Download Filters and Ultrafilters

Filters and Ultrafilters (continued)
We would like to describe the maximal ideals
of the ring
R=
Y
k
i∈I
where k is a field and I is an index set.
This is a special case of describing the maximal
ideals of a ring of functions on a topological
space. We can view R as the ring M ap(I, k) of
all functions from I to k. For instance, I could
be the unit interval [0, 1] on the real line, and
k could be the real numbers.
Later in the course we will try to determine
the maximal ideals of particular subrings of
M ap(I, k).
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Here is one example:
Theorem: (Hilbert’s Nullstellensatz) Suppose
A is the ring of polynomials C[x1, · · · , xn], so
that A ⊂ M ap(Cn, C). Then every maximal
ideal M of A has the form
Ma = {f ∈ A : f (a1, · · · , an) = 0}
for some point a = (a1, · · · , an) of Cn. (The
same is true if C is replaced by any algebraically
closed field.)
Q
Returning to the case of R = i∈I k = M ap(I, k),
we discussed last time how if I is infinite, there
are maximal ideals M not of the form
Mi0 = {f ∈ R : f (i0) = 0}
for some i0 ∈ I. One can construct such an M
by taking a maximal ideal containing the ideal
of all f ∈ R which are zero at all but finitely
many elements of I.
This shows that M ap(I, k) doesn’t satisfy the
Nullstellensatz when I is infinite.
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To describe the maximal ideals of M ap(I, k),
let’s recall the definition of filters and ultrafilters on the index set I.
Def: Suppose I is a set. A filter for I is
a collection J of subsets J of I with these
properties:
1. If j ∈ J and J ⊂ J 0 ⊂ I then J 0 ∈ J .
2. If J1, J2 ∈ J then J1 ∩ J2 ∈ J .
In other words, J is closed under taking oversets and intersections.
Def: We say J is an ultrafilter if it is a maximal proper filter, in the sense that it is not the
power set of I, and it is not properly contained
in any filter for I other than the power set.
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Example: Suppose i0 ∈ I. Let’s show that
the collection Ji0 of all subsets J ⊂ I such
that i0 ∈ J is an ultrafilter.
Clearly Ji0 is closed under oversets and finite
intersections, so its a filter. It doesn’t contain
the emptyset, so it is a proper filter.
Suppose Ji0 is properly contained in some other
filter J 0. Then there would be an element J 0 ∈
J 0 such that i0 6∈ J 0. However, {i0} ∈ Ji0 ⊂ J .
So by the axioms of filters, ∅ = {i0} ∩ J 0 ∈ J 0.
which implies J 0 is the powerset. This implies
J is an ultrafilter.
Example: It’s a homework problem this week
to use Zorn’s Lemma to show that every proper
filter is contained in an ultrafilter. So for example, if I is infinite, then there is an ultrafilter
which contains the filter of all complements of
finite subsets of I. This ultrafilter can’t be of
the form Ji0
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Def: Suppose F is an filter for the set I. SupQ
Q
pose α = i∈I αi is an element of R = i∈I k.
Let
zer(α) = {i ∈ I : αi = 0}.
Thus zer(α) is a subset of I. Define MF to be
the set of all α ∈ R such that zer(α) ∈ F .
Theorem: (See this week’s Homework) The
set MF is an ideal. The map
F → MF
defines a bijection between the set of ultrafilters on I and the set of maximal ideals of
R.
Example: The filter of all complements of finite subsets of I corresponds the ideal of all
elements of R which have only finitely many
non-zero components.
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Prime ideals
Let R be a commutative ring.
Def: A ideal P of R is a prime ideal if P 6= R,
and whenever a, b ∈ R have the property that
ab ∈ P , then one of a or b is in P .
Comment: An equivalent definition is that
R/P is an integral domain, i.e. that R/P is a
commutative non-zero ring which has no zero
divisors.
Example: All maximal ideals M are prime,
since R/M is a field.
Example: The zero ideal {0} is prime if and
only if R itself is an integral domain. This is
so, for example, if R = Z.
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Def: The spectrum Spec(R) of a commutative
ring is the following topological space.
1. The elements of Spec(R) are the prime ideals of R.
2. The closed subsets of Spec(R) (which are
the complements of the open subsets) have
the form
VA = {P ∈ Spec(R) : A ⊂ P }
as A ranges over all ideals of R. ( This is
called the Zariski topology.)
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Example: Suppose R = Z. The prime ideals
of Z are the zero ideal {0} together with the
maximal ideals Zp as p ranges over all prime
numbers. Each ideal A has the form Zn for
some (possibly zero) integer n. One has
V{0} = Spec(R)
and for n 6= 0,
VZn = {P = Zp : p|n}
It follows that the closed sets of Spec(Z) are
just the finite subsets of Spec(Z) − {{0}} together with all of Spec(Z).
Picture:
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One does have to check that the Zariski topology is a topology. For this one needs to show:
1. ∅ and Spec(R) are closed. This is so because ∅ = VR and Spec(R) = V{0}.
2. The intersection ∩A∈S VA is closed if S is
any collection of ideals of R. This is true
since
∩A∈S VA = VB
when B is the R-ideal generated by all the
A ∈ S. (A prime ideal P contains B if and
only if it contains each A ∈ S.)
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3. The union ∪A∈S VA is closed if S is any
finite set of ideals of R. We claim
∪A∈S VA = VC
when
C = ∩A∈S A.
In one direction, if P ∈ VA for some A ∈ S,
then C ⊂ P so P ∈ VC .
If P ∈ VC we want to show P ∈ VA for some
A ∈ S. Suppose this is not the case. Then
for each A ∈ S, it is not true that A ⊂ P .
Choose for each A ∈ S an element pA ∈ A
not in P . Then
Y
A∈S
pA ∈
Y
A⊂C
A∈S
but A∈S pA 6∈ P since P is prime. (Note
that the product is finite.) This contradicts P ∈ VC .
Q
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Picture of Spec(C[t]):
Picture of Spec(C[x, y]).
We will later show that the prime ideals P of
Spec(C[x, y]) are either
(i) P = {0} (the generic point);
(ii) P = C[x, y] · f for some irreducible
polynomial f = f (x, y);
(iii) maximal ideals, which by the Nullstellensatz have the form Ma = {h : h(a) = 0} for
some a = (a1, a2) ∈ C.
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Localization
Idea: Generalize the construction of Q as the
ring of quotients ab for which a, b ∈ Z and b 6= 0.
Def: Suppose R is a commutative ring. A
multiplicatively closed subset D of R is a subset
which contains 1 and such that d1d2 ∈ D if
d1, d2 ∈ D.
Theorem: There is a commutative ring D−1R
and a ring homomorphism φ : R → D−1R with
the following properties:
1. φ(D) is contained in the group of units of
D−1R.
2. The homomorphism φ is universal in the
following sense. If φ0 : R → R0 is any other
ring homomorphism for which φ0(D) ⊂ (R0)∗,
there is a unique ring homomorphism h :
D−1R → R0 such that φ0 = h ◦ φ.
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Example: Suppose R = Z and D = Z − {0}.
Then D−1R = Q and the inclusion φ : Z → Q
have the right universal property. To construct
h : Q → R0 given φ0 : R → R0, one sets
a
h( ) = φ0(a) · φ0(b)−1.
b
We would like to construct D−1R by formalizing the definition of ab for a ∈ R and d ∈ D. To
do this we need:
Lemma: There is an equivalence relation ˜
on the product set R × D defined by
(r, d) ˜ (r0, d0)
d”(rd0 − r0d) = 0
iff
d” ∈ D
for some
Comment: To motivate this, one has formally
d”(rd0 − r0d) = d”d0d
r0
r
− 0
d d
!
So (r, d) ˜ (r0, d0) should correspond to the right
side being 0.
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Proof: Reflexivity and symmetry are easy, since
R is a commutative ring. For transititivity,
one has to show that if (r, d) ˜ (r 0, d0) and
(r0, d0) ˜ (r 00, d00) then (r, d) ˜ (r00, d00).
The principle in proving this sort of statement
is to treat (r, d) as the formal fraction dr . One
then takes an identity to be proved and puts
the terms over a common denominator.
In the present situation, one has formally that
r” r
− =
d” d
r0
r”
− 0
d” d
!
+
r0
r
−
d0 d
!
Putting the left and right hand sides over the
common denominator dd0d” and then multiplying by dd0d” gives the identity
d0(r”d − d”r) = d(r”d0 − d”r0) + d”(r0d − rd0)
which holds because R is commutative.
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The hypotheses that
(r, d) ˜ (r 0, d0)
and
(r 0, d0) ˜ (r00, d00)
imply that we can multiply the right hand side
of
d0(r”d − d”r) = d(r”d0 − d”r0) + d”(r0d − rd0)
by d0 · d1 for some d0, d1 ∈ D to get 0. Since
d0 · d1 · d ∈ D because D is multiplicatively
closed, the left hand side of the resulting equality shows (r, d) ˜ (r00, d00).
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Corollary: We can define D−1R to be the set
of equivalence classes [(r, d)] of pairs (r, d) ∈
R × D with respect to the equivalence relation
˜ . This set becomes a ring if we define
addition and multiplication by
1. [(r, d)] + [(r0, d0)] = [(rd0 + r 0d, dd0)].
2. [(r, d)] ∗ [(r 0, d0)] = [(rr0, dd0)].
One defines φ : R → D−1R by φ(r) = [(r, 1)].
The proof of the Corollary consists of first
checking that the addition and multiplication
are well defined; this is necessary because we
used representatives for equivalence classes in
the definitions. Then one has to check + is
an abelian group law, ∗ defines a monoid with
multiplicative identity (1, 1), and the distributive laws hold.
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To prove any of these facts, one treats [(r, d)]
as dr and then puts terms over a common denominator in the identities to be proved.
The check the universal property of D−1R,
suppose φ0 : R → R0 is a ring homomorphism
with φ(D) ⊂ (R0)∗. The ring homomorphism
h : D−1R defined by h([(r, d)]) = φ(d)−1φ(r)
has the property we want.
In the following examples, we’ll write dr for the
equivalence class [(r, d)] ∈ D−1R.
Example: Suppose R an integral domain. Then
D = R−{0} is multiplicatively closed. The ring
D−1R is a field, since dr · dr = 1. Furthermore,
φ : R → D−1R is injective. This is because
r = 0 = 0 implies d(r · 1 − 1 · 0) = dr = 0 for
1
1
some d ∈ D, and d 6= 0 then forces r = 0 since
R has no zero divisors. Once calls D−1R the
fraction field of R.
Subexample: The field of p-adic numbers Qp
is the fraction field of Zp.
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Example: Suppose P is a proper ideal R. The
set D = R − P is multiplicatively closed if and
only if P is a prime ideal. Let’s suppose this
is the case. The ring D−1R is called the localization of R at P , and is denoted RP .
Subexample 1: If R = Z, primes P ∈ Spec(Z)
are either P = {0} or P = Zp = (p) for some
prime p. When P = {0}, one gets D = Z − {0},
so R{0} is the fraction field Q of Z. Suppose
now that P = (p) for some prime number p.
We have D = Z − P = {n ∈ Z : p 6 |n}. So
r
Z(p) = D−1Z = { : r, n ∈ Z, p 6 |n}
n
is the subring of rationals which have denominators prime to p.
There is an embedding
Z(p) → Zp = the p − adic numbers.
This is because each rational integer r ∈ Z not
in Zp is a unit in Zp, since it can be inverted
mod pn for all n. But Zp is much bigger than
Z(p), e.g. its uncountable.
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Subexample 2: Suppose R = C[x]. We now
the prime ideals of R are either {0} or the ideal
(x − α) for some α ∈ C.
As in subexample 1, the localization R{0} is the
fraction field C(x) of R. This is the field of all
(x)
rational functions fh(x)
in one variable (where
h(x) 6= 0).
Suppose now that P = (x − α). Then R − P
is the set of h(x) for which h(α) 6= 0. This
means
C[x](x−α) = {
f (x)
: f (x), h(x) ∈ C[x], h(α) 6= 0}
h(x)
Thus C[x](x−α) consists of those rational functions which take on well-defined values in some
small open disc around α. This is where the
term ”localization” comes from; one has functions defined locally near α.
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Local Rings
Def: A commutative ring R is called local if
it has a unique maximal ideal mR .
Proposition: A commutative ring R is a local
ring if and only if R −R∗ is an ideal; in this case
R − R∗ = mR is the unique maximal ideal of R.
Proof: If R is local, and β ∈ R − mR , then Rβ
can’t be a proper ideal of R, since this would
imply Rβ ⊂ mR because proper ideals have to
be contained in some maximal ideal. Hence
Rβ = R, so β ∈ R∗ = R − mR . Conversely, no
proper ideal of R can contain an element of
R∗. So if R − R∗ is an ideal, this must be the
unique maximal ideal of R, and R is local.
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Example: Each field F is a local ring, since
{0} is the unique proper ideal of F and hence
the unique maximal ideal.
Example: The ring Z is not a local ring, since
it has many distinct maximal ideals Zp.
Example: Suppose R is a commutative ring
with prime ideal P . Let’s check that the localization RP is a local ring.
Define an ideal of RP by
r
mRP = { ∈ RP : r ∈ P, d ∈ R − P }.
d
Using the fact that P is an ideal of R, one finds
that mRP is an ideal of RP .
We claim mRP 6= RP . Otherwise,
r
1
=
1
d
for some r ∈ P and d ∈ R − P , there is a d0 ∈
R − P such that d0 · (d − r) = 0. This would
force d0d = d0r ∈ P , contradicting the fact that
d0d 6∈ P since P is a prime ideal.
1=
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We have
r
R − mRP = { ∈ RP : r, d ∈ R − P }
d
But such elements are units in RP , since dr ∈ RP
if r, d ∈ R − P . Thus
R − mRP ⊂ R∗.
Because mRP is a proper ideal,
R∗ ⊂ R − mRP .
So
R − mRP = R∗
and R is local.
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Example: Let X be a topological space, and
let F be the presheaf which over each open
subset U ⊂ X has sections F (U ) the ring of real
valued continuous functions on U . Suppose z
is a point of X. The stalk
Fz = lim F (U )
−→
z∈U
is a local ring. To see why this is so, define
m(Fz ) to be the kernel of the ring homomorphism
Fz → R
which sends each f ∈ Fz to its value at z. Then
m(Fz ) is an Fz -ideal, and it is not all of Fz since
it does not contain 1. If h is an element of
Fz − m(Fz ), then h(z) 6= 0, so h is non-zero in
some neighborhood U of z. This means 1/h is
a continuous function on U , so 1/h ∈ Fz and h
is a unit of Fz . Thus Fz is a local ring.
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