be a noetherian topological space. The
dimension of
Definition
Let
P be a prime ideal in a ring
R (as always, commutative with 1). The height of
P is defined to be
| ht(P) = sup
| { n : there exists a chain of
distinct prime ideals |
| P0⊂ P1⊂…⊂ Pn =
P⊂ R} |
Definition
The
Krull dimension of a ring
R
is defined to be
dim(R) = sup{ht(P) : P⊂ R is
prime}.
Lemma
Let
X = Spec(R) be an affine scheme. Then
dim(X) = dim(R).
Proof:
There is a one-to-one correspondence between the irreducible
closed subsets of X and the prime ideals of
R.
Proposition
Let
X be a noetherian topological space. Then
- dim(X) = sup(dim(Xi)), where the
Xi range through the irreducible components of
X.
- If U⊂ X is a dense open set and
if dim(U) is finite, then dim(X) =
dim(U).
- If Y-⊂ X, then
dim(Y)≤ dim(X). Moreover, if
X is irreducible and the inclusion is proper, then
the inequality is strict.
Proof:
(i) Let
Z0⊂…⊂ Zn be a
proper chain of irreducible closed subsets of
X. Since
Zn = ∪i(Xi∩ Zn)
and
Zn is irreducible, there exists an
i such that
Zn⊂
Xi. But then the entire chain is contained inside
Xi.
(ii) By (i), we may assume that X is
irreducible. If Z0⊂…⊂
Zn is a proper chain of closed irreducible subsets of
U, then their closures
Z-0⊂…⊂Z-n form a
proper chain of closed irreducibles of
X. (Properness follows because Zi =
U∩Z-i, and closures always stay irreducible.) So
dim(U)≤ dim(X).
Since dim(U) is finite, we can choose a chain
Z0⊂… Zn in U
of maximal length. Now let W be any irreducible
closed subset of X such that
Z-i⊂
W⊂Z-i+1. Intersecting back with
U yields
Zi ⊂ W∩ U ⊂ Zi+1.
Since the original chain had maximal length, one of these inclusions
must be an equality. But
W∩ U is dense in the
irreducible space
W, hence one of the earlier
inclusions in
X was already an equality. It follows
that
Z-0⊂…⊂Z-n is
a maximal chain in
X, and the dimensions are equal.
(iii) Without loss of generality, we may assume that
Y=Y-. Let
Z0⊂…⊂ Zn be a proper
chain in Y. Then it is also a proper chain in
X. If the containment is proper, then we can add
X to the top of the chain to lengthen it.
Proposition
Let X be a scheme that is a noetherian topological
space. Then dim(X)=0 if and only if
X is a finite set with the discrete topology.
Proof:
It is clear that any set with the discrete topology has dimension
zero, since the only irreducible subsets are singeltons. On the other
hand, we know that every scheme contains at least one closed point
(coming from a maximal ideal inside some ring defining an open affine
subset). If X contained a non-closed point, then
its closure would be irreducible, forcing the dimension of
X to be greater than zero. So, any
zero-dimensional scheme has the property that every point is
closed. Thus, in any affine open subset Spec(R) of
X, the ring R must have the
property that every prime ideal is maximal. This forces
Spec(R), and hence X to have the
discrete topology. Now use the finite decomposition into irreducibles
for the noetherian topological space X to obtain
the desired conclusion.
Theorem
(Dimension Theorem) Let
X be an algebraic
variety over a field
k. Then
dim(X) = tr.degk(k(X)).
Proof:
Since every variety contains an open, dense, affine variety, the
result will follow if we can establish the following three facts.
- The coordinate ring A(X) is an integral
extension of a polynomial ring.
- If B⊂ A is an integral extension
of noetherian rings, then dim(B)=dim(A) and
tr.deg(B)=tr.deg(A).
- If k is a field, then
dim(Ank) = n.
Let's start with the easiest one first.
Proposition
If k is a field, then dim(Ank) =
n.
Proof:
Since
(0) ⊂ (x1) ⊂ (x1, x2)
⊂…⊂ (x1,…, xn) is a
proper chain of prime ideals, we have
dim(An)≥
n.
On the other hand, we know that
n = tr.deg(k[x1,…,xn]). Moreover, if
A is an integral domain and if P
is any nonzero prime ideal in A, then
tr.deg(A/P)<tr.deg(A). For, take any
algebraically independent set x-1, …,
x-r∈ A/P. Lift them arbitrarily to elements
x1, …, xr∈ A, and choose a nonzero
element x∈ P. Then {x1,
…, xr, x} is an algebraically independent
set in A. To see this, suppose
F(t1,…, tr, tr+1)∈
k[t1,…,tr+1] is any polynomial such that
F(x1,…,xr,x) = 0 in
A. Since A is an integral
domain, we can assume that F is an irreducible
polynomial. Then F(x-1,…,x-r,0) = 0
in A/P. Since there are no nonzero algebraic
relations between these elements in A/P, the
polynomial F(t1,…,tr,0) must be
identically zero. But then F is divisible by
tr+1; by irreducibility,
F=tr+1. Since x≠0 in
A; this is a contradiction.
Now let P0⊂…⊂ Pt be a
chain of prime ideals in k[x1, …,
xn]. Then
n > tr.deg(A/P0) > tr.deg(A/P1) >…>
tr.deg(A/Pt),
so
n≥ t. Thus,
dim(An) ≤
n, and the result follows.
In order to establish the other two parts, we need some preliminaries
from commutative algebra; in particular, we need to know how prime
ideals behave in integral extensions of rings. To understand that, we
introduce one of the essential tools of commutative algebra:
localization.
Definition
A ring is called local if it contains exactly one
maximal ideal.
Definition
If A is a ring and if P is a
prime ideal in A, we define the
localization of A at
P to be the ring AP obtained by
inverting all elements of A\ P.
Definition
Let f:B⊂ A be an extension of rings. If
P⊂ A is a prime ideal, then
Q=f-1(P) = P∩ B is a prime ideal in
B. In this circumstance, we say that
P lies over
Q.
Lemma
Let B be a local ring with maximal ideal
Q. Let B⊂ A be an
integral extension. Then the set of prime ideals of
A lying over Q is just the set
of maximal ideals of A.
Proof:
We show first that every maximal ideal
M of
A lies over
Q. Define
N=M∩ B. Then
A- = A/M
is a field that is integral over the subring
B- =
B/N. Now, let
0≠ x∈
B-. Since
1/x∈A-, we have a
monic polynomial equation
(1/x)n + b1(1/x)n-1+…+bn = 0
with coefficients in
B-. Now multiply by
xn-1 to get
(1/x) = - (b1 + b2 x +…+ bnxn-1)∈ B-.
So,
B- is a field and
N=Q is
the unique maximal ideal in
B.
Next, we must show that every ideal P in
A that lies over Q is
maximal. This time, A- = A/P is an integral
domain that is integral (thus algebraic) over the field
B-=B/Q. Take 0≠ y ∈
A-. There is an algebraic dependence relation
b0yn + b1yn-1 +…+bn = 0
of minimal degree with coefficients in
B-. By
minimality,
bn≠0. Since
B- is a field, we can divide the polynomial by
bn, and assume that
bn=1. Now
we have
[1 / y]
= -(b0yn-1 + b1yn-2 +…+bn-1) ∈ A-.
Hence,
A- is a field and
P is
maximal.
Proposition
(The Lying-Over Theorem) Let B⊂ A be an
integral extension of rings. If Q⊂ B is
any prime ideal, then there exists a prime ideal P⊂
A lying over Q.
Proof:
Let BQ be the localization of B
at Q. Then AQ = (B\
Q)-1A is an integral extension of
BQ, and contains it as a subring. The prime ideals
of A lying over Q correspond to
the prime ideals of AQ lying over
QBQ, which are necessarily the maximal ideals of
AQ. Since BQ≠0, we know
that AQ is nonzero, so it has maximal ideals.
Proposition
(The Going-Up Theorem) Let B⊂ A be an
integral extension of rings. Let Q⊂ Q'⊂
B be prime ideals and let P⊂ A
be a prime ideal lying over Q. Then there exists an
ideal P' lying over Q' such that
P⊂ P'.
Proof:
The quotient A/P contains and is integral over
B/Q. By the Lying Over Theorem, there exists a
prime P'/P in A/P lying over the
prime ideal Q'/Q. By the isomorphism theorems,
P' is a prime ideal of A lying
over Q'.
Lemma
Let B be an integral domain that is integrally
closed in its field of
fractions L. Let K/L be a normal
extension with Galois group G, and let
A be the integral closure of B
in K. If Q is any prime ideal of
B, then G acts transitively on
the set of prime ideals lying over Q.
Proof:
One can assume (with some work that I'm omitting) that
K/L is a finite Galois extension. Suppose now that
P' and P are two primes lying
over Q, and that P' is not
contained in any of the conjugates Pi=
σi(P) of P for
σi∈ G. Then there is an element
x∈ P' that is contained in no
Pi. But y=NK/L(x)∈ B is
not in Q (because all
σi(x)∉ P) and is in
P'∩ A. This is a contradiction.
Proposition
(The Going-Down Theorem) Let B⊂ A be an
integral extension of integral domains, and assume that
B is integrally closed in its field of
fractions. Let Q⊂ Q'⊂ B be prime
ideals and let P'⊂ A be a prime ideal
lying over Q'. Then there exists an ideal
P lying over Q such that
P⊂ P'.
Proof:
Let K be the field of fractions of
A, let L be the field of
fractions of B, and let F be the
Galois closure of the field extension K/L. Inside
F, let C denote the integral
closure of B (and hence also of
A). By the Lying Over Theorem, there exists a prime
ideal Q0 in C lying over
Q. By the Going Up Theorem, there exists a prime
ideal Q'0⊃ Q0 lying over
Q'. Also by the Lying Over Theorem, there exists a
prime ideal Q'1⊂C lying over
P'. Now the ideals Q'0 and
Q'1 lie over the same ideal Q'
in B. Because the extension is Galois, there exists
an automorphism σ ∈ Gal(F/L) such that
σ(Q'0) = Q'1. Now the result follows by
taking P= σ(Q0)∩ A.
Now we can carry out part (ii) of the proof of the dimension theorem.
Proposition
Let B⊂ A be an integral extension of
noetherian rings. Then dim(B)=dim(A) and
tr.deg(B) = tr.deg(A).
Proof:
Since integral extensions are algebraic, the equality for
transcendence degrees is trivial.
The Lying Over and Going Up Theorems show that any chain of prime
ideals in B can be lifted to a chain in
A of the same length; thus, dim(A)≥
dim(B). Conversely, if P⊂ P'
are distinct prime ideals in A, then the ideals
P∩ B ⊂ P'∩ B must also be
distinct. (Localize at the prime ideal P∩ B;
since the ideals lying over its maximal ideal are precisely the
maximal ideals of the localization of A, there
cannot be any proper containments between them.) Thus, any chain of
primes in A produces a chain of the same length in
B, and dim(A)≤ dim(B).
Finally, we can carry out step (i).
Theorem
Let
A=k[x1, …, xn] be a polynomial ring
over a field
k, and let
P be an
ideal of
A of height
h. Then
there exist elements
v1, …, vn∈ A
such that
- A is integral over k[v], and
- P∩ k[v] = (v1, …, vh).
Proof:
The proof is by induction on the height
h=ht(P). The case
h=0 is
trivial, for then
P=0 and we can take
vi=xi. A better base for the induction, however,
is provided by the case
h=1. Choose a nonzero
polynomial
v1=f(x)∈ P. Write
f(x) =
∑i=1s ci Mi(x) where
ci∈
k and each
Mi(x) is a monomial. Given
any positive integers
d1,…,dn, we define
the
d-weight of a monomial
M(x)=∏
xiai to be
∑ aidi. Now
choose weights
d1=1, d2, …, dn so that
all the monomials appearing in
f(x) have distinct
weights. (We can achieve this by using large prime powers for the
weights.) Put
vi=xi-x1di for
i=2,…,n. Then
| v1 | = f(x) = f(x1, v2+x1d2, …, vn +
x1dn) |
| = ajx1e + g(x1, v2, …, vn) |
where
g is a polynomial whose degree in
x1 is strictly less than
e, and
aj is the coefficient of the term of largest
weight in
f. It follows that
x1
is integral over the ring
k[v1, …, vn],
and hence so are the elements
xi= vi+x1di.
We're not done yet, however, since we only know that
k[v]⊂ k[x] is an integral extension. We
still need to mod out the prime ideal P and its
pullback P∩ k[v]. We clearly have a
containment (v1) ⊂ P∩ k[v]. In
fact, however, both these ideals are primes of height 1, so they must
be equal. (Note: that P∩ k[v] is height 1
uses the fact that k[v] is integrally closed, so
the Going Down Theorem applies.)
Now suppose h>1. Choose an ideal
Q⊂ P of height h-1. By
induction, there exist w1, …, wn such
that k[x] is integral over k[w]
and Q∩ k[w] = (w1, …,
wh-1). Define P'=P∩ k[w]. Then
P' also has height h, so it
contains (w1,…,wh-1) properly. Choose a
nonzero polynomial of the form f(wh, …, wn)∈
P' and repeat the weight argument of the case
h=1.
Theorem
(Noether Normalization Theorem) Let A be an
integral domain that is finitely generated as an algebra over a field
k. Then there exist elements y1,
…, yr∈ A that are algebraically independent
over k such that A is integral
over k[y1, …, yr].
Proof:
Write A=k[x1, …, xn]/P where
P is a prime ideal of height
n-r. By the previous theorem, there are elements
v1, …, vn in k[x1, …,
xn] such that k[v]⊂ k[x] is
integral and such that P∩ k[v] = (vr+1, …,
vn). The result follows by taking yi ≡
vi mod P.
Corollary
Let
A be an integral domain that is finitely
generated as a
k-algebra. Then for any prime
P in
A, we have
ht(P) + dim(A/P) = dim(A).
Proof:
Reduce to the case A=k[x1,…,xn] and use
the proof of the Noether Normalization Theorem.
Corollary
If X and Y are algebraic varieties,
then dim(X× Y) = dim(X) + dim(Y).
Proof:
The transcendence degree of a tensor product satisfies the
corresponding relation.
Proposition
Let X be a hypersurface in
Ank. Then every irreducible component of
X has dimension n-1.
Proof:
We may assume that X is irreducible, and hence of
the form Z(f) for a nonconstant irreducible
polynomial f∈ k[x1,…,xn]. By
renumbering the variables, we can assume that xn
actually occurs in f. Now let ti∈
A(X) be the image of xi. We claim that
{t1, …, tn-1} is an
algebraically independent set, and therefore dim(X)≥
n-1. For, suppose that G(t1, …, tn-1)
=0 is an algebraic relation. Then G∈ I(X)=
(f), so f divides
G. But this is impossible, since
xn occurs in f but not in
G. On the other hand, the proof that
An has dimension n shows
that dim(X)≤ n-1.
Theorem
Let X⊂Ank be an algebraic set all of
whose irreducible components have dimension
n-1. Then X is a hypersurface.
Proof:
Without loss of generality, we may assume that X is
irreducible. Let f∈ I(X) be a nonzero
polynomial. By irreducibility, some irreducible factor
h of f must vanish on
X. So, X⊂ Z(h). But
this is an inclusion of irreducible sets of the same dimension, so it
must be an equality.
Comments on this web site should be addressed to the
author:
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210