Let R be a noetherian ring or valuation ring. Then
dim R [ x ] = dim R + 1. 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 the more precise result. For any prime ideal p in R,
ht ( p R [ x ] ) = ht ( p ) .
ht ( q ) = ht ( p ) + 1 for any prime ideal
q ⊋ p R [ x ] in
R [ x ] that contracts to
p .
This can be shown within basic ring theory (cf. Kaplansky, commutative rings). By the way, it says in particular that in each fiber of Spec R [ x ] → Spec R , one cannot have a chain of primes ideals of length ≥ 2 .
Since an artinian ring (e.g., a field) has dimension zero, by induction, one gets the formula: for an artinian ring R,
dim R [ x 1 , … , x n ] = n . Let ( R , m ) be a noetherian local ring and I a m -primary ideal (i.e., it sits between some power of m and m ). Let F ( t ) be the Poincaré series of the associated graded ring gr I R = ⊕ 0 ∞ I n / I n + 1 . That is,
F ( t ) = ∑ 0 ∞ ℓ ( I n / I n + 1 ) t n where ℓ refers to the length of a module (over an artinian ring ( gr I R ) 0 = R / I ). If x 1 , … , x s generate I, then their image in I / I 2 have degree 1 and generate gr I R as R / I -algebra. By the Hilbert–Serre theorem, F is a rational function with exactly one pole at t = 1 of order d ≤ s . Since
( 1 − t ) − d = ∑ 0 ∞ ( d − 1 + j d − 1 ) t j ,
we find that the coefficient of t n in F ( t ) = ( 1 − t ) d F ( t ) ( 1 − t ) − d is of the form
∑ 0 N a k ( d − 1 + n − k d − 1 ) = ( 1 − t ) d F ( t ) | t = 1 n d − 1 d − 1 ! + O ( n d − 2 ) . That is to say, ℓ ( I n / I n + 1 ) is a polynomial P in n of degree d − 1 . P is called the Hilbert polynomial of gr I R .
We set d ( R ) = d . We also set δ ( R ) to be the minimum number of elements of R that can generate a m -primary ideal of R. Our ambition is to prove the fundamental theorem:
δ ( R ) = d ( R ) = dim R .
Since we can take s to be δ ( R ) , we already have δ ( R ) ≥ d ( R ) from the above. Next we prove d ( R ) ≥ dim R by induction on d ( R ) . Let p 0 ⊊ ⋯ ⊊ p m be a chain of prime ideals in R. Let D = R / p 0 and x a nonzero nonunit element in D. Since x is not a zero-divisor, we have the exact sequence
0 → D → x D → D / x D → 0 .
The degree bound of the Hilbert-Samuel polynomial now implies that d ( D ) > d ( D / x D ) ≥ d ( R / p 1 ) . (This essentially follows from the Artin-Rees lemma; see Hilbert-Samuel function for the statement and the proof.) In R / p 1 , the chain p i becomes a chain of length m − 1 and so, by inductive hypothesis and again by the degree estimate,
m − 1 ≤ dim ( R / p 1 ) ≤ d ( R / p 1 ) ≤ d ( D ) − 1 ≤ d ( R ) − 1 .
The claim follows. It now remains to show dim R ≥ δ ( R ) . More precisely, we shall show:
Lemma:
R contains elements
x 1 , … , x s such that, for any
i, any prime ideal containing
( x 1 , … , x i ) has height
≥ i .
(Notice: ( x 1 , … , x s ) is then 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 ( R , m ) be a noetherian local ring and put k = R / m . Then
dim R ≤ dim k m / m 2 , since a basis of m / m 2 lifts to a generating set of m by Nakayama. If the equality holds, then R is called a regular local ring. dim R ^ = dim R , since gr R = gr R ^ .(Krull's principal ideal theorem) The height of the ideal generated by elements x 1 , … , x s in a noetherian ring is at most s. Conversely, a prime ideal of height s can be generated by s elements. (Proof: Let p be a prime ideal minimal over such an ideal. Then s ≥ dim R p = ht p . The converse was shown in the course of the proof of the fundamental theorem.)Proof: Let x 1 , … , x n generate a m A -primary ideal and y 1 , … , y m be such that their images generate a m B / m A B -primary ideal. Then m B s ⊂ ( y 1 , … , y m ) + m A B for some s. Raising both sides to higher powers, we see some power of m B is contained in ( y 1 , … , y m , x 1 , … , x n ) ; i.e., the latter ideal is m B -primary; thus, m + n ≥ dim B . The equality is a straightforward application of the going-down property. ◻
Proof: If p 0 ⊊ p 1 ⊊ ⋯ ⊊ p n are a chain of prime ideals in R, then p i R [ x ] are a chain of prime ideals in R [ x ] while p n R [ x ] is not a maximal ideal. Thus, dim R + 1 ≤ dim R [ x ] . For the reverse inequality, let m be a maximal ideal of R [ x ] and p = R ∩ m . Clearly, R [ x ] m = R p [ x ] m . Since R [ x ] m / p R p R [ x ] m = ( R p / p R p ) [ x ] m is then a localization of a principal ideal domain and has dimension at most one, we get 1 + dim R ≥ 1 + dim R p ≥ dim R [ x ] m by the previous inequality. Since m is arbitrary, it follows 1 + dim R ≥ dim R [ x ] . ◻
Proof: First suppose R ′ is a polynomial ring. By induction on the number of variables, it is enough to consider the case R ′ = R [ x ] . Since R' is flat over R,
dim R p ′ ′ = dim R p + dim κ ( p ) ⊗ R R ′ p ′ .
By Noether's normalization lemma, the second term on the right side is:
dim κ ( p ) ⊗ R R ′ − dim κ ( p ) ⊗ R R ′ / p ′ = 1 − t r . d e g κ ( p ) κ ( p ′ ) = t r . d e g R R ′ − t r . d e g κ ( p ′ ) . Next, suppose R ′ is generated by a single element; thus, R ′ = R [ x ] / I . If I = 0, then we are already done. Suppose not. Then R ′ is algebraic over R and so t r . d e g R R ′ = 0 . Since R is a subring of R', I ∩ R = 0 and so ht I = dim R [ x ] I = dim Q ( R ) [ x ] I = 1 − t r . d e g Q ( R ) κ ( I ) = 1 since κ ( I ) = Q ( R ′ ) is algebraic over Q ( R ) . Let p ′ c denote the pre-image in R [ x ] of p ′ . Then, as κ ( p ′ c ) = κ ( p ) , by the polynomial case,
ht p ′ = ht p ′ c / I ≤ ht p ′ c − ht I = dim R p − t r . d e g κ ( p ) κ ( 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. ◻
Let R be a noetherian ring. The projective dimension of a finite R-module M is the shortest length of any projective resolution of R (possibly infinite) and is denoted by pd R M . We set g l . d i m R = sup { pd R M | 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,
pd R M ≤ n ⇔ 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 n = 0 . But then, by the local criterion for flatness, Tor 1 R ( M , k ) = 0 ⇒ M flat ⇒ M free ⇒ pd R ( M ) ≤ 0. Now,
g l . d i m R ≤ n ⇒ pd R k ≤ n ⇒ Tor n + 1 R ( − , k ) = 0 ⇒ pd R − ≤ n ⇒ g l . d i m R ≤ n , completing the proof. ◻
Remark: The proof also shows that pd R K = pd R M − 1 if M is not free and K is the kernel of some surjection from a free module to M.
Proof: If pd R M = 0 , then M is R-free and thus M ⊗ R 1 is R 1 -free. Next suppose pd R M > 0 . Then we have: pd R K = pd R M − 1 as in the remark above. Thus, by induction, it is enough to consider the case pd R M = 1 . Then there is a projective resolution: 0 → P 1 → P 0 → M → 0 , which gives:
Tor 1 R ( M , R 1 ) → P 1 ⊗ R 1 → P 0 ⊗ R 1 → M ⊗ R 1 → 0 .
But Tor 1 R ( M , R 1 ) = f M = { m ∈ M | f m = 0 } = 0. Hence, pd R ( M ⊗ R 1 ) is at most 1. ◻
Proof: If R is regular, we can write k = R / ( f 1 , … , f n ) , f i a regular system of parameters. An exact sequence 0 → M → f M → M 1 → 0 , some f in the maximal ideal, of finite modules, pd R M < ∞ , gives us:
0 = Tor i + 1 R ( M , k ) → Tor i + 1 R ( M 1 , k ) → Tor i R ( M , k ) → f Tor i R ( M , k ) , i ≥ pd R M . But f here is zero since it kills k. Thus, Tor i + 1 R ( M 1 , k ) ≃ Tor i R ( M , k ) and consequently pd R M 1 = 1 + pd R M . Using this, we get:
pd R k = 1 + pd R ( R / ( f 1 , … , f n − 1 ) ) = ⋯ = n . The proof of the converse is by induction on dim R . We begin with the inductive step. Set R 1 = R / f 1 R , f 1 among a system of parameters. To show R is regular, it is enough to show R 1 is regular. But, since dim R 1 < dim R , by inductive hypothesis and the preceding lemma with M = m ,
g l . d i m R < ∞ ⇒ g l . d i m R 1 = pd R 1 k ≤ pd R 1 m / f 1 m < ∞ ⇒ R 1 regular . The basic step remains. Suppose dim R = 0 . We claim g l . d i m 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 M with 0 < pd R M < ∞ and thus in fact we can find M with pd R M = 1 . By Nakayama's lemma, there is a surjection F → M from a free module F to M whose kernel K is contained in m F . Since dim R = 0 , the maximal ideal m is an associated prime of R; i.e., m = ann ( s ) for some nonzero s in R. Since K ⊂ m F , s K = 0 . Since K is not zero and is free, this implies s = 0 , which is absurd. ◻
Proof: Let R be a regular local ring. Then gr R ≃ k [ x 1 , … , 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. ◻
Let R be a ring and M a module over it. A sequence of elements x 1 , … , x n in R is called an M-regular sequence if x 1 is not a zero-divisor on M and x i is not a zero divisor on M / ( x 1 , … , x i − 1 ) M for each i = 2 , … , 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 m and put k = R / m . Then, by definition, the depth of a finite R-module M is the supremum of the lengths of all M-regular sequences in m . For example, we have depth M = 0 ⇔ m consists of zerodivisors on M ⇔ m is associated with M. By induction, we find
depth M ≤ dim R / p for any associated primes p of M. In particular, depth M ≤ 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 k [ x 1 , … , x d ] 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 x 1 , … , x n in m ,
(*)
Ext R n ( N , M ) ≃ Hom R ( N , M / ( x 1 , … , x n ) M ) . The basic step n = 0 is trivial. Next, by inductive hypothesis, Ext R n − 1 ( N , M ) ≃ Hom R ( N , M / ( x 1 , … , x n − 1 ) M ) . But the latter is zero since the annihilator of N contains some power of x n . Thus, from the exact sequence 0 → M → x 1 M → M 1 → 0 and the fact that x 1 kills N, using the inductive hypothesis again, we get
Ext R n ( N , M ) ≃ Ext R n − 1 ( N , M / x 1 M ) ≃ Hom R ( N , M / ( x 1 , … , x n ) M ) ,
proving (*). Now, if n < depth M , then we can find an M-regular sequence of length more than n and so by (*) we see Ext R n ( N , M ) = 0 . It remains to show Ext R n ( N , M ) ≠ 0 if n = depth M . By (*) we can assume n = 0. Then m is associated with M; thus is in the support of M. On the other hand, m ∈ Supp ( N ) . It follows by linear algebra that there is a nonzero homomorphism from N to M modulo m ; 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 pd R M , the basic case (i.e., M free) being trivial. By Nakayama's lemma, we have the exact sequence 0 → K → f F → M → 0 where F is free and the image of f is contained in m F . Since pd R K = pd R M − 1 , what we need to show is depth K = depth M + 1 . Since f kills k, the exact sequence yields: for any i,
Ext R i ( k , F ) → Ext R i ( k , M ) → Ext R i + 1 ( k , K ) → 0. Note the left-most term is zero if i < depth R . If i < depth K − 1 , then since depth K ≤ depth R by inductive hypothesis, we see Ext R i ( k , M ) = 0. If i = depth K − 1 , then Ext R i + 1 ( k , K ) ≠ 0 and it must be Ext R i ( k , M ) ≠ 0. ◻
As a matter of notation, for any R-module M, we let
Γ m ( M ) = { s ∈ M | supp ( s ) ⊂ { m } } = { s ∈ M | m j s = 0 for some j } . One sees without difficulty that Γ m is a left-exact functor and then let H m j = R j Γ m be its j-th right derived functor, called the local cohomology of R. Since Γ m ( M ) = lim → Hom R ( R / m j , M ) , via abstract nonsense,
H m i ( M ) = lim → Ext R i ( R / 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). ◻
Let R be a ring and x an element in it. We form the chain complex K(x) given by K ( x ) i = R for i = 0, 1 and K ( x ) i = 0 for any other i with the differential
d : K 1 ( R ) → K 0 ( R ) , r ↦ x r . For any R-module M, we then get the complex K ( x , M ) = K ( x ) ⊗ R M with the differential d ⊗ 1 and let H ∗ ( x , M ) = H ∗ ( K ( x , M ) ) be its homology. Note:
H 0 ( x , M ) = M / x M ,
H 1 ( x , M ) = x M = { m ∈ M | x m = 0 } .
More generally, given a finite sequence x 1 , … , x n of elements in a ring R, we form the tensor product of complexes:
K ( x 1 , … , x n ) = K ( x 1 ) ⊗ ⋯ ⊗ K ( x n ) and let H ∗ ( x 1 , … , x n , M ) = H ∗ ( K ( x 1 , … , x n , M ) ) its homology. As before,
H 0 ( x _ , M ) = M / ( x 1 , … , x n ) M ,
H n ( x _ , M ) = Ann M ( ( x 1 , … , 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
H m i ( M ) ≃ lim → H i ( K ( x 1 j , … , 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
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, Tor s R ( k , k ) ≠ 0 and thus g l . d i m R ≥ s . On the other hand, as g l . d i m R = pd R k , the Auslander–Buchsbaum formula gives g l . d i m R = dim R . Hence, dim R ≤ s ≤ g l . d i m 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
ϵ 1 ( R ) = dim k H 1 ( x _ ) where x _ = ( x 1 , … , x d ) is a system of parameters. By definition, R is a complete intersection ring if dim R + ϵ 1 ( R ) is the dimension of the tangent space. (See Hartshorne for a geometric meaning.)
Injective dimension and Tor dimensions
Let R be a ring. The injective dimension of an R-module M denoted by id R M is defined just like a projective dimension: it is the minimal length of an injective resolution of M. Let Mod R be the category of R-modules.
Proof: Suppose g l . d i m R ≤ n . Let M be an R-module and consider a resolution
0 → M → I 0 → ϕ 0 I 1 → ⋯ → I n − 1 → ϕ n − 1 N → 0 where I i are injective modules. For any ideal I,
Ext R 1 ( R / I , N ) ≃ Ext R 2 ( R / I , ker ( ϕ n − 1 ) ) ≃ ⋯ ≃ Ext R n + 1 ( R / I , M ) , which is zero since Ext R n + 1 ( R / I , − ) is computed via a projective resolution of R / I . Thus, by Baer's criterion, N is injective. We conclude that sup { id R M | M } ≤ n . 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:
w . g l . d i m = inf { n | Tor i R ( M , N ) = 0 , i > n , M , N ∈ 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, w . g l . d i m R ≤ g l . d i m R .
Let A be a graded algebra over a field k. If V is a finite-dimensional generating subspace of A, then we let f ( n ) = dim k V n and then put
gk ( A ) = lim sup n → ∞ log f ( n ) log n .
It is called the Gelfand–Kirillov dimension of A. It is easy to show gk ( A ) is independent of a choice of V.
Example: If A is finite-dimensional, then gk(A) = 0. If A is an affine ring, then gk(A) = Krull dimension of A.
See also: Goldie dimension, Krull–Gabriel dimension.
