Lecture 4 discussed Spec A and its topology, and was mostly
taken from [UCA], Section 5.
The definition I gave said
V(I) is closed.
Xf is basis for topology.
If X = union Xfi then 1 in (fi), so that we only need finitely many,
and X is quasi-compact. (compact but of course not Hausdorf.)
It is natural to associate A[1/f] to Xf and A_P to P in X.
Lecture 4.
The case that we understand fairly well:
A = k[X]/J(X) is the affine coordinate ring of an algebraic set
X in AA^n_k over an algebraically closed field k.
The maximal spectrum m-Spec A corresponds to k-valued points
of X. Every prime ideal is J(Y) for an irreducible subvariety
Y in X; in particular every prime ideal P is the intersection
of maximal ideals at points of Y, so that m-Spec A determines
Spec A.
X has a Zariski topology, with closed subsets the subvarieties
Y in X. The sets Xf (the set of points with f <> 0) is the
complement of the variety V(f). Every closed set is
V(I) = intersect_{f in I} V(f)
so that its complement if the union of V(f) taken over f in I.
This means that the Xf form a basis for the Zariski topology
of X. In fact Xf is itself isomorphic to an affine variety
with coordinate ring k[Xf] = k[X][1/f].
A prime ideal P in A = k[X] corresponds to an irreducible closed
subvariety Y = V(P) in X. This has its own Zariski topology (the
subspace topology of Y in X or Y in AA^n). The localisation A_P
consists of fractions f = g/h of k[X] having one possible
denominator h that does not vanish on Y. This means that the
fraction f restricts to a rational function on Y. We know A_P is
a local ring with maximal ideal PA_P, so that the quotient ring
A_P/PA_P is a field, the function field k(Y) of the irreducible
variety Y. The fraction f = g/h has domain of definition dom f
(in X) that intersections Y in a dense open set of Y, and
restricts to a rational function f_|Y on Y whose domain includes
(dom f) intersect Y.
For a general ring A, we can't rely on any useful relation
between prime ideals and maximal ideals, so m-Spec A is not
especially useful. We work with Spec A. It has a Zariski topology
with closed sets V(I) = { primes P containing I }. This means
that f in I maps to 0 in the quotient ring A/P, so we can still
view this as saying the f evaluates to 0 at P.
The Zariski topology on X = Spec A has a basis by the open sets
Xf = { P in X | f notin P }. This topology is far from Hausdorff
or separated, but it is compact (without any finitess conditions
on A). The ring of fractions A[1/f] has Spec A[1/f] = Xf. Each
point P in X has the local rings A_P with maximal ideal PA_P
and residue field A_P/PA_P.
What is missing here is the idea that f is a function on Spec A
in any conventional sense. Instead of evaluating f at a point
P in Spec A, the operation that preserves all the information
is mapping f to its class in A_P. The notion of variety is
replaced by Grothendieck's definition of affine scheme, that puts
together:
X = Spec A,
its Zariski topology,
the partial localisation A[1/f] associated to Xf,
the local rings A_P over P in Spec A.
New chapter
Chain conditions on modules and on rings:
M Noetherian modules: every increasing chain terminates
<=> every nonempty set of submodules has a maximal submodule
<=> every submodule is f.g.
You know lots of examples: If A is f.g. as a ring over k or
over ZZ, or a localisation of such, a finite A-module is
Noetherian. Another class of examples: if A is Noetherian,
a formal power series ring A[[x1,x2,..xn]] is Noetherian.
(later ex. -- these are of course not f.g. rings).
M Artinian module: every decreasing chain of submodules terminates
<=> every nonempty set of submodules has a minimal submodule
This is not a frivolous issue.
e.g. ZZ is not Artinian: take integer n not a unit (say n = 10),
then the chain of ideals (n^i) is an infinite decreasing chain.
Likewise k[x] is not Artinian: take nonunit polynomial f, then
the chain of ideals (f^i) is infinite decreasing.
What examples?
Inverse polynomials: Let k[x,x^-1] be the ring of Laurent power
series, and consider the quotient module
M = k[x,x^-1] / k[x].
As a vector space, this is countable dimensional, with basis
{ x^-i }.
As a k[x]-module, multiplication by x does
x * x^-i |-> x^{-i-1}.
Whereas in k[x] multiplication by x is injective, in M,
multiplication by x is surjective. For any m in M, we can find a
predecessor m' with x*m' = m.
M is Artinian: every proper submodule N in M only involves
finitely many x^-n, and is the k-vector space based by x^i for i
in the range -n < i < -1. The longest chain of these is
M subsetneq .. Nn subsetneq N{n-1} subsetneq .. N{-1} subsetneq N0 = 0.
On the other hand, it is not finitely generated as module, and
not Noetherian. As n gets bigger n, the modules (x^-n) get bigger,
so infinite ascending sequences are the order of the day.
You should think of M as a tempered dual of the polynomial ring k[x].
Since k[x] is infinite dimensional, the dual vector space
k[x] = Hom(k[x], k)
would be uncountable dimensional. Instead, think of k[x] as the union
of the spaces k[x]_{<=n} of polynomials of degree <= n. Then M is the
union of their duals
Hom(k[x]_{<=n}, k).
The duality between M and k[x] is the analog of Cauchy residue of a
meromorphic function -- take inverse polynomial q and polynomial f
into the residue of product q*f, that is, the coefficient of x^-1.
M is called the module of Macaulay inverse systems. It contains a unique
submodule N = k*x^-1 that is isomorphic to the residue field k[x]/(x),
and M is injective as k[x] module in a sense that we will come to later.
The same trick applies to the localisation ZZ_(p) of ZZ at a prime p.
The module M = QQ/ZZ_(p) is generated by the negative powers of p. Any
proper ZZ_(p)-submodule N in M has only finitely many p^-n, and the
chains of modules have the same shape as above.
The module QQ/ZZ is the sum of the QQ/ZZ_(p) taken over all p. It is
Artinian but not Noetherian.
Jordan-Hoelder filtration and modules of finite length
Theorem Let M be an A-module. Then the following are equivalent:
(1) M is both Artinian and Noetherian
(2) There exists a finite filtration
0 = M0 inneq M1 inneq .. M_{n-1} inneq Mn = M.
such that for each i, there is no A-module strictly intermediate
between Mi and M{i+1}.
The last condition is: if Mi inneq N inneq M_{i+1} then
either Mi = N or N = M_{i+1}. You can also say that the module
M_{i+1}/M_i is simple (has no nontrivial submodule).
(1) => (2) is straightforward. M is Noetherian so (if <> 0) there
is a maximal submodule M' in M. The inclusion M' in M has not
no strictly intermediate property by construction. Apply the same
to M' (assuming <>) and get a descending chain. Since M is
Artinian, this must terminate, giving (2).