Re: Prime ideals in the ring of continuous functions

In article <18749928.1161357471779.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Sasha P <abp83@xxxxxxxxxx> wrote:
No, it clearly is maximal. Take an ultrafilter F on
P([0,1]). Then the
set of functions f such that the zero set of f is in
F is a prime ideal.

Why the set of such functions is ideal?


An ultrafilter F on P([0,1]) is a collection of subset of [0,1] such
that:

(i) If A is in F, and B is a subset of [0,1] that contains A, then B
is in F.

(ii) If A and B are in F, then A/\B is in F.

(iii) F does not contain the empty set.

(iv) F is maximal among collection of subsets of [0,1] satisfying
(i)-(iii).

(A collection that satisfies (i)-(iii) is called a filter).

Examples of ultrafilters are "all subsets that contain {x}", where x
is an element of [0,1]. These are called "principal
ultrafilters". However, if we assume the Axiom of Choice, then there
are nonprincipal ultrafilters.

Now: let F be any ultrafilter on [0,1], and let S be the collection
of all continuous functions f on [0,1] such that the zero set of f is
an element of F.

If f and g are two continuous functions on [0,1], and Z(f),Z(g) are
both in F, then Z(f)/\Z(g) is also in F. It is also clear that
Z(f-g) contains Z(f)/\Z(g), hence Z(f-g) is also in F, since it is a
subset of [0,1] that contains a set in F. Thus, if f and g are in S,
then f-g is in S.

Now let f be a continuous function on [0,1] such that Z(f) lies in F,
and let g be any continuous function on [0,1]. Then Z(g*f) contains
Z(f), hence lies in F. Thus, g*f is also in S.

So the set S = {f in C[0,1] | Z(f) is in F}

is an ideal of C[0,1].

Notice that we never used the fact that F is an ->ultra<-filter, just
the fact that it is a ->filter<-. So, if F is any filter on [0,1],
then Z(F) = {f in C[0,1] : Z(f) is in F} is an ideal of C[0,1].

If F is a principal ultrafilter, then we get the usual maximal ideals
of all continuous functions that vanish on a point.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

Relevant Pages

  • Re: Prime ideals in the ring of continuous functions
    ... Examples of ultrafilters are "all subsets that contain ", ... of all continuous functions f on such that the zero set of f is ... If f and g are two continuous functions on, and Z,Zare ... If F is a principal ultrafilter, then we get the usual maximal ideals ...
    (sci.math)
  • Re: Noetherian
    ... chain of ideals. ... of continuous functions vanishing on that set is an ideal in your ring, ... Any member of I times any member of the ring is a member of I. ... What can one say about the set of all those continuous functions defined ...
    (sci.math)