutative rings

Let A be a graded algebra over a field k. If V is a finite-dimensional generating subspace of A, then we let ${\displaystyle f(n)=\dim _{k}V^{n))$ and then put

$${\displaystyle \operatorname {gk} (A)=\limsup _{n\to \infty }{\log f(n) \over \log n}.}$$

${\displaystyle \operatorname {gk} (A)}$

${\displaystyle {gk}(M)}$

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

Example: If ${\displaystyle A_{n}=k[x_{1},...,x_{n},\partial _{1},...,\partial _{n}]}$ is the n-th Weyl algebra then ${\displaystyle \operatorname {gk} (A_{n})=2n.}$