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In article <18235534.1161365641076.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Sasha P <abp83@xxxxxxxxxx> wrote:
Quote the message you are replying to. Otherwise, there is no context
and I don't know what you are talking about without doing a bunch of
work.
In Mathforum, use the "Quote Original" button.
>Thank you very much! But why ideals corresponding nonprincipal
>ultrafilters is not maximal?
You are asking: is the ideal corresponding to a nonprincipal
ultrafilter (perhaps with the added condition that the filter contains
some closed subset of [0,1] other than the entire interval) prime? Is
it maximal?
First: suppose F is a principal ultrafilter on [0,1] (of any kind). Is
V(F), the set of all functions in C[0,1] which vanish on some element
of F, a prime ideal?
The key is the following property of ultrafilters:
THEOREM. If F is an ultrafilter on the set X, then for every subset Y
of X, either Y is in F, or X-Y is in F.
(Take an ultrafilter; if Y/\A is empty for some A in F, then A is
contained in X-Y, and therefore X-Y is in F. If Y/\A is nonempty for
all A in F, then you can consider the set of all elements of F and all
intersections of these elements of F, and make a filter out of it;
since F is an ultrafilter, this new filter will have to be equal to F,
and since this new filter contains Y, that means that Y is in F. So,
either X-Y is in F, or else Y is in F).
(In fact, some sources ->define<- an ultrafilter as a filter that
satisfies this condition, instead of as a filter which is maximal)
Now, let F be an ultrafilter on [0,1], and suppose that f*g lies in
V(F). That means that Z(f*g) is an element of F. Note that
Z(f*g)=Z(f)\/Z(g).
Now, either Z(f) lies in F (in which case f lies in Z(F) and we are
done), or else [0,1]\Z(f) lies in F. In this latter case, the
intersection of Z(f*g) and [0,1]\Z(f) lies in F; this intersection is
Z(g)\Z(f). So Z(g)\Z(f) lies in F, hence Z(g) lies in F (as it
contains an element of F), hence g is in V(F) (since its zero set is
an element of F). This proves that if f*g is in V(F), then f is
in V(F) or g is in V(F), and therefore V(F) is a prime ideal whenever
F is an ultrafilter.
Is it maximal? Depends. We know that if the ultrafilter is "all
functions that vanish on {x}" (i.e., a principal ultrafilter), then
V(F) is maximal. So suppose F is not principal; is V(F) maximal then?
I believe the argument is as follows, if I did not screw up somewhere:
Given any ideal I of C[0,1], we can consider
Z(I) = {x in [0,1] : f(x) = 0 for all x in I}.
If Z(I) is empty, then I=C[0,1]: for, given any x in [0,1],
there exists f_x in I such that f_x(x)<>0. Find a nbd of U_x of x
such that f(x) is not zero on U_x. Then {U_x : x in X} form an open
cover for [0,1], which is compact. Thus there exist x_1,...,x_n such
that U_{x_1},...,U_{x_n} cover [0,1]. Letting f_i = f_{x_i}, we have
that f_1^2 + ... + f_n^2 is a function in I which does not vanish on
[0,1], hence is a unit of C[0,1], and therefore I = C[0,1].
So if I is a PROPER ideal of C[0,1], then Z(I) is nonempty.
Now, if F is a nonprincipal ultrafilter, then V(F) is not equal to
C[0,1]; if it contains a unit, then we would have the empty set in F
(the zero set of a unit is empty), which is not the case. Thus,
Z(V(F)) is nonempty. Let x be in Z(V(F)). Then every function in V(F)
vanishes on {x}, so V(F) is contained in {f in C[0,1] : f(x) = 0}. But
it is PROPERLY contained in V(F): for there certainly is a function g
in C[0,1] which vanished only at {x} and is nonzero elsewhere. This g
is in {f in C[0,1]: f(x)=0} but is not in V(F) (for Z(g) = {x} cannot
be an element of F, because F is nonprincipal). So V(F) is properly
contained in the maximal ideal {f in C[0,1]: f(x)=0}, hence V(F) is a
nonmaximal prime ideal of C[0,1].
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
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