Dimension theory (algebra)


In mathematics, dimension theory is the study in terms of commutative algebra of the notion dimension of an algebraic variety. The need of a theory for such an apparently simple notion results from the existence of many definitions of the dimension that are equivalent only in the most regular cases. 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.
Throughout the article, denotes Krull dimension of a ring and the height of a 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
If R is noetherian, this follows from the fundamental theorem below, but it is also a consequence of a more precise result. For any prime ideal in R,
This can be shown within basic ring theory. In addition, in each fiber of, one cannot have a chain of primes ideals of length.
Since an artinian ring has dimension zero, by induction one gets a formula: for an artinian ring R,

Local rings

Fundamental theorem

Let be a noetherian local ring and I a -primary ideal. Let be the Poincaré series of the associated graded ring. That is,
where refers to the length of a module. 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 find that the coefficient of in is of the form
That is to say, is a polynomial in n of degree. P is called the Hilbert polynomial of.
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. 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:
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 be a noetherian local ring and put. 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.
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.

Nagata's altitude formula

Proof: First suppose is a polynomial ring. By induction on the number of variables, it is enough to consider the case. Since R is flat over R,
By Noether's normalization lemma, the second term on the right side is:
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.
See also: Quasi-unmixed ring.

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 and is denoted by. We set ; 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,
By dimension shifting, it is enough to prove this for. But then, by the local criterion for flatness,
Now,
completing the proof.
Remark: The proof also shows that if M is not free and is the kernel of some surjection from a free module to M.
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.
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. 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.
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.

Depth

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
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
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.
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.
The Auslander–Buchsbaum formula relates depth and projective dimension.
Proof: We argue by induction on, the basic case 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
As a matter of notation, for any R-module M, we let
One sees without difficulty that is a left-exact functor and then let be its j-th right derived functor, called the local cohomology of R. Since, via abstract nonsense,
This observation proves the first part of the theorem below.
Proof: 1. is already noted 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.

Koszul complex

Let R be a ring and x an element in it. We form the chain complex K given by for i = 0, 1 and for any other i with the differential
For any R-module M, we then get the complex with the differential and let be its homology. Note:
More generally, given a finite sequence of elements in a ring R, we form the tensor product of complexes:
and let its homology. As before,
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
Another instance would be
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,.
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
where is a system of parameters. By definition, R is a complete intersection ring if is the dimension of the tangent space.

Injective dimension and Tor dimensions

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.
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.
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.
Remark: for any ring R,.

Multiplicity theory

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 and then put
It is called the Gelfand–Kirillov dimension of A. It is easy to show is independent of a choice of V.
Example: If A is finite-dimensional, then gk = 0. If A is an affine ring, then gk = Krull dimension of A.
See also: Goldie dimension, Krull–Gabriel dimension.