Lect 2 (summary)

In Lect 1 I talked about the Dedekind-Weber paper of 1882
in colloquial or propaganda terms. To be more formal, I
should have stated the theory as a definition and theorem.

Definition A _Dedekind domain_ is a Noetherian integral
domain A that is 1-dimensional and normal (integrally closed
in Frac A).

You already know what integral domain means. Krull dimension
refers to the maximum length of a strictly increasing chain
of prime ideals P1 in P2 in .. Since A is an integral domain
0 is prime. Krull dimension 0 means 0 is maximal, which
happens iff A is a field. For an integral domain, Krull
dimension 1 means that A is not a field, but any prime ideal
0 <> P is maximal.

As an integral domain A is in its field of fractions
A in Frak A, and normal means that any element x in K
that is integral over A is actually in A.

Exa. You already know ZZ and k[x]. These are basic results
that I will prove later:

Theorem. Let A be a Dedekind domain, and K in L a finite
extension of its field of fractions. Write B for the
integral closure of A in L, giving

     B  in  L
     |      |
     A  in  K

Integral closure means that B consists of all elements of the
extension field L that satisfy a monic equation over A. (Exactly
the same definition as ring of integers of a number field.)

Then B is finite as an A-module, and is again a Dedekind domain.

Theorem. Let A be a Dedekind domain and P in Spec A a prime
ideal. Then the localisation A_P = S^-1A (where S = A - P)
is a discrete valuation ring (DVR).

A DVR A_P is the simplest kind of UFD: there is a unique prime
element z (up to multiplication by units), and every element f of
the field of fractions is f = z^n*(unit of A_P), where the
exponent n = v_P(f) is the valuation of f at P (like the order of
zeros or poles of a meromorphic function).

Theorem. Let A be a Dedekind domain with K = Frak A. Write
K^x for the multiplicative group of nonzero elements of K.

(I) For every nonzero f in K^x, there are only finitely many
P in Spec A for which v_P(f) <> 0.

(II) Any nonzero ideal of A is uniquely expressed as a product
of prime ideals. More precisely, let I in A be a nonzero ideal,
and write v_P(I) for the minimum
  v_P(I) = min_{f in I} v_P(f).
Then
  I = Product P^v_P(f).

====
Existence of prime ideals.

Baby result: a maximal ideal is prime.

m in A maximal means that m <> A  but any strictly bigger
ideal is the whole of A. Thus for x in A - m, m+Ax = A.
Let x1, x2 in A-m. Then I can write
  1 = a1*x1 + m1 and 1 = a2*x2 + m2
for some m1,m2 in m and a1,a2 in A. Expanding the product
  1 = (a1*x1 + m1)*(a2*x2 + m2)
gives 1 = a1*a2*x1*x2 + (sum of terms in m)
Therefore the ideal m + A*(x1*x2) is strictly bigger
than m, so product x1*x2 notin m.

This result illustrates:

Principle. If an ideal I is _maximal in some restricted class_
or "conditionally maximal", then we can hope that I is prime.

This is a main technique for producing prime ideals.

====
Zorn's axiom ("Zorn's lemma").
Let Si be a nonempty partially ordered set. Assume that
every totally ordered subset S0 in Si has an upper bound
in Si. [_Totally ordered_ means that for all x,y in S0,
either x<y or x=y or y<x.]
Then Si has a maximal element.

This is similar to the completeness axiom for the reals (there
exists a sup). Much as any result in analysis asserting the
existence of a limit necessarily uses the completeness axiom,
algebraists use Zorn's lemma for practically any infinite
induction task in algebra. See for example the Wikipedia page.

Application. Let A be a ring, S a multiplicative set, and I an
ideal disjoint from S. [Multiplicative set means 1 in S and
s1,s2 in S implies s1*s2 in S.]

Then there exists an ideal P that is maximal among the ideals
containing I and disjoint from S. And any such is prime.

Let Si be the set of ideals containing I and disjoint from S.
Si <> emptyset because I in Si. If {I_j} is a totally ordered
subset of Si, write Q = Union_j I_j. Then this is an element
of Si: It obviously contains I. It is an ideal (because
if x_j in I_j is a finite set of elements, there is a biggest
among the I_j, and so they are all in I_j, which is an ideal).
Also it is disjoint from S because all the I_j are.

So we can use Zorn's lemma to assert that P exists. P is an ideal
disjoint from S, but maximal in that class. Copy and paste the
above baby argument:

Let x1, x2 in A-P. Then P + A*xi is a bigger ideal, so intersects
S by the maximal assumption. So I can write
  s1 = a1*x1 + m1 and s2 = a2*x2 + m2
for some m1,m2 in P and a1,a2 in A. Expanding the product
  s1*s2 = (a1*x1 + m1)*(a2*x2 + m2)
gives s1*s2 = a1*a2*x1*x2 + (sum of terms in P).

Therefore the ideal P + A*(x1*x2) is strictly bigger than P, so
product x1*x2 notin P, and P is prime.

This is all basic stuff from [UCA, Chap. 1], but the same
principle applies in many more advanced settings.

====
Ring of fractions S^-1A and modules of fractions S^-1M.