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, denotes Krull dimension of a ring and 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.
If R is noetherian, this follows from the fundamental theorem below (in particular, Krull's principal ideal theorem), but it is also a consequence of a more precise result. For any prime ideal in R,
for any prime ideal in that contracts to .
This can be shown within basic ring theory (cf. Kaplansky, commutative rings). In addition, in each fiber of , one cannot have a chain of primes ideals of length .
Since an artinian ring (e.g., a field) has dimension zero, by induction one gets a formula: for an artinian ring R,
where refers to the length of a module (over an artinian ring ). If generate I, then their image in have degree 1 and generate as -algebra. By the Hilbert–Serre theorem, F is a rational function with exactly one pole at of order . Since
We set . We also set to be the minimum number of elements of R that can generate an -primary ideal of R. Our ambition is to prove the fundamental theorem:
Since we can take s to be , we already have from the above. Next we prove by induction on . Let be a chain of prime ideals in R. Let and x a nonzero nonunit element in D. Since x is not a zero-divisor, we have the exact sequence
The degree bound of the Hilbert-Samuel polynomial now implies that . (This essentially follows from the Artin-Rees lemma; see Hilbert-Samuel function for the statement and the proof.) In , the chain becomes a chain of length and so, by inductive hypothesis and again by the degree estimate,
The claim follows. It now remains to show More precisely, we shall show:
Lemma — The maximal ideal contains elements , d = Krull dimension of R, such that, for any i, any prime ideal containing has height .
(Notice: is then -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.
, since a basis of lifts to a generating set of by Nakayama. If the equality holds, then R is called a regular local ring.
, since .
(Krull's principal ideal theorem) The height of the ideal generated by elements 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 be a prime ideal minimal over such an ideal. Then . The converse was shown in the course of the proof of the fundamental theorem.)
Theorem — If is a morphism of noetherian local rings, then
Proof: Let generate a -primary ideal and be such that their images generate a -primary ideal. Then for some s. Raising both sides to higher powers, we see some power of is contained in ; i.e., the latter ideal is -primary; thus, . The equality is a straightforward application of the going-down property. Q.E.D.
Proposition — If R is a noetherian ring, then
Proof: If are a chain of prime ideals in R, then are a chain of prime ideals in while is not a maximal ideal. Thus, . For the reverse inequality, let be a maximal ideal of and . Clearly, .
Since is then a localization of a principal ideal domain and has dimension at most one, we get by the previous inequality. Since is arbitrary, it follows . Q.E.D.
Next, suppose is generated by a single element; thus, . If I = 0, then we are already done. Suppose not. Then is algebraic over R and so . Since R is a subring of R', and so
since is algebraic over . Let denote the pre-image in of . Then, as , by the polynomial case,
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 . We set ; it is called the global dimension of R.
Assume R is local with residue field k.
Lemma — (possibly infinite).
Proof: We claim: for any finite R-module M,
By dimension shifting (cf. the proof of Theorem of Serre below), it is enough to prove this for . But then, by the local criterion for flatness,
completing the proof. Q.E.D.
Remark: The proof also shows that if M is not free and is the kernel of some surjection from a free module to M.
Lemma — Let , f a non-zerodivisor of R. If f is a non-zerodivisor on M, then
Proof: If , then M is R-free and thus is -free. Next suppose . Then we have: as in the remark above. Thus, by induction, it is enough to consider the case . Then there is a projective resolution: , which gives:
But Hence, is at most 1. Q.E.D.
Theorem of Serre — R regular
Proof: If R is regular, we can write , a regular system of parameters. An exact sequence , some f in the maximal ideal, of finite modules, , gives us:
But f here is zero since it kills k. Thus, and consequently . Using this, we get:
The proof of the converse is by induction on . We begin with the inductive step. Set , among a system of parameters. To show R is regular, it is enough to show is regular. But, since , by inductive hypothesis and the preceding lemma with ,
The basic step remains. Suppose . We claim 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 with and thus in fact we can find M with . By Nakayama's lemma, there is a surjection from a free module F to M whose kernel K is contained in . Since , the maximal ideal is an associated prime of R; i.e., for some nonzero s in R. Since , . Since K is not zero and is free, this implies , which is absurd. Q.E.D.
Corollary — A regular local ring is a unique factorization domain.
Proof: Let R be a regular local ring. Then , 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.
Let R be a ring and M a module over it. A sequence of elements in is called an M-regular sequence if is not a zero-divisor on and is not a zero divisor on for each . 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 and put . Then, by definition, the depth of a finite R-module M is the supremum of the lengths of all M-regular sequences in . For example, we have consists of zerodivisors on M is associated with M. By induction, we find
for any associated primes of M. In particular, . 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 is universally catenary since it is regular and thus Cohen–Macaulay.
Proposition(Rees) — Let M be a finite R-module. Then .
More generally, for any finite R-module N whose support is exactly ,
Proof: We first prove by induction on n the following statement: for every R-module M and every M-regular sequence in ,
The basic step n = 0 is trivial. Next, by inductive hypothesis, . But the latter is zero since the annihilator of N contains some power of . Thus, from the exact sequence and the fact that kills N, using the inductive hypothesis again, we get
proving (⁎). Now, if , then we can find an M-regular sequence of length more than n and so by (⁎) we see . It remains to show if . By (⁎) we can assume n = 0. Then is associated with M; thus is in the support of M. On the other hand, It follows by linear algebra that there is a nonzero homomorphism from N to M modulo ; hence, one from N to M by Nakayama's lemma. Q.E.D.
Theorem — Let M be a finite module over a noetherian local ring R. If , then
Proof: We argue by induction on , the basic case (i.e., M free) being trivial. By Nakayama's lemma, we have the exact sequence where F is free and the image of f is contained in . Since what we need to show is .
Since f kills k, the exact sequence yields: for any i,
Note the left-most term is zero if . If , then since by inductive hypothesis, we see If , then and it must be Q.E.D.
As a matter of notation, for any R-module M, we let
This observation proves the first part of the theorem below.
Theorem(Grothendieck) — Let M be a finite R-module. Then
If R is complete and d its Krull dimension and if E is the injective hull of k, then
is representable (the representing object is sometimes called the canonical module especially if R is Cohen–Macaulay.)
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).
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, and thus . On the other hand, as , the Auslander–Buchsbaum formula gives . Hence, .
Let R be a ring. The injective dimension of an R-module M denoted by is defined just like a projective dimension: it is the minimal length of an injective resolution of M. Let be the category of R-modules.
Theorem — For any ring R,
Proof: Suppose . Let M be an R-module and consider a resolution
where are injective modules. For any ideal I,
which is zero since is computed via a projective resolution of . Thus, by Baer's criterion, N is injective. We conclude that . 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:
It was originally called the weak global dimension of R but today it is more commonly called the Tor dimension of R.
Serre, Jean-Pierre (1975), Algèbre locale. Multiplicités, Cours au Collège de France, 1957–1958, rédigé par Pierre Gabriel. Troisième édition, 1975. Lecture Notes in Mathematics (in French), vol. 11, Berlin, New York: Springer-Verlag
Weibel, Charles A. (1995). An Introduction to Homological Algebra. Cambridge University Press.