====
Lecture 4.
Discuss Spec A and its topology, mostly following [UCA], Chapter 5.
Spec of a ring and Zariski topology. Spec A = set of prime ideals.
If phi: A -> B is a ring homomorphism, sending P |-> phi^-1(P) defines a
map of set Spec B -> Spec A (note that is is contravariant).
The same does not work for maximal ideals, because phi^-1(m) is prime
but usually not maximal.
Proof: Let P in Spec B, and f,g in A with f,g notin phi^-1(P). Then
phi(f), phi(g) notin P, so the product
phi(f)*phi(g) = phi(f*g) notin P ==> f*g notin phi^-1(P).
As a formal and rather trivial example, let A be an integral domain and
A -> K = Frac A the inclusion to its field of fractions. The ideal 0 in
K is maximal, but 0 in A is only a maximal ideal if A = K. Less
trivially, a partial ring of fractions such as
k[x,y] into k(x)[y],
the localisation S^-1 where S is the multiplicative set (k[x]\0). Then
(y) or (y-a) in k(x)[y] is maximal, but (y) or (y-a) in k(x)[y] is
obviously not
The Zariski topology on Spec A is the topology having closed sets V(I)
for I an ideal of A, where
V(I) = { P in Spec A | I in P }.
This is a topology.
Proof. The empty set is V(A) and the whole set V(0).
Arbitrary intersection V(sum I_i) = intersect V(I_i)
Finite union:: V(I1) union V(I2) = V(I1 intersect I2) = V(I1*I2)
[UCA] Chapter 5 is mostly about the Zariski topology. I cover it in more
detail later, together with some motivation based on algebraic varieties
in AA^n.
Notice that the topology on Spec A is formal, and has rather little
content. In number theory, you often has "except for finitely many bad
primes", reflecting Spec ZZ = { 0 } union { (p) primes } where open
means exclude finitely many of the primes. Every non-empty open contains
0 <-> localisation QQ, the generic point.
As less trivial examples, you might like to consider a number field K, its
ring of integers O_K, the polynomial ring O_K[X] and the Zariski topology
on Spec O_K[X].
The definition I gave said: the closed sets of X = Spec A are V(I), which
is intersect V(f) for f in I. Therefore if we write
Xf = primes not containing f,
then Xf is open, and the set of {Xf} is a basis for the Zariski 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.) This does not
use Noetherian. 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).
It is natural to associate A[1/f] to Xf and A_P to P in X.
A case we understand fairly well:
A = k[X]/J(X) the affine coordinate ring of an algebraic set
X in AA^n_k over an algebraically closed field k = kbar.
The maximal spectrum m-Spec A or Specm 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 so 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 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 the basic opens Xf,
the local rings A_P over P in Spec A.
====
Exercises on the Zariski topology:
Hint: Remind yourself of the relation between radical and prime ideals,
that I discussed under "plenty of primes by Zorn's lemma". This is a
key results, but in my experience not particularly memorable.
By definition, a closed set V in Spec A is V = V(I) for an ideal I. This
I is not uniquely specified (for example V(I) = V(I^2)).
(1) Prove there is a unique biggest J. Namely, the ideal
J = intersection of all primes P in V.
Moreover, this is J = rad(I) for any I with V = V(I). [See the hint.]
A closed set V in Spec A is _irreducible_ (by definition) if it not the
union of two smaller closed sets.
(2) Prove that this happens if and only if V = V(prime ideal P).
(3) If A is a Noetherian ring, prove that the closed sets of the Zariski
topology satisfy the d.c.c. (any descending chain eventually
stabilises). This is called a Noetherian topology.
(4) In a Noetherian topology, every closed set is a union of finitely
many irreducibles.
Please give these a serious try. If you have difficulties with any of
this, see [UCA], 5.12--5.13.
====
Lecture 5
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
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.
Topics.
(0) For modules over a given ring A, the Artinian condition is a kind of
categorical dual of Noetherian.
(1) Examples: Macaulay inverse systems as injective modules and the
injective hull of residue fields A/m.
(2) Modules that are both Artinian and Noetherian are of finite length,
and have Jordan-Hoelder sequences with simple quotients.
(3) In complete contrast to modules, when applied to rings, Artinian is
an extremely strong condition, that implies Noetherian. For a k-algebra
over a field k, Artinian is equivalent to a ring that is a finite
dimensional vector space. Artinian rings play a foundational role in the
dimension theory of rings. They are characterised as zero-dimensional
Noetherian rings.
(1) Inverse polynomials: Let A = k[x] and consider the ring of Laurent
polynomials k[x,x^-1], and the quotient module
M = k[x,x^-1] / k[x].
As a vector space, this is countable dimensional, with basis
{ x^-i }.
M is Artinian: every f.g. A-submodule of k[x,x^-1] only involves
finitely many x^-n. The same applies to every f.g. A-submodule N in M,
so if x^-n is the last of these,
N = Nn = k-vector space of M based by x^i for i in [-n,-1].
The k[x]-module multiplication by x does
x * x^-i |-> x^{-i-1}, and in particular, x^-1 |-> 0.
A decreasing chain is a chain of f.d. vector spaces, so must terminate.
All the Nn are in the chain
.. Nn subsetneq N{n-1} subsetneq .. N{-1} subsetneq N0 = 0.
Whereas multiplication by x in A = k[x] is injective, in M it is surjective,
so that M is infinitely divisible: for any m in M, we can find a
predecessor m' in M with x*m' = m.
Multiplication by x is nilpotent when restricted to any A-submodule of Nn.
Nn has the single associated prime m, that is Ass Nn = m the annihilator
of x^-1. Whereas A = k[x] has 1 as the single generator, and the only
unit element, and bases the residue field k = A/m, in Nn, the element
x^-1 is the _socle_, the last element to go under nilpotent
multiplication. That is, it the submodule of Nn annihilated by m, which
is the same thing as Hom_A(A/m, Nn).
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. The free
module k[x] has basis {x^i}, whereas M = k[x,x^-1] / k[x] has basis
{ x^-{i+1} }, which is the dual basis under the residue pairing.
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, and the map k[x]/(x) -> M
given by the basis { x^-1 } of the socle is the injective hull of the
residue field k = k[x]/(x) in a sense that I introduce later.
We can do the same thing with other rings. For example k[x1,..xn] and
the module with basis { prod xi^{-ai-1} } (Laurent monomials with
strictly negative exponents) so the socle is based by { prod xi^-1 }.
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). Each is a
1-dimensional vector space of a field k(P) for some P in Spec A.
(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).
Jordan-Hoelder filtration is unique. Any two filtrations have successive
simple quotients M_{i+1}/M_i that up to permutation are isomorphic.
In particular, the length of the chain is well defined, and is called
the length of M.
====
Lecture 6
The previous lecture discussed a.c.c. and d.c.c. and the related notion
of Noetherian and Artinian modules.
This lecture has 2 aims:
(I) to treat modules that satisfy both chain conditions, and
characterise them as modules of finite length.
(II) To discuss Artinian rings, that are analogous to k-algebras that
are finite dimensional as vector spaces over a field k. As opposed to
modules, Artinian rings are necessarily also Noetherian.
Definition: Simple module. M is a simple A-module if its only submodules
are 0,M. One sees that then M iso A/m = k(m) for some maximal ideal k.
Obviously, if M is a simple module and N any module,
any homomorphism M -> N is 0 or injective
any homomorphism N -> M is 0 or surjective
If M, N are both simple, any M -> N is 0 or an isomorphism.
Definition: A a ring, and M an A-module. A Jordan-Hoelder filtration is
a chain of submodules
0 = M0 in M1 in .. M_{n-1} in M_n = M (*)
with no strictly intermediate modules between M_{i-1} and M_i. The
latter condition holds if and only if M_i/M_{i-1} is simple.
Theorem Equivalent conditions on M:
(1) M is Artinian and Noetherian.
(2) M has a Jordan-Hoelder filtration.
Moreover, if they hold, the set of simple quotient modules Mi/M_{i-1} in
any JH filtration is unique up to isomorphism and up to permutation. This
means that M is just a bunch of fields k(m) tied together in a successive
extension module.
If the conditions hold we say M has _finite length_, or has length n as
an A-module where n is in (*). (This of course depends on A -- if we
view M as a module over a smaller ring, it might lose the finite length
property, and its length will usually increase.)
(1) => (2) is the obvious follow-your-nose argument:
First, define Si = set of all nonzero submodules. If M <> 0 this is
nonempty, so by the Artinian condition it has a minimal element M1.
Next, set Si = set of all submodules strictly bigger that M1. If this is
empty, M1 = M and we are finished. If nonempty, it has a minimal element
M2. Continue by induction: we construct
0 = M0 < M1 < .. Mi <= M
where each step cannot be refined. At any stage, either we have reached
M or we can take another step. So far, this has only used Artinian. Now
since M is also Noetherian the increasing chain must terminate, so at
some point Mn = M.
You can also do this from the other end, working down from M and taking
maximal nontrivial M' < M.
The proof of (2) => (1) and the "moreover" final clause involve standard
arguments using the isomorphism theorems. Given one JH filtration (*),
and any chain of submodules {Ni}, if none of the Ni contain M1, they are
part of a chain for M/M1, which has a JH chain of length n-1. In the
contrary case, there is some i s.t. N_{i-1} does not contain M1 but Ni
does, so necessarily Ni/N{i-1} iso M1. etc. (Clean up the proof for
yourself.)
====
Artinian rings
For modules, the a.c.c. and d.c.c. conditions are in a sense logically
and categorically dual, and there are duality relations (so far only
vaguely hinted at) between the two classes of Noetherian and Artinian
modules. At least the modules and the classes of modules are "the same
size".
For rings, however, Artinian is a much more restrictive condition: an
Artinian ring A is also Noetherian, and has finite length as A-module.
A particular case is that A is an algebra over a field k. Then A is
finite dimensional as k-vector space, so k in A is similar to the finite
field extensions k in K you studied in Galois theory.
As a preview, consider the example A = k[x]/(F) with k a field. Suppose
F = prod (fi)^ni with fi irreducible and coprime. Then the maximal
ideals of A are (fi), and
A = directsum (k[x]/(fi^ni))
each of the summands has a JH filtration with ni steps and each quotient
isomorphic the field k[x]/(fi), which is a finite extension field of k
and a simple A-module.
More generally, you could bear in mind this picture: an Artinian ring is
a disjoint union of nilpotent blobs surrounding finitely many points
that are the residue fields at its maximal ideals.
Proposition (I) If an integral domain A is Artinian, it is a field.
(II) If A is an Artinian ring, every prime ideal is maximal (that
is, A has Krull dimension = 0).
(I) Pick x in A, and consider the descending chain of ideals
(x) > (x^2) > .. > (x^n) ..
By the d.c.c., eventually (x^n) = (x^{n+1}), so that x^n is a multiple
of x^{n+1}, say x^n = y*x^{n+1}. Now either x = 0 or x*y = 1. This
condition characterises a field.
(II) follows by passing to the quotient A/P, which is still Artinian.
Theorem. An Artinian ring A is Noetherian
Step 1 A has only finitely many maximal ideals m_i.
Proof: An exercise that is "easy", but I always have to do again from
first principles. (Please do it for yourself before reading the given
solution at the end of this file.)
Step 2 Write J = product mi = intersection mi. (The Jacobson radical.)
Then every x in J is nilpotent.
Since every prime ideal is maximal, J is also the intersection of all
prime ideals of A, which is its nilradical.
Step 3
Claim. J^n = 0.
J = intersection mi = intersect of all primes,
so every x in J is nilpotent. Consider the descending chain
J > J^2 > .. > J^n.
This must terminate (again by the d.c.c.), say in N = J^n with
J*N = N^2 = N.
We prove that N = 0 by contradiction.
If N <> 0 the set Si of ideals b such that b*N <> 0 is nonempty, so it
has a minimal element c (again by the d.c.c.). Now some x in c
has x*N <> 0, and c = (x), else (x) would be smaller than minimal.
Now (x*N)*N = x*N^2 = x*N <> 0. So x*N is an ideal contained in (x), and
again by minimality, x*N = (x). It follows that x = x*y for some y in N,
and therefore
x = x*y = x*y^2 = .. x*y^n.
But y is an element of N, and every element of J is nilpotent, so
x = x*y^n = 0. This is a contradiction, hence N = 0.
Step 4. An Artinian ring A has a JH filtration, and so is Noetherian as
A-module.
In fact we proved that the finitely many prime ideals m_i have
prod m_i^n = 0.
The Artinian condition on A implies that m_i^n/m_i^{n+1} is an Artinian
module; but it is a vector space over the residue field k(m_i) = A/m_i,
so finite dimensional. Therefore taking prod m_i^{ni} with one of the
exponents increasing by 1 at a time, we get a decreasing sequence with
each a finite dimensional vector space over one of the k(m_i). Q.E.D.
The above argument divides into a number of fairly tricky steps, making
repeated use of minimality of ideals in a sequence. Is there an improved
argument with fewer appeals to d.c.c.? [A&M] and [Matsumura] give
essentially the same proof (possibly cribbed from a common source? [Ma,
p.16] refers to Akizuki 1935 and Hopkins 1939), and as far as I know,
no-one seems to have found a shorter argument.
This result in the converse direction is much easier:
Theorem: Noetherian and dim A = 0 => Artinian.
As in the preceding argument, all primes are maximal, so the
intersection J of all maximal ideals is the nilradical, the set of all
nilpotent elements. However, Noetherian gives that J is finitely
generated, so J nilpotent is clear. Now A/J is a product of fields, and
Noetherian implies there are only finitely many of them. etc.
Exc.1: m1, m2 distinct maximal ideals => m1+m2 = A (Easy). Also m1
intersect m2 = m1*m2.
Suppose e1 in m1 and e2 in m2 satisfy e1+e2 = 1_A. Then for
x in m1 intersect m2 we get x = e1*x + e2*x. The first term e1*x is in
m1*m2 (because x in m2) and e2*x similarly.
The e1,e2 map to complementary idempotents of A/(m1 intersect m2).
Exc. Now m1,m2,m3 distinct maximal ideals. Claim: there is an x in m1
interect m2 such that x notin m3.
Because: take y in m1 notin m3 and z in m2 notin m3. Then y*z does what
I claim.
Exc. Similar for m1,..mn, m_{n+1}. Compare for example [A&M] Lemma 1.11.