Lecture 12-13 Dimension
Introduction
We use throughout the Noetherian condition as a finiteness
assumption for rings and modules: it is convenient to use, and in
practice it covers most of the cases we need.
I treat dimension theory (originally due to Krull) following
[Matsumura, Chap 5]. Intuitively, there are 3 different basic points
of view, and we set these up in rigorous algebra, and show that they
coincide in appropriate situations.
(1) Dimension n means that a ball of radius r has volume ~ r^n, or
the number of lattice points in a simplex of size is
(r + bit choose n) ~ (const.)*r^n/n! + l.o.t.
(2) A n-dimensional body is something you can cut by a hyperplane
(setting a single equation equal to zero) to get down to n-1
dimensions, and so on to dimension 0, that is, a point or a finite
set of points.
(3) Krull dimension n is the longest chain of prime ideals
P0 in P1 in .. Pn or the longest chain of irreducible subvarieties
pt in curve in surface in .. n-dimensional subvariety in a variety.
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Graded ring
I work mostly with NN-graded rings and modules NN = {0,1,2,..}
(gradings by more general semigroups are also useful in different
contexts)
A graded ring is R = directsum_{n>=0} R_n where R is a ring, and on
the right, the R_n are additive subgroups with ring multiplication
doing R_n1 x R_n2 -> R_{n1+n2}.
It follows of course that R0 is a subring, with 1_R in R0, and each
Rn is an R0-module. Also, J = directsum_{n>=0} R_n is an ideal such
that R/J = R0. Elements of Rn are homogeneous of degree n. Any f in
R is uniquely expressible as a finite sum f = sum_n fn with fn
homogeneous of degree n (the homogeneous _component_ or homogeneous
_piece_ of f).
In dimension theory, we often restrict to the case that R0 is
an Artinian ring, or even just R0 = k is a field. (There will
eventually be a way around this.)
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Lemma [Standard elementary fact]
A graded ring R is Noetherian if and only if
R_0 is a Noetherian subring and
R is generated as an R_0 algebra by finitely many homogeneous elements
xi in R_di
both implications are straightforward (please think them through).
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Why graded rings?
grading ~ representation of CC^x ~ equivalence relation defining PP^n
The equivalence relation defining PP^n is an action of the algebraic
group Gm(k) = GL(1,k), vulgarly known as CC^x.
Projective space PP^n has homogeneous coordinates (x0,..n), with a
point P in PP^n the ratio (x0 : x1 :.. xn). The xi are not meaningful,
but the ratio xi : xj or the rational function xi/xj is well defined
if xj(P) <> 0.
If f in k[x0,..xn] is homogeneous of deg d, changing all
xi |-> la*xi then changes f(x0..n) |-> la^d*f, so the condition
f(P) = 0 is well defined.
Implicit in this is the key point that the algebraic group Gm is
_reductive_. Every representation of Gm splits into 1-dimensional
representations, and every 1-dimensional representation of Gm is
given by a homomorphism Gm -> Gm, which is an nth power map
z |-> z^n for some n in ZZ.
Making a vector space V into a representation of Gm is exactly the
same as putting a ZZ-grading on V: V = sum_{n in ZZ} Vn, where Gm
acts on Vn by v |-> la^n*v.
====
Graded modules and graded ideals
Given R as above, a graded module M over R is an R-module with a
grading M = directsum_n M_n, with the ring operation doing
R_a x M_n -> M_{a+n}. For some purposes, we might want to allow the
grading of M to have negative pieces M_n (win n<0), but a finite
module M only has finitely many, and the textbooks assume n >= 0
for ease of notation when setting up the theory.
A homogeneous ideal or graded ideal I in R is a particular case of
graded submodule. It is an ideal I in R that is of the form
I = sum In where In = I intersect Rn. Another way of saying this:
for every element f in I, write f = sum fn in the graded structure
of R; then the homogeneous pieces fn of f are also in I.
In the I <-> V correspondences between subvarieties V in PP^n_k and
ideals I in k[x0.. xn] (the homogeneous coordinate ring of PP^n),
note that the I taking part are the homogeneous ideals with
nilrad(I) not contained in (x0,.. xn),
meaning that they define a nonempty subvariety
Cone(V) in k^{n+1} \ 0
that is invariant under the Gm action, the _affine cone_ over V.
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Hilbert series:
For the present, consider an NN-graded ring R = sum Rn with the
additional condition that R0 is an Artinian ring (e.g., simply a
field). Write l(P) = length_R0(P) for the length of a finite R0
module; if R0 = k is a field, l(P) is just the dimension of P as a
k-vector space. Suppose also that R is Noetherian, so generated over
R0 by finitely many homogeneous elements {xi in R_di}, that is:
R = R0[x1,.. xr]/IR (where IR is the ideal of relations) (1)
When there are different di, we use _weight_ (or weighted degree) as
another word for degree, to avoid the ambiguity between homogeneous
and weighted homogeneous monomial: a monomial prod xi^ai has weight
sum ai*di.
For given weight n, there are only finitely many monomials prod xi^ai
of degree sum ai*di = n, so each homogeneous piece Rn of R is finite
over R0, and of finite length. This gives us something to count.
The same applies to a finite R-module M: each Mn is a finite R0
module.
The Hilbert series of M is defined by setting
Pn(M) = length(Mn) for n >= 0, and
P(M,t) = sum Pn(M)*t^n.
This is a formal power series in t, a generating series for the
infinitely many coefficients. Its big selling point is that it takes
a closed form, giving lots of useful information for little effort.
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Theorem Let R be as in (1). Then P(M,t) is a rational function of t
with denominator product_{i=1}^r (1-t^di), where di = weight xi.
That is
P(M,t) = H(M,t) / prod_{i=1}^r (1-t^di).
Here H(M,t) is a polynomial with integer coefficients (called in
different contexts the Hilbert numerator).
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The "straight" homogeneous case. The case all di = 1 is important.
It is known (and argued below) that
1/(1-t)^{m+1} = sum_{n >= 0} (n+m choose n)*t^n. (*)
To get the hang of it, write it out for m = 0, 1, 2.
Corollary. When all di = 1, suppose that the numerator H(M,t) has
degree D. Then P_n(M) is a polynomial in n for n >= D.
Proof of corollary. In fact if the numerator
H(M,t) = a0 + a1*t + .. aD*t^D,
then P(M,t) = H(M,t)/(1-t)^{m+1}
= H(M,t) * RHS of (*),
so that when n >= D the coefficient of t^n is
a0*(n+m choose n) + a1*(n+m-1 choose n) + .. + aD*(n_m-D choose n),
which is a polynomial.
Proof of Theorem. The method of proof is induction over the number r
of generators of R. If r = 0 then M is a finite graded module over
the Artinian ring R0, so is a sum of finite length R0-modules Mn in
finitely many degrees n, and P(M,t) is just a polynomial.
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Inductive step. The main step is the following induction: consider
multiplication by xr
mu_{xr}: M -> M.
This increases degrees by dr.
Now the kernel K = ker mu_{xr} in M is the submodule annihilated by
xr, so is a graded module over Rbar = R/(xr). Also, by construction,
N = coker mu_{xr} = M/(xr*M) is a graded R-module annihilated by xr.
Write the resulting exact sequence
0 -> K -> M -xr-> M -> N -> 0. (2)
Note that M is a graded module over R, whereas by what I just said,
K and N are graded modules over the quotient ring Rbar = R/(xr),
which is a graded R0-algebra of the same shape as (1), but now with
only r-1 generators.
Thus by induction on r, I can assume that the result is already
known for both K and N: the Hilbert series for K and N have the form
P(K,t) = H(K,t) / prod_{i=1}^{r-1} (1-t^di), and (3)
P(N,t) = H(N,t) / prod_{i=1}^{r-1} (1-t^di).
where the numerators are polynomials with integer coefficients.
Since mu_{xr} increases degrees by dr, the exact sequence (2) breaks
up according to weighted degrees as exact sequences
0 -> K_{n-dr} -> M_{n-dr} -> M_n -> N_n -> 0 (4)
of R0-modules of finite length, one for each n.
This gives the equality
l(M_n) - l(M_{n-dr}) = N_n - K_{n-dr} for each n. (5)
Now sum (3) * t^n gives
(1-t^dr) * P(M,t) = P(N,t) - t^{-dr} * P(K,t). (6)
This is the result we wanted. In fact both terms on the r-h.s.
are of the form polynomial (or possibly Laurent polynomial)
divided by prod^{r-1} (1-t^di). Therefore P(M,t) equals
polynomial / prod^r (1-t^{di}). QED
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Explicit cases
The classic case is R0 = k and all the di = 1. This means discussing
rings of the form k[x0,.. xn] with the ordinary homogeneous grading,
or its quotients k[x0,.. xn]/I with I a homogeneous ideal.
k[x0,.. xn] is the homogeneous coordinate ring of projective
space PP^n_k over k.
For a prime ideal I (over algebraically closed k for simplicity),
k[x0,.. xn]/I is the homogeneous coordinate ring of the projective
variety X = V(I) in PP^n. To avoid complications, I only treat the
case X = V(f) where f is an irreducible homogeneous polynomial of
degree m, so that X in PP^n is a hypersurface of degree m.
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Proposition (1) Set R = k[x0,.. xn]. Its rth homogeneous component
is the vector space
S^r(x0,.. xn) = {x0^r, x0^{r-1}*x1,.. xn^r}
with basis the (n+r choose n) monomials of degree d in R. Therefore
R has Hilbert series
P(R,t) = sum_{r >= 0} (n+r choose n)*t^r = 1/(1-t)^{n+1},
and dim R = n+1. Notice that the dimension
Pr = (n+r choose n) = r^n/n! + l.o.t.
(Here the denominator n! is the volume of the n-dimensional simplex
as a fraction of the n-dimensional unit cube.)
(2) Set Rf = R/(f) with f as above. Then
P(Rf,t) = (1-t^m) / (1-t)^{n+1} = (1+t+.. t^{m-1}) / (1-t)^n.
Its order of growth has the leading term m * r^{n-1} / (n-1)!.
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Proof Start from the formal expansion for a single variable:
1/(1-x) = 1 + x + x^2 + .. x^n + ..
If there are 2 variables, 1/((1-x)*(1-y)) is the formal product
of the two power series
= (1 + x + x^2 + .. x^n + ..)*(1 + y + y^2 + .. y^n + ..)
= sum_{i,j} x^i*y^j,
or in other words, the sum of all monomials in k[x,y] each with
coefficient 1.
In the same way
1/(prod (1-xi)) = sum m m in k[x0..xn] a monomial
Now substitute all xi = t. The left-hand side is 1/(1+t)^{n+1}.
The r-h.s is a sum of t^r taken over all monomials prod xi^ai
with sum ai = t^r. In other words, it is a power series in t,
with the coefficient of t^r equal to the number of monomials
of degree r. That is statement (1).
The number of monomials is (n+r choose n): think of the rule for
Pascal's triangle: on multiplying the series for 1/(1-t)^{n-1} by
the series for 1/(1-t), the coefficient of t^r is the cumulative sum
of the coefficients of all t^i in 1/(1-t)^{n-1} from i = 0..r. Thus
the coefficient of t^r equals the coefficient of t^{r-1} in the
product plus the coefficient of t^r in the first factor.
Or in formulas
(n+r-1 choose n) + (n+r-1 choose n-1) = (n+r choose n)
LHS is (n+r-1)! / n!.(r-1)! + (n+r-1)! / (n-1)! r!
In the first term mult top and bottom by r; in the second term mult
top and bottom by n. This gives
(r+n) * (n+r-1)! / n! * r! = RHS.
For (2), Rf = R/R*f (divided by the principal ideal (f)). The
exact sequence
0 -> R*f -> R -> Rf -> 0
splits into homogeneous terms
0 -> R_(r-m) -> R_r -> Rf_r -> 0.
Or the monomials of degree r in R map to a generating set of
Rf_r, with multiples of f giving linear dependant relations.
So
dim Rf_r = (n+r choose r) - (n-m choose r)
or = (n+r choose r) if n < m.
This gives
P(Rf,t) = (1-t^m)*P(R,t) = (1-t^m)/(1-t)^(n+1)
= (1+t+..+t^(m-1))/(1-t)^n.
The conclusion is dimension one less, but degree *m.
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Weighted homomogeneous case.
When the di are not all equal to 1, you don't expect
Pn(M) to be a polynomial in n.
[A&M] and [Ma] say nothing about the general case, but move
on to the next topic. I mention the following:
Theorem [Zariski] R = R0[x1,..xk]/I and M as above, with
deg xi = di.
Separate the n into the different congruence classes
i mod N = prod di. Then there are polynomials p_i such that
Pn(M) = p_i(n) for n == i mod N and n >> 0.
In other words, P_n is not given by a polynomial, but
by a "periodic choice of polynomial": there exists N,
and polynomials p0,p1,.. p_{N-1} s.t.
for n >> 0, Pn(M) = p_i for n == i mod n.