Re: F-algebra homomorphism

On Aug 7, 7:52 pm, leiko <leikomats...@xxxxxxxxx> wrote:
On Aug 7, 7:13 pm, Mariano Suárez-Alvarez



<mariano.suarezalva...@xxxxxxxxx> wrote:
On Aug 7, 7:37 pm, leiko <leikomats...@xxxxxxxxx> wrote:

On Aug 7, 5:38 pm, Mariano Suárez-Alvarez

<mariano.suarezalva...@xxxxxxxxx> wrote:
On Aug 7, 5:16 pm, leiko <leikomats...@xxxxxxxxx> wrote:

On Aug 7, 3:56 pm, A N Niel <ann...@xxxxxxxxxxxxxxxxxxxxx> wrote:

What are all algebra-homomorphisms F-> F{X}? "F{X}" stands for F
valued functions on the set X. We can drop the condition f(1)=1 for
this question.

F is a field?  Call the map u : F -> F{X}. Every element of F is an
F-multiple of 1, so if u is an algebra-homomorphism, then once you know
u(1), you know all of u. What could u(1) be?  Since u is a
homomorphism, it must be a function equal to its own square.  That is,
a function taking values 0 and 1.  These correspond to the subsets of
X.

How can you show that any algebra homomorphism F{X}->F is evaluation
at some x in X, i.e, f->f(x) for some x in X? Again, "F{X}" stands for
F valued functions on the set X.

If F is the two-element field, then I think the algebra homomorphisms v
: F{X} -> F correspond to the ultrafilters on X.  If X is infinite,
then (using AC) there are ultrafilters other than the fixed ones, so
there are homomorphisms other than the evaluations.

Thanks for you answer. I understand the first part. However, the
second one I find very confusing because I don't know about
ultrafilters..Is there a way to avoid them and show directly that the
homomorphisms are evaluations at some x?

Well, the whole point of ultrafilters is that they
enable you to show that there do in fact exist
homomorphisms which are *not* evaluations...

-- m

Now I am even more confused, because I am supposed to show that EACH
homomorphism is an evaluation...

Then you are going to have lots and lots of trouble...

Let's do the details:

Let us fix a set X. An _ultrafilter_ U on X is simply
a subset of P(X), the set of all subsets of X, such that

  (1) the empty set is not in U;
  (2) U is closed under finite intersections;
  (3) If A is an element of U and B is a subset of X
      such that A is contained in B, then B is an element
      of U; and, finally,
  (4) if A is a subset of X, then either A is in U
      or X-A is in U.

The last property is what makes the ultrafilter "ultra".

We say that an utrafilter U is _principal_ if there exists
an element x in X such that U coincides with the set of
all subsets of X which have x as an element. If U is not
principal, we say that it is _free_

One can show, as soon as X is an infinite set, that there
exist free ultrafilters on X.

Now, let us fix an infinite set and an ultrafilter U on X.
Let F be a field, and let F(X) be the F-algebra of all
functions X --> F, with pointwise operations. When f : X --> F
is an element of F(X), let us write Z(f) = {x in X : f(x) = 0}.

Consider the set I(U) = {f in F(X) : Z(f) is in U}.
This is an ideal in F(X), as you can easily check.
It is a proper ideal because the function which has
value 1 on all elements of X is not in I(U).

I claim that I(U) is, moreover, maximal.

Indeed, let J be an ideal in F(X) which properly contains
I(U). Pick an element f in J which is not in I(U). Since f
is not in I(U), the set Z(f) is not in U. Axiom (4), then,
tells us that the set X - Z(f) is an element of U, so its
characteristic function, which we write g, is an element of
I(U). It follows that f + g is an element of J. But this
function has non-zero values on all points of X, so that it
is invertible in F(X). We thus see that J contains
invertible elements, so that J = F(X) is not a proper ideal.
This substantiates my claim.

Let K be the quotient F(X) / I(U).

Let us suppose now that F = F_2 is the field with two
elements. I claim that in this case that K is also F_2.
To see this, let f be in F(X), and let [f] be its class
in K. Then if Z(f) is in U, we have that f is in I(U) and
that [f] = 0. On the other hand, if Z(f) is not in U,
then X - Z(f) is and the function 1 - f, which is such that
Z(1 - f) = X - Z(f), is an element of I(U), so that
1 - [f] = [1 - f] = 0. We thus see that either [f] = 0,
or [f] = 1, so that K has at most two elements.

Up to isomorphimsms, then, we have constructed a map
p : F(X) --> F of rings, which is automatically of
F-algebras.

I'll leave you the enjoyment of proving that this map
is an evaluation map exactly when U is a principal
ultrafilter.

Since free ultrafilters exists, we thus obtain examples
which are not evaluations.

-- m

Thank you so much for the detailed answer. Does the last part work
only for F_2, or does it hold in general? I am interested in the
general case. I'll try to understand the details better and prove the
claim...

Thanks again :)

Hi, leiko:

What you are calling F[X] is the direct product
of copies of F indexed by some set X. The maps
F[X] -> F which are ring homomorphisms can be
expressed as quotients of F[X] modulo ideals;
in the case F is a field and the maps are onto,
the ideals must be maximal. In particular if
the map is an F-algebra homomorphism, the kernel
is a maximal ideal of F[X], excluding the zero
map as a possibility.

The earlier discussion of free (nonprincipal)
ultrafilter for F = Z/2Z can be replaced by a
construction of a maximal ideal containing the
direct _sum_ of copies of F indexed by X. This
subring of F[X] is an ideal and can therefore be
extended to a maximal ideal via Axiom of Choice.
(The direct sum consists of maps from X to F with
all but finitely many of the "component values"
zero.)

Indeed the only hope for showing what you ask
is the case X is a finite set.

regards, chip
.

Relevant Pages

  • Re: Algebra with gaussian integer.
    ... the order of the cokernel of the map ...   ... and the latter module has obvious norm 4*2, ... Ideals in quadratic number fields.http://www.fen.bilkent.edu.tr/~franz/ant/ant02.pdf- Hide quoted text - ...
    (sci.math)
  • Re: What the Dylan/Beatles/Sixties Generation stood for
    ...   The work ethic has been co-opted and warped to make slaves. ... the ideals of those earlier decades themselves ... revolution and mass urbanisation in the Christian west. ...
    (rec.music.dylan)
  • Re: About characters on H^{oo}
    ... unit disk with the natural product, ... a homomorphicm of H^into C, i.e. character. ... maximal ideals ... from A onto A/I is actually a complex homomorphism ...
    (sci.math)
  • Re: OT A personal view
    ... I thank God I saw that pathetic hate for what it was. ... My country is, of course, it's geography and it's people. ...  is ideals. ...
    (rec.outdoors.fishing.fly)
  • Re: Algebra with gaussian integer.
    ... the order of the cokernel of the map ... Or one may use the standard criterion for a module to be an ideal, ... and the latter module has obvious norm 4*2, ... Ideals in quadratic number fields. ...
    (sci.math)