Notes on Week 1 lectures
A main aim is to indicate the prerequisites, with some illustrative
examples, arguments and exercises. All of this is in [UCA] or [AM] or
any number of other standard texts.
== Section 1 ==
Let's check out some of the prerequisites by considering the
following sequence of utterances:
Let A be a Noetherian ring,
M a finite A-module,
S a multiplicative set in A,
A -> S^-1A the corresponding ring of fractions,
M -> S^-1M the corresponding module of fractions,
S^-1M = S^-1A tensor_A M,
The operation S^-1 is exact.
Ring = commutative ring with a 1
Noetherian = ideals of A have the a.c.c. (any ascending chain eventually
terminates)
or if you prefer, every ideal of A is f.g.
equivalent (assumes axiom of choice): every set of ideals has a
maximal element
Finite A-module. M is generated by finitely many elements m1,.. mk
Exc: It follows that the A-module M is Noetherian. Write out a
detailed statement and proof.
Rmk: Noetherian enables arguments by finite induction (see below for an
illustration).
Noetherian for modules and for rings behave in exactly the same way --
Later on in the course, we do Artinian rings and modules, with d.c.c. in
place of a.c.c., and then the situation for rings and modules are quite
different.
Ring of fractions: an integral domain A <> 0 is contained in its field
of fractions K = Frac A, and you already know what S^-1A is: the subring
of K with denominators in S. However, if S has zerodivisors, you need to
be more careful with the equivalence relation:
a1/s1 ~ a2/s2 iff there exists t in S such that t*(s1*a2 - s2*a1).
A -> S^-1A is the universal ring homomorphism such that every element of
S mapsto a unit of S^-1A.
Exc: if you have not seen this before, check that ~ is an
equivalence relation. State and prove the universal mapping property.
M -> S^-1M. This is the same construction for modules. You have seen
this before whenever we go from a finitely generated Abelian group L (as
ZZ-module) to the QQ-vector space L_QQ = L tensor QQ. Obviously the
torsion elements map to zero. As for rings, M -> S^-1M has the UMP for
maps of A-modules such that each s in S maps to an invertible
homomorphism.
Exc: State the nec. and suff. condition for S^-1M = 0.
Treating tensor products of modules over a ring is moderately tedious.
Once we have this on board, it is straightforward to show that
S^-1M = S^-1A tensor_A M.
S^-1 is a functor from A-modules to S^-1A modules. That is, given
A-linear f: M -> N, there is a canonical way of defining
g = S^-1f: S^-1M -> S^-1N.
Exc: elaborate on this. Prove that f injective implies g injective,
the same for surjective, and that S^-1(M/N) = (S^-1M)/(S^-1N).
Remark on learning: there is nothing hard about this. Figure out what
the statement means, and what you have to do to prove it. It takes time,
but no thought or choices or originality is involved, like navigating a
maze without any turnings.
The result of the exercise can be stated as: S^-1 is an exact functor:
M1 -> M2 -> M3 an exact complex implies that S^-1M1 -> S^-1M2 -> S^-1M3
is again exact. In view of S^-1M = S^-1A tensor_A M this means that
S^-1A is a _flat_ A-module. [Reassurance: In general, flat is a scary
notion, but nothing here is difficult.]
What I tried to do in Week 1
Prerequisites:
Noetherian rings and finite modules
S^-1A and S^-1M partial rings and modules of fractions
UMP A -> S^-1A and M -> S^-1M are universal for
ring/module homomorphisms making S action invertible
S^-1M = S^-1A tensor_A M
(this is easy if you know what tensor product is)
S^-1 is an exact functor.
Reminder that a multiplicative set contains 1_A <> 0.
By definition, a prime ideal P in A does not contain
1_A, and an integral domain A/P has 0 <> 1.
Definition of local ring A or A,m or A,m,k. For any
prime ideal A, the complement S = A\P is a multiplicative
set, and S^-1A = A_P is a local ring, with maximal ideal
PA_P and residue field A_P/PA_P = k(P) = Frac(A/P).
Enough primes by Zorn's lemma.
Prime P is an _associated prime_ of module M, written
P in Ass M, if (P is a prime ideal and) there exists
element m in M such that P = Ann(m). Equivalently,
M contains an A-submodule isomorphic to A/P.
Existence of associated prime: an ideal I that is
maximal among the annihilators of m in M is prime/
If A is Noetherian and M <> 0, there exists a
maximal element, so Ass M <> 0. If also M is
finite then Ass M is a finite set (not proved in
lectures, see [UCA, 7.6]).
== Section 2 ==
Prime ideals. P is a prime ideal of A iff
P is an ideal of A, and S = A\P is multiplicative
That is, 1 notin P, and x,y notin P => x*y notin P.
The set of primes of A is written Spec A. This is a central object
both in this course and later in scheme theory, and it has a lot more
structure, but at first I treat it simply as a set.
Local ring: A ring is local if it has a unique maximal ideal.
The maximal ideal is the set of all nonunit elements, that is, m = A\A^x.
One says A,m is a local ring to mean A is a ring with m its unique maximal
ideal or A,m,k is a local ring to mean ditto, and the residue field is
A/m = k.
Note that in the expression A,m,k, the ring A determines m and the
surjective homomorphism A ->> A/m = k. The m and k are not additional
data, just a convenient way of setting out the terminology.
Overall slogan. The key strategy of commutative algebra is to reduce
problems on rings and modules to local arguments.
Prop. A ring, P a prime ideal. Set AP = S^-1A where S = A\P.
Then A_P is a local ring with max ideal PA_P, and residue field
A_P/PA_P = Frac(A/P) the field of fractions of the integral domain
A/P.
Exc. Prove this.
Rmk. The set of maximal ideals in Spec A (sometimes written Specm A) is
identified with the set of surjective ring homomorphisms A -> k to a
field. In the case A = k[X] is the coordinate ring of an affine
algebraic variety over an algebraically closed field, we can identify
the subring k in A and the quotient ring k = A/m, so that all the k
are one and the same field. The more general construction is _affine
scheme_, where the quotient field varies from point to point (think
of ZZ or ZZ[x]), and we will see more of this later.
Existence of prime ideals
You know a maximal ideal is prime. One way of finding a maximal ideal is
to take any ideal (not the whole of A), then a bigger one and so on. Use
Zorn's lemma (or work under the Noetherian assumption) to get a maximal
one. You should have seen this in the result:
Proposition (Plenty of primes). If S in A is a multiplicative set and I
in A an ideal disjoint from S then there exists a prime ideal containing
I and disjoint from S.
Exc. Find your own proof using Zorn's lemma.
This includes as a corollary the characterisation of nilpotents, and
radical of an ideal: an element x in A is nilpotent if and only if
it is in every prime P, that is
nilradical(A) = intersect P taken over all P in Spec A
radical(I) = intersect P taken over all primes P containing I,
Exc. Prove this. [Hint: if x not nilpotent, apply the proposition,
starting from S = {1,x,.. x^n} and I = 0.]
There is a more general point of view. Slogan: If an ideal I is "maximal
in some class of ideals"", we may hope to find a way of proving I is
prime. I give an important illustration, and at the same time introduce
the important idea of associated prime:
Definition. Let M be an A-module. For nonzero m in M, write Ann(m), the
_annihilator_ of m for the ideal
Ann(m) = { f in A | f*m = 0 in M }.
A prime ideal P of the form Ann(m) for m in M is an _associated prime_
of M. We write Ass M for the set of associated primes of M.
Proposition (i) Suppose that I = Ann(m) is maximal among the class of
annihilators of nonzero m in M. Then I is prime.
(ii) If A is Noetherian then any nonzero module M has Ass M <> 0.
(iii) If A is Noetherian and M is finite then Ass M is finite.
Proof (i) Let I = Ann(m) be maximal among annihilators, and suppose
x,y in A and x*y in I. Required to prove either x or y in I.
Clearly I = Ann(m) in Ann(y*m). If y in I, we're home, so suppose
y notin I. Then y*m <> 0 and Ann(y*m) is another ideal in the class of
annihilators. But then I maximal implies they are equal, so
Ann(m) = Ann(y*m). In this case, x*y*m = 0 means that x in Ann(y*m),
so x in I. In conclusion, x*y in I implies either x in I or y in I.
(ii) Noetherian induction. If I is not maximal, choose a bigger one
I2 also in the class of annihilators, and continue.
NB. Appplying Zorn's lemma does not work: if {I\la}_{la in La} is a
totally ordered infinite set of annihilators, there is no particular
reason why their union should still be an annihilator. [Question. It
would be nice to have a counter-example, that is, an ascending chain of
annihilators whose union is not an annihilator of a nonzero element. I
also don't know whether Ass M = emptyset is possible for a module over a
non-Noetherian ring. For example, something like this might work? A =
ring of germs of continuous functions on RR near 0, and f_n a sequence
of functions that are zero outside [-1/n,1/n]. Functions that are zero
on [-1/n,1/n] obviously annihilate these, but the union of these is the
whole ring.]
(iii) This is a dÃ©vissage argument [UCA, 7.6], that I leave as an
exercise. QED
====
Discussion of Ass M.
It is clear that P is an associated prime of M if and only if M contains
a submodule A*m isomorphic to A/P. The point to note here is that A/P is
an integral domain. If A/P is contained in M, then every nonzero element
of A/P has Ann = P.
Any zerodivisor of M is contained in an annihilator ideal, so (if A is
Noetherian) in a maximal annihilator P, so in a P in Ass M. Therefore
the zerodivisors of M = union P taken over P in Ass M.
Exa 1. A = k[x,y], I the ideal (y^2, x*y) and M = A/I.
"The main example in primary decomposition".
The ideal I is given by generators. We can alternatively describe is as
the set of polynomials that vanish on the x-axis, AND vanish twice at
the origin (that is, I = (y) intersect (x,y)^2).
We can also view this as the ideal of polynomials f in (y)
having partial df/dx = 0 (tangent direction the y-axis).
The element y is not in I.
The class of y in M = A/I is nonzero, and has annihilator m = (x,y)
so that Ass M = { (y), m }.
For a module M and a maximal ideal m, we have m in Ass M if and only
if M contains the field A/m as a submodule, which holds if an only
if Hom_A(A/m, M) <> 0. This will be important later when we work with
a local ring (A,m), and try to find an element s in m that is a
nonzerodivisor for m.
Proposition Let (A,m,k) be a Noetherian local ring and M a
nonzero A-module. Then equivalent conditions:
(1) every element s in m is a zerodivisor for M
(2) the maximal ideal m in Ass_A M
(3) Hom_A(k, M) <> 0.
This can be restated: there exists s in m that is a nonzerodivisor
for M if and only if Hom_A(k, M) <> 0.
Exa 1 was obviously set up artificially. However, exactly the same
phenomenon appears in geometrically important cases.
Exa 2. A = k[x,y,z]/(x*z - y^2) is the coordinate ring of
the ordinary quadratic cone Q: (x*z = y^2) in AA^3.
Its generating lines, such at the z-axis L : (x = y = 0), need
2 defining equations: the ideal I_L = (x,y). It is "not locally
principal." On the other hand, the ideal (z) consists of
polynomial function on Q that vanish twice on L, so that
I_{2L} = (z) is principal.
It might look reasonable to hope that vanishing twice on L should
be the same as belonging to (I_L)^2. However,
z notin (I_L)^2 = (x^2,x*y,y^2) = (x^2,x*y,x*z).
In other words, (I_L)^2 = (z) intersect (x,y,z)^2. The elements
of (I_L)^2 all vanish twice at the origin, as well as vanishing
twice on L.
====
Characterisation of DVR
I assume that you know the definition of discrete valuation rings (see
the textbooks or later in these notes). Informally, a DVR is the
simplest kind of UFD, with just one prime element. The main theorem says
DVR is characterised by
-> Noetherian
-> Spec A = {0,m} -- that is, 1-dimensional local integral domain
-> normal (integrally closed in its function field)
Sketch proof. Let A be a ring with those properties, and take x in
m\m^2. I prove that x generates m.
If m = (x) then we're home, else Ass m/x <> empty by the existence of
associated primes discussed in Lecture 2. This means there exist y in m
such that y notin (x) but m*y in (x). Consider the nonzero fraction
y/x in Frac(A). Since m*y in (x), it has the property that m*(y/x) in A.
Now case division: Either m*(y/x) = A ==> it contains 1, so x in m*y, which
contradicts x notin m^2.
Or m*(y/x) subset m. This means that m is a finite A-module (by the
Noetherian assumption), and multiplication by y/x is a A-linear
homomorphism m -> m. The determinant trick then give an integral
dependence relation for y/x over A.
Therefore y/x is integral over A, so by the assumption that A is normal,
also y/x in A. But this contradicting y notin (x). QED
====
DVRs and Dedekind domains
One important cultural point about this material is that it applies
directly both in number theory and in geometry.
In algebraic geometry, a point P in C on an algebraic curve has a local
ring O_{C,P} (rational functions on C that are regular at P), and
nonsingular <=> normal <=> DVR.
More generally, for a higher dimensional variety X, a codimension 1
(irreducible) subvariety Y in X has a local ring O_{X,Y} (rational
functions on X that are regular at the general point, or at a dense open
set of Y), and normal <=> DVR. This is necessary for X to be nonsingular
near Y, but not sufficient. (The argument says nothing about possible
singular locus of X in codimension 2.)
In number theory, ZZ itself is a PID so a UFD, so each localisation
ZZ_(p) at a prime is a DVR. However, as soon as you pass to an algebraic
number field K (an extension of QQ with [K : QQ] < infty), its ring of
integers O_K is normal by definition, but no longer has unique
factorisation in general.
Write K = QQ[al] (alpha), and f in QQ[x] for the minimal polynomial of
al. Multiply by common denominator d to get d*f in ZZ[x]. Now O_K / ZZ
is a finite ZZ-algebra, and a prime ideal P of O_K outside d is given by
taking a prime p of ZZ, then a root of the restriction fbar mod p in
FF_p[x]. The prime P of O_K can always be written (p, x-al), but is
almost never generated by 1 element. The local ring O_{K,P} only looks
near P, and is a DVR by the above main theorem.
In 2nd year algebra you saw factorization in ZZ and in a polynomial ring
K[x] treated in terms of division with remainder and the Euclidean
algorithm. In each case, the ring is a PID, and so a UFD. The striking
point is that we are dealing with objects of a completely different
nature: integers and polynomials. Nevertheless, the methods of argument
are practically identical.
The ring of integers O_K of a number field and the coordinate ring k[C]
of a nonsingular affine curve C continue this analogy. They are
Noetherian domains, 1-dimensional (the only prime ideals are 0 and
maximal ideals), and normal. Localising at 0 gives the field K or the
function field k(C) = Frac(k[C]). Localising at a nonzero prime ideal
gives a DVR. A ring with these properties is a Dedekind domain.
=====
What is commutative algebra?
The 1882 paper [DW] extended this analogy in elementary algebra to a
theory that encompasses both the ring of integers of a number field and
the ring of functions on an algebraic curve. Their paper is a landmark
in the development of modern algebra, and marks the starting point of
commutative algebra.
[DW] Richard Dedekind and Heinrich Weber, Theorie der algebraischen
Funktionen einer VerÃ¤nderlichen, J. reine angew. Math. 92 (1882),
181--290
I explain this briefly (don't worry about the details -- I will return
to the full arguments later).
==
Ring of integers of an algebraic number field
An _algebraic number field_ is a finite extension field QQ in K.
Corresponding to the ring of integers ZZ in QQ, the field K also has a
subring O_K of integers, the subset of K of integral elements (details
later). In any fairly complicated case, the division with remainder that
we used for ZZ does not work for O_K, and it is _not_ a UFD.
==
Integral closure of an algebraic function field
You know the polynomial ring k[x] over a field k (say k = CC to be
definite). Its field of fractions k(x) consists of rational functions
f(x)/g(x) with f,g polynomials and g <> 0. An _algebraic function field_
in one variable is a finite extension field k(x) in K (where x is
transcendental).
Corresponding to the polynomial ring k[t] in k(t), the same definition
as the number field case gives the integral closure A of k[x] in K: A is
the subset of elements of K that are _integral_ over k[x] (satisfy a
monic equation with coefficients in k[x] -- no denominators allowed, and
leading coefficient 1). This integral closure A = k[C] is the coordinate
ring of a nonsingular affine algebraic curve C over k. (I am not saying
that this is obvious.) In any fairly complicated case, this A does not
have division with remainder, and is _not_ a UFD.
==
Dedekind and Weber's synthesis
The preceding paragraphs set up the ring of integers O_K of a number
field K, and the coordinate ring k[C] of a nonsingular affine curve C.
These objects are major protagonists of algebraic number theory and
algebraic geometry, and are clearly very different in nature. However,
Dedekind and Weber [DW] say that these two rings can be studied using
the same algebraic apparatus. As I said, they are usually not UFDs.
The good news: if A is a ring of either type (a Dedekind domain), the
ideals of A have _unique factorisation into prime ideals_.
The key method of argument is _localisation_ (partial ring of
fractions). If P is a prime ideal of A, the localisation of A at P is
A_P = S^-1A where S = multiplicative set S = A - P.
(I will go through this in detail later.) In arithmetic, A_P in K is the
algebraic numbers that have an expression f/g with g notin P. For a
point P of and algebraic curve C, A_P consists of the rational functions
in k(C) that have an expression f/g with denominator g not vanishing at
P in C.
For either kind of ring A_P is a discrete valuation ring (DVR). Although
when the ring A is not a UFD, its localisation A_P is the simplest
possible UFD: it has a single prime element z (up to units), and every
nonzero element h in K has the factorisation
h = z^n*(unit), where n = v_P(h) is the valuation of h at P.
Valuations then determine everything about A in K and the ideals of A:
an element h in K is in A if and only if it has valuation >= 0 at every
P. Moroever, every ideal I in A also has a valuation at P (namely, min
v_P(i) taken over i in I). For any given nonzero ideal I of A, there are
just finitely many primes P such that v_P(I) > 0, and I equals the
product of P^v_P(I).
==
Modern abstract algebra
Notice the breakthrough aspect of Dedekind and Weber: modern algebra has
axioms and abstract arguments, and you often work with objects in a
symbolic way. In this case, without reference to what the elements of
the ring actually are.