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.