Download 23. Dimension Dimension is intuitively obvious but - b

58
RICHARD BORCHERDS
23. Dimension
Dimension is intuitively obvious but surprisingly hard to define rigorously and
to work with.
There are several different concepts of dimension
• It was at first assumed that the dimension was the number or parameters
something depended on. This fell apart when Cantor showed that there is
a bijective map from R → R2 . The Peano curve is a continuous surjective
map from R → R2 .
• The Lebesgue covering dimension: a space has Lebesgue covering dimension
at most n if every open cover has a refinement such that each point is in
at most n + 1 sets. This does not work well for the spectrums of rings.
Example: dimension 2 (DIAGRAM) no point in more than 3 sets. Not
trivial to prove that n-dim space has dimension n. No good for commutative
algebra as A1 has infinite Lebesgue covering dimension, as any finite number
of non-empty open sets intersect. .
• The ”classical” definition.
Definition 23.1. (Brouwer, Menger, Urysohn) A topological space has
dimension ≤ n (n ≥ −1) if all points have arbitrarily small neighborhoods
with boundary of dimension < n.
The empty set is the only space of dimension −1. This definition is
mostly used for separable metric spaces. Rather amazingly it also works
for the spectra of Noetherian rings, which are about as far as one can get
from separable metric spaces.
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Definition 23.2. The Krull dimension of a topological space is the supremum of the numbers n for which there is a chain Z0 ⊂ Z1 ⊂ . . . ⊂ Zn of
n + 1 irreducible subsets.
DIAGRAM pt ⊂ curve ⊂ A2 For Noetherian topological spaces the Krull
dimension is the same as the Menger definition, but for non-Noetherian
spaces it behaves badly. Any Hausdorff space has dim 0 with this definition,
as only irreducible sets are points.
Various forms of (real-valued) Hausdorff dimension of fractals: these mostly
require a metric so are of no use for commutative rings.
Transfinite dimensions: we can extend the definition of dimension to allow
the dimension to be a transfinite ordinal. This seems to be of little use in
ring theory.
The deviation of a poset. A poset has deviation ≤ α (an ordinal) if for any
descending chain all but a finite number of the intervals between element of
the chain have deviation less than α. For a Noetherian ring the dimension is
the deviation of its poset of ideals. This definition has the advantage that it
works well for noncommutative Noetherian rings and modules. (Note that
the poset of rationals does not have a deviation, so some non-Noetherian
rings do not either.)
Algebraic definition: main idea is that higher dimensional spaces have more
functions on them.
Example 23.3. dim B/k = tr. deg. quotient field of B over k.
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This works fine for ”geometric” domains such as coordinate rings of
affine varieties (and used to be used as the definition) but does not work
well on ”arithmetic” rings such as Z: SpecZ should have dimension 1, but
its field of fractions has transcendence degree 0. (Even for geometric rings
the definition needs some fudging for things like rings with zero divisors or
complete local rings.)
• The Hilbert polynomial.
Example 23.4. Easiest definition to calculate uses Hilbert polynomial of
a local ring. Look at dim(A/mk ): this is a polynomial in k for large k of
degree d, and d is the dimension of the local ring. It is also the same as the
dimension of the completion as these have the same Hilbert polynomials.
For example, k[x1 , . . . , xn ] has dimension n. k[x, y]/(y 2 − x3 − ax − b)
has dimension 1.
• The Gelfand-Kirillov dimension for possibly non-commutative finitely generated algebras R over a field is
(23.1)
d = lim sup log dim Rn / log n
n
where Rn is the subspace generated by monomials in the generators of
length at most n. (Informally Rn has dimension very roughly nd ). When
R is commutative this is the same as the dimension defined using Hilbert
polynomials. For non-commutative rings the dimension can be 0, 1, or any
real number at least 2, or infinity.
• Dimension of a ring: Dimension usually defined as the Krull dimension
of its spectrum, in other words sup n such that P0 ⊂ P1 ⊂ . . . ⊂ Pn are
distinct prime ideals. Dim R = sup dim Rm , m maximal, so can be reduced
to dimension of local rings. Problem: What are irreducible subsets of (say)
A4 ? Hard to describe, and one needs nontrivial commutative algebra to
calculate dimension.
• The dimension of the tangent space at a point works for non-singular
schemes but fails when there are singularities. This is the number of generators of a maximal ideal of a local ring. A variation that gives a good
definition of dimension is the minimal number of elements of a system of
parameters (generating an ideal containing some power of m).
• Homological dimension. There are several ways to define dimension as ”the
number beyond which certain homology or cohomology groups vanish”.
These definitions work well, but require developing a lot of homology theory.
Most of the topological definitions are ”local” in the sense that one can define
the dimension of a space at a point, and the dimension of the space is the sup of
the dimensions at a point.
Theorem 23.5. The Krull dimension is at most equal to the dimension. It is at
least the dimension for Noetherian topological spaces.
Proof. Suppose the Krull dimension is at least n. Let I0 , I1 , . . . , In be a chain of
n + 1 closed irreducible subsets such that Ii is in Ii+1 . Then by induction on i we
see that Ii is of dimension at least i, because for any point in Ii+1 the set Ii is in
the boundary of any neighborhood not containing Ii . So the space has dimension
at least n.
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RICHARD BORCHERDS
Now suppose the space is Noetherian. We show that for every closed subset
the Krull dimension is at least the dimension. Otherwise there is a minimal closed
subset that does not have this property, which we may as well assume is the whole
space. The space is irreducible, otherwise we can write it as the union of two
proper closed subsets, for which the Krull dimension is at least the dimension by
Noetherian induction. Suppose the space has dimension at least n. We can find a
point p and a neighborhood of it whose boundary has dimension at least n − 1, so
there is a chain of n irreducible subsets not containing p. But then we can add the
whole space to get a chain n + 1 irreducible subsets, so the Krull dimension is at
least n.
In particular for any Noetherian rings, the Krull dimension of the spectrum is
the Menger dimension. This also coincides with the dimension defined using Hilbert
polynomials, or deviations of posets, or systems of parameters, and for varieties over
fields is the same as the definition using transcendence degree. For non-Noetherian
rings the dimension does not seem to be such a useful concept: the Krull dimension
is usually infinite and behaves in strange ways: for example the dimension of R[x]
can be larger than the ”expected” value 1 + dim R. There seems to be no really
satisfactory concept of dimension for non-Noetherian rings, and in any case many
of them tend to be ”infinite dimensional”.
We want to show the following are equal, for Noetherian local rings:
• The Krull dimension
• The degree of the Hilbert polynomial
• The smallest cardinality of a system of parameters
We will prove this by proving 3 inequalities between these numbers. This will take
some time and is somewhat technical.
The Hilbert polynomial.
Suppose that R = ⊕Rn is a graded ring finitely generated over R0 and M = ⊕Mn
a finitely generated graded module over it. We want to study the growth of Mn .
Usually we will take R0 to be a field, in which case we measure the size of Mn by its
dimension lambda(Mn ) (as a vector space). It is sometimes useful to allow R0 to be
an Artinian ring, in which case we can measure the size of Mn by its length λ(Mn ),
which generalized the dimension of a vector space. We recall that the dimension
(or lenght) is additive for short exact sequences of modules.
We put all these lengths together in the POINCARE SERIES
X
P (t) =
λn tn
Theorem 23.6. The Poincare series is a rational function of t of the form
Y
f (t)/ (1 − tki )
for some polynomial f .
Proof. We take the elements xi to be homogeneous generators of R over R0 of
degrees ki . We will prove the theorem by induction on the number of generators.
If the number of generators is 0 so the R = R0 then the therom is trivial as Mn = 0
for n large.
Now look at the exact sequence
0 → Kn → Mn →×xs Mn+ks → Ln+ks → 0
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where K and L are both modules where xs acts as 0.
This shows that
λ(Kn ) − λ(Mn ) + λ(Mn+ks − λ(Ln+ks ) = 0
so
PK (t) − PM (t) + t−sk PM (t) − tsk PL (t) = 0
so
PL (t) − tks PK (t)
1 − tks
The result now follows by induction on the number of generators, as both K and L
can be regarded as modules over the quotient R/(xs ).
PM (t) =