New chapter: Completion
See [A&M] Chapter 10
[Matsumura] Section 8
[Schlichting] Chapter 2

Informal discussion

Why modules? To study A, we may need to do linear algebra inside
A, and in all kinds of structures related to A: its ideals I, how
the I are generated, the quotients A/I, the relations between the
generators of I, eventually tensor products A x A, derivations and
differentials, and much more. We might as well go the whole hog
and do linear algebra systematically in modules over A.

Why completion? Let A be a ring and M an A-module. Suppose we are
told M = I*M. Can we deduce that M = 0?

Take m in M. Then m = sum ai*mi. On the other hand, the same goes
for each mi. If mi = sum bij mj then m = sum_{i,j} ai*bij*mj, so
that M = I^2*M, then M = I^3*M. This is getting ridiculous!
Surely continuing the argument gives M = 0? Not so. For example,
it may happen that I contains invertible elements, and M = I*M
tells us nothing.

That's not the right way to go. I remind you of a basic result.

Nakayama's lemma. Suppose M is finite (that is, finitely
generated as A-module) and M = I*M. Then

  there exists a in A with a-1 in I such that a*M = 0.

Proof ("determinant trick") Choose generators m1.. m_n, so that
  M = sum A*mi.
Then each mi in M, so mi in I*M. Hence there exists aij in I with 
mi = sum a_ij*mj. Rewrite this as
  sum (de_ij - a_ij)*mj = 0  (Kronecker delta).  (*)
Write N for the nxn matrix N = de_ij - a_ij. Recall the standard
linear algebra formula  N^dag*N = det N * Id_n  (here N^\dag is
the adjugate matrix made up of (n-1)x(n-1) cofactors).

Multiply (*) by N^dag_{jk} and sum over j to get (det N)*mj = 0
for all j, hence (det N)*M = 0. This is what we wanted:
  a = det N has a*M = 0 and a == 1 mod I.

A similar failed argument: Let A subset B be a finite extension
ring. Required to prove that b in B is integral over A. The
argument you used in Galois theory was easy: since B is finite
over A there is a linear dependence relation between the powers
{ 1,b,b^2,.. b^n }, and you can divide through by the leading
coefficient to make it monic.

That doesn't work with A an integral domain, because you may not
be able to divide through. But it goes through in a
straightforward way if you can apply the determinant trick.

Completion

The idea of _completion_ is to work with formal power series in
place of polynomials. For example, k[[x1.. xn]] as a substitute
for k[x1.. xn] or p-adics ZZ_p in place of the subring
ZZ_(p) in QQ. The word "formal" reflects that we allow all
infinite power series, ignoring convergence -- like replacing a
differentiable function by its Taylor series to all orders.
These formal rings are bigger (usually uncountably so), but
much simpler in structure. Any nonsingular point P in X of any
algebraic variety or complex analytic space (independently of
X, or P in X) has a small neighbourhood isomorphic to a ball
around 0 in CC^n, and formal functions on it make up the
completed ring CC[[x1.. xn]]. Completion is thus a much more
drastic form of localisation.

As an algebraic process, completion passes from a filtration such
as the I-adic filtration M > I*M >.. I^n*M to the inverse limit
lim_{<-} M/I^n*M. I run through the theory in the following
lecture. For now, I want to discuss the finished product and the
advantages of working with it.

Definition (first attempt) Let A be a ring and I an ideal. We say
that A is I-adically complete to mean that 
  A = lim_{<-} A/I^n.
This means 
 (I) an element f in A is uniquely determined by its class in
A/I^n for every n.
 (II) If {f_n in A/I^n} is a compatible sequence of elements
mod I^n then there is f in A that maps to f_n for every n.

_Compatible_ means that for m > n, the element f_m in A/I^m
reduces modulo I^n to f_n in A/I^n. An alternative way of stating
(II) is as a sequence {F_n} of elements of A with the Cauchy
sequence property:
  for every N > 0, there exists n0 such that
    for all n,m >= n0 the difference  F_n-F_m  is in I^N.
This puts I-adic completion on a footing that is similar in
overall logic to completion in a metric space.

The real motivation for completion is to solve problems in
A[[t]] by term-by-term calculations. e.g. is a0 is invertible in
A, you can find the inverse of a0 + a1*t + .. by calculating
successive coefficients. Or if a0 is a perfect square in
A (and n! is invertible), then you can take the square root
of a0 + a1*t + .. using the binomial theorem and term-by-term
approximation. 

The highpoint is Hensel's Lemma: this says that, under appropriate
conditions, if you can solve a polynomial equations modulo m (so
over the field k = A/m), you can solve it over A.

Hensel's lemma. Let (A,m,k) be a local ring, and assume that A is
m-adically complete.

Let F(x) in A[x] be a monic polynomial, and set Fbar = f in k[x].
(This means reduce the coefficients of f in A[x] modulo m.)

Suppose f factors as f = g*h with g,h in k[x] monic and coprime.

Then F = G*H where G,H in A[x] are such that
  Gbar = g and Hbar = h.

====
Applying this with a linear factor g(x) = (x-r) gives the
corollary that if the reduction of f(x) in k[x] has a _simple_
root r in k (a root such that (x-r) is coprime to f(x)/(x-r)),
then F(x) in A[x] has a root R in A that reduces to r mod m.

For example, if a polynomial f in ZZ[x] has a simple solution when
viewed as a congruence modulo p, it has a root in the ring ZZ_p of
p-adic integers. This version of Hensel's lemma is popular with
algebraic number theorists.

====
A preliminary step. First, suppose deg g = n and deg h = m. Then
g, h coprime in k[x] means I can choose polynomials a,b with
  deg a <= m-1 and deg b <= n-1 such that a*g + b*h = 1.

In fact  (1,x,.. x^{m-1})*g, (1,x,.. x^{n-1})*h  are n+m linearly
independent elements in the (n+m)-dimensional vector space of
polynomials of degree n+m-1, so a basis.

====
Setting up the induction step.

Start from the assumption f = g*h, choose lifts G1,H1 in A[x]
of g,h in k[x] that are still monic of the same degree. Then
reducing modulo m gives
  F - G1*H1 in m*A[x], that is F - G1*H1 = sum mi*Ui  (*)
with mi in m, and Ui in k[x] polynomials with deg Ui < deg F.

I show how to modify G1,H1 to G2,H2 by adding corrections in m
to achieve
  F - G2*H2 in m^2*A[x].

This is elementary algebra in k[x]: for each i, write ui in k[x]
for the reduction of Ui mod m, and use the above a,b to give
  g*a*ui + h*b*ui = ui.
The sum is obviously unaffected by subtracting a multiple of h
from a*ui and adding the same multiple of g to b*ui:
  g*vi + h*wi = ui, where
      vi = a*ui-c*h  and  wi = b*ui+c*g.  (**)
I choose c to reduce a*ui to 
  vi = a*ui - c*h with deg vi < deg h.
Then since ui and g*vi are both of degree < deg f, the same
goes for h*wi.

Now choose lifts Vi, Wi in A[x] of vi and wi of (**), of the
same degrees, and modify G1, H1 by setting:
  G2 = G1 + sum mi*Wi and H2 = H1 + sum mi*Vi
using the same coefficients mi as in (*). Then comparing
with (*) gives
  F - G2*H2 = F - G1*G2 - sum mi*(G1*V1 + H1*W1) - mi^2*V1*W1
           in m^2*A[x].
In each term of the sum, I have subtracted off a term that
modulo m^2 cancels the mi*Ui of (*) in view of (**), and the
final cross term is in mi^2*A[x].

The inductive step from Gn,Hn satisfying F - Gn*Fn in m^n*A[x]
to G_{n+1}, H_{n+1} repeats the above argument point by point.

Each step only modifies the Gn, Hn by terms in m^n*A[x], so that
both sequences are Cauchy sequences for the m-adic topology. Q.E.D.