Lectures 15-17
Krull's dimension theorem

Finish Hilbert series
  in straight homogeneous case (graded in deg 1)

Hilbert series R = R0[x1..xm]/I with R0 Artinian.
Restrict now to straight homogeneous case deg xi = 1. We have
seen P(R,t) = N(t)/(1-t)^m where N is a polynomial with integer
coefficients.
There is a similar result for a finite module N.

By clearing common factors, assume that N(t) does not have t=1 as
a root, that is, P(R,t) of P(M,t) has pole of order n. Then for
n >> 0, Pn(M) is a polynomial function of n growing as n^d/d!

P(R,t) = N(t)/(1-t)^d, and assume that the fraction is
in normal form, that is (1-t) not | N(t) or N(1) <> 0.
Then Pn(R) = length(Rn) as R0 module
  is a polynomial in n for n >= D (so for all N >> 0)
  and the leading term is N(t) * n^d / d!, where
  N(t) = sum ai.

Mention Zariski's cyclic polynomials.
In the weighted homogeneous case, R is a graded ring with generators in
possibly different degrees (weights) deg di. Write N = prod di.

The result concerning the shape  N(t)/Prod (1-t^di)  of the
Hilbert series then does not imply that Pn = l(Rn) is a
polynomial function of n. Instead, Zariski asserts that there are
different polynomials Q0, Q1,.. Q_{N-1} such that
  Pn = Q_i(n) for n == i mod N.
So Pn has values that take a choice of polynomial values.

The weighted polynomial rings are important in various branches
of higher dimensional algebraic geometry, but we leave them for
now. Some examples are given in the exercises. The material of
this chapter passes to associated graded rings, that are straight
graded in degree 1.

====
Move on to filtration M = M0 > M1 > ..
esp. I-adic filtration for suitable I.

When discussing completions up to now, we were interested in the
inverse system R/I^n. and lim_{<-} R/I^n to get the I-adic
filtration. Here the focus is on the _associated graded_ ring
Gr_I R = sum I^n/I^{n+1} and its modules Mi/M_{i+1}. Under
appropriate assumptions, this is straight homogeneous (all di=1).

The appropriate assumptions, can arrange that all the quotients
I^i/I^{i+1} have finite length, and the Hilbert series still
expresses the "size" or dimension of R or of M -- and moreover,
that dimension is independent of I up to a scaling factor. The
dimension controls the order of growth of the length or "number
of monomials" of R/I^n as n-> infty.

====
Hilbert-Samuel function

If A,m,k is a Noetherian local ring then each m^n / m^{n+1} is a
finite dimensional vspace over k = A/m.

Ex. If A = OX,P is local ring of point p in X then
  m/m^2 = vspace dual to tangent space T_X,P.

But you could also consider ZZ[x] and its localisation A,m,FFp at
(x,p), which is a ring of mixed characteristic. Then A/m^n
contains both ZZ/p^n and FFp[x]/x^n, so is a curious mix incl.
both p-adic and formal power series in x. This is not a vector
space over any field, but it has a finite length, with all the
quotients Mi/M^{i+1} in any Jordan-Hoelder sequence is to FFp.

====
[Ma] works slightly more generally, saying semilocal ring and I
an "ideal of definition". Take A a semilocal ring with maxl ideals
mi and Jacobson radical J = intersect mi. The mi are strongly
coprime, so that
  J^n = intersect mi^n = product mi^n
and A/J^n = sum A/mi^n. So the whole story happens locally as a
direct sum of what happens at each of the mi separately. It is
clearer to work only with the local case.

An ideal I with J^n in I in J is called an "ideal of definition".
The point of the definition is that A/I^n or M/(I^n*M) is still a
module of finite length, which still gives us something to count.
I restrict myself to the local case, so ideal of definition boils
down to m-primary ideal (ideal sandwiched between m^n and m).

====
What is dimension? For an algebraic variety V,
  dim V = Tr dim k(V)/k . 
One see that this equals the max length of a chain of irreducible
varieties V, so equals the Krull dimension of an affine
coordinate ring k[V] or of its local ring O_{V,P}. The dim in
terms of k(V) is important in geometry as a birational invariant.

But comm alg is mostly about rings after localisation, and [Ma]
is concerned with doing this properly.

For A,m,k a local ring, there are 3 definitions via the
quantities:

d = order of growth of Hilbert-Samuel series
   (like the number of monomials needed to base A/m^n)
   I need to go through some work to define this
   and establish its main properties

de = smallest number of generators of an m-primary ideal
   m-primary ideal means rad(I) = m or m^n < I < m
   A/I does not have any associated primes P other than m.
That is, every J-H sequence for A/I^n has every successive
quotient iso to A/m = k.

dim = Krull dimension = max length of chain of prime ideals.

===
A local, I an m-primary ideal
(that means m^n < I < m or A/I Artinian. The case I = m is most
important)

length of I-primary module is number of occurrences of the field
k = A/m among M_i/M_{i+1} in a JH filtration. This does not
depend on whether we use I or m as the basic object.

I-adic filtration is A > I > I^2 ..
Associated graded ring is

 Gr_I A = direct sum_{n\ge0} I^n/I^{n+1}

is an NN-graded ring with degree 0 term A/I Artinian, and is
generated in degree 1, so it has Hilbert-Samuel function

 HS_A(n) = l(A/I^{n+1}) = sum_{i in 1..n} l(I^i/I^{i+1})

======
Lecture 16

Recall Hilbert series and function:
R = sum_{i>=0} R_i with R_0 Artinian and R finitely generated by
xi in R_di. From now on, assume all di = 1. The Hilbert series is
defined as the generating series
  P_R(t) = sum l(Ri)*t^i
for the length of the graded pieces Ri. The result is that this
power series is a rational function of t of the form
  P_R(t) = N(t)/(1-t)^d
with N a polynomial in ZZ[t] and sum of coeffs = N(1) <> 0.

Corollary: for n >= deg N, the length
  l(Rn) is a polynomial function of n of degree d-1
  with leading term N(1)*n^{d-1}/(d-1)!
The same kind of result applies to a finite graded module M over R.

====
The Hilbert-Samuel function applies this result to a ring A with
I-adic filtration via the associated graded ring
  Gr_I(A) = direct sum_{i>=0} I^i/I^{i+1}.
For this to work, I need the 0th graded piece A/I to be Artinian,
and I a finitely generated ideal (automatic since A is
Noetherian).

Let A,m,k be a Noetherian local ring

An ideal I is m-primary if m^n < I < m for some n.

For example: If P in X is a point on an affine algebraic variety
over k, and A = Oh_{X,P} the local ring of X at P, then
  I in A is m-primary <=>
        every f in I has f(P) = 0, and
        for every xi in mP, some xi^ni in I.
Geometrically, this means P in X is an isolated point component
of the algebraic set V(I).

It follows directly from the definition that if I, J are both
m-primary then there are a,b such that
  I^a < m^a < J and J^b < m^b < I.

Thus the I-adic topology on a finite module M is independent of
I, and likewise the order of growth of the Hilbert function and
(below) the Hilbert-Samuel function.

====
Define the Hilbert-Samuel function of A wrt I by
  HS^I_A(n) = length A/I^{n+1} = sum_{i=0}^n l(I^i/I^{i+1}).
([A&M] and [Ma] both use \chi^I_A.) The RHS makes clear that
this is determined in a simple way from the Hilbert series of
the associated graded ring Gr_I(A).
The HS function of a finite A-module M is defined in the same
way in terms of the I-adic filtration I^n*M and the associated
graded module Gr_I(M) = (I^i*M)/(I^{i+1}*M).

====
Proposition-Definition For n >> 0, HS^I_A(n) is a polynomial
of degree d in n. Its leading term is N(1)*n^d/d! where the
Hilbert series of Gr_I(A) is N(1)/(1-t)^d.

The same statement holds for modules: HS^I_M(n) is a polynomial
for n >> 0, of degree d and with leading term N(1)*n^d/d! where
P(Gr_I(M)) = N(1)/(1-t)^d.

The d appearing here is the HS dimension d(A), respectively d(M),
one of the equivalent definitions of dimension in the discussion
below.

The small point to note here is that the Hilbert series
relates to the length of the graded piece I^n/I^{n+1}
and its leading term is N(1)*n^{d-1}/(d-1)! For example,
the dimension of the space k[x1,..xn]_d of polynomials of
degree = d. The HS function is the cummulative total of
these length of A/I^{n+1} = sum{i<=n} length(I^i/I^{i+1}),
like the space of polynomials of degree <= d.

In the same context, for a finite A-module M, (and I an m-primary
ideal), the graded module of the I-adic filtration
   Gr_I(M) = direct sum I^n*M/(I^{n+1}*M)
has Hilbert series defined in terms of the lengths of the
graded terms, and Hilbert-Samuel series
  HS^I_M(n) = length M/I^{n+1}M = sum_{i=0}^n l(I^iM/I^{i+1}M).
As for the ring, define d(M) so that
  HS^I_M(n) grows as a polynomial in n of degree d for n >> 0.

====
Lemma (Theorem 13.3 of [M]) A,I as above, M a finite module,
N < M a submodule with quotient M/N.

Then the HS dimensions of the 3 modules satisfy
  d(M) = max(d(N), dim(M/N)).
Morever, in the case d(M) = d(N) = dim(M/N), the leading
coefficient of the HS function of M equal the sum of those of N
and M/N.

An interesting point is that the proof involves the Artin-Rees
lemma in its full form.

====
Proof. The idea is to take the I-adic filtration of the exact
sequence. For the quotient,
  (M/N)/I^n(M/N) = M/(N+I^nM).   (1)
Consider N+I^nM as an intermediate submodule
  I^nM < N + I^nM < M.
This gives M/I^nM as an extension of successive quotients
  (N+I^nM)/I^nM  and  M/(N+I^nM),
with the second term as in (1). The second isomorphism theorem
turns the first term into N/(N intersect I^nM).

Thus
   l(M/I^nM) = l(N/(N intersect I^nM)) + l((M/N)/I^n*(M/N)).  (2)

This is of the general shape we want, but we still need to relate
the first summand to invariants of the I-adic filtration of the
submodule N. For this, apply the Artin-Rees lemma: there exists c
such that
   N intersect I^nM  =  I^{n-c}(N intersect I^cM)
for all n >= c.

Hence
   I^nN < N intersect I^nM < I^{n-c}N.
This implies that the first term of (2) has the same order of
growth as the polynomial HS^I_N(n), and the same leading term: in
fact it is sandwiched between HS^I_N(n-c) and HS^I_N(n).

This proves the lemma.

====
Corollary For A a local integral domain and x in A nonzero, the
HS dimensions satisfy
 d(A/(x)) <= d(A) - 1 for all nonzero x in A.

Proof. We assume that A is an integral domain and x <> 0, so the
principal ideal (x) = x*A is a submodule of A isomorphic to A.
Apply the lemma to A, x*A, and A/(x).

Then A and x*A are isomorphic, so have the same HS dimension, and
the same order of growth of HS(n). Therefore by the lemma A/(x)
has smaller order of growth, that is d(A/(x)) <= d(A) - 1.

====
The hard part of the main theorem: the Krull dimension of a Noetherian
ring A is <= its HS dimension:
 dim A <= d(A).

The proof is by induction on d(A).

If d(A) = 0 then l(A/m^n) is eventually constant. Therefore
m^{n+1} = m^n, and Nakayama's lemma gives m^n = 0. Then m is the
only prime ideal of A so dim A = 0. [For example, because if
P in m is prime then any x notin P maps to a nonzero element of
the integral domain A/P, which contradicts nilpotent.]

Now suppose d(A) > 0. If dim A = 0 we are home, so assume
that dim A > 0. Let
   P0 < P1 < .. Pe
be any strictly increasing chain of prime ideals of A.

The inductive step picks x in P1 notin P0. Then x maps to
a nonzero element y = xbar of the integral domain A/P0.
Now apply the most recent corollary to y in A/P0.

Set B = A/(P0+xA) and write p: A -> B for the quotient map. The
corollary states that d(B) <= d(A) - 1 and therefore we can use
the inductive assumption to conclude dim(B) <= d(B). Consider the
rest of the chain
  P1 < .. < Pe
of primes in A. Then for i > 0, the p(Pi) are ideals of B with
B/p(Pi) = A/Pi, so they are prime, and form a chain of prime
ideals of B of length e-1.

Therefore e-1 <= dim B <= d(B) <= d(A)-1, so e <= d(A).

This applies to every strictly increasing chain of prime ideals
of A, therefore dim A <= d(A).

====
The same result applies to a finite module M (but I haven't had
time to think it through, and will not use it -- the argument
works with Ass M, so submodules isomorphic to A/P, and applies
the result for A itself -- see [Ma], p. 99, where he says "it is
easy to see").

====
Main theorem. The 3 definitions of dimension coincide

dim A = Krull dimension: maximum length of chain of prime ideals.

d(A) = Hilbert-Samuel definition: order of growth of HS(n) = l(A/I^n).

de(A) = minimal number of generators of
            an m-primary ideal I = (x1,..x_de).

We have already done  dim A <= d(A).

====
The other implications are more straightforward and formal.

d(A) <= de(A).

Saying de(A) = 0 means the 0 ideal is m-primary, so m
is nilpotent and A itself has finite length. So its
Hilbert-Samuel function HS_A(n) < l(A) and is bounded,
so d(A) = 0.

Suppose de(M) = s. Then there are x1,..xs in m that generate an
m-primary ideal I = (x1,..xs). Then A/I has finite length l(A/I),
and monomials in I of degree <= n generate I^i/I^{i+1} as
A/I-module for all i < n, so that
   HS^I_A(n) <= (n+s choose s)*l(A/I) ~ const * n^s/s!
and d(A) <= s. [Of course the case s=0 is included in this.]

====
de(A) <= dim A.

If dim A = 0 then m is nilpotent so that 0 is m-primary,
and de(A) = 0.

Assume dim A = s > 0. There are only finitely many minimal prime
ideals of A. Consider all the minimal prime ideals Pi that can
take part at the bottom of a chain of length s. These are not
maximal so Pi is strictly contained in m. Therefore there exists
x in m with x notin union Pi.

I claim that A/x has Krull dimension <= dim A - 1. [This
statement could have been given together with the definition of
Krull dimension.] In fact any prime ideal of A/x pulls back to a
prime of A that contains x, which by construction excludes all
the Pi. Therefore any chain of prime ideals of A/x pulls back to
a chain of prime ideals of A of length <= s-1. 

So by the inductive assumption de(A/x) <= dim(A/x) <= s-1.
Therefore there exists a finite set x2,..xs in m
so that (x,x2,..xs) is primary.


====
If A is an Abelian group under + and I,J1,J2 subgroups, then
I in J1 union J2 implies I in J1 or I in J2.

Proof. If possible choose x1 in I - J2 and x2 in I - J1. Then
x1 in J1 and x2 in J2 by the assumption that I in J1 union J2.
Also x1+x2 in I because I is a subgroup. So again x1+x2 is in
either J1 or J2. Either case is a contradiction: x1 in J1 and
x1+x2 in J1 (and J1 a subgroup) implies x2 in J1.

Now the "prime avoidance" of [A&M], 1.11

If ideal I is contained in Union P1 .. Pn and the Pi are
prime ideals, then I is contained in one of the Pi.

By induction on n, we can assume that I is not in the union of
any n-1. So for each i, choose xi in I that is not in Pj for any
other i. Necessarily xi in Pi because I is in the union of all
the Pi. Now take the product of all the xj with j <> i. None of
the xj are in Pi, so the product is not in Pi (this uses Pi
prime). Now the sum of the products
  sum_i prod_{j<>i} xj
is a sum of element in I (in fact in I^(n-1)) but is not in any
Pi. (All but one term is in Pi, and the nth definitely not.)
This contradicts the assumption that I in Union Pi.