Re: Field isomorphism

On 30 Apr 2007 04:09:33 -0700, Jose Capco
<cliomseerg@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

Dear NG,

Let K be a field (with char=0, Im not sure I will require this) and
consider the ring R=K^N

There's been some question regarding exactly what you mean by this.
I'm going to assume you meant exactly what you wrote, that R is
the space of _all_ sequences of elements of K.

(N being the natural numbers), with pointwise
addition and multiplication. What is the easiest way to show that for
any maximal ideal
M in R, R/M is isomorphic to K (or is this true at all?) ?

I tend to suspect this is false. I can't prove that, but I _can_
prove that there exist more maximal ideals than I tend to
suspect you had in mind when you conjectured there might
be a simple proof that it's true:

Let K = Q, the rationals. Let I be the ideal consisting
of all r = (k_1, k_2, ...) such that lim_j k_j = 0.
I suspect that if M is a maximal ideal containing I then
R/M must be uncountable.

(For every _real_ number b consider I_b, the set of
all elements of R which tend to b at infinity. The
I_b correspond to distinct elements of R/I...)

In general, for r = (k_1, ...) in R let Z(r) be the
set of j such that k_j = 0. Let Z(I) be the intersection
of the Z(r) for r in I. If Z(I) is nonempty then it's
not hard, I think, to show that Z(M) must be a single
point and that R/M is isomporphic to K. But there exist
I with Z(I) empty.

Sincerely,
Jose Capco


************************

David C. Ullrich
.

Relevant Pages

  • Re: Field isomorphism
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  • Re: Field isomorphism
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  • Re: Field isomorphism
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  • Re: Field isomorphism
    ... R/M is isomorphic to K? ... However not all maximal ideals of R have the form M_j. ... The idea of the proof is to show that R/M has a subfield isomorphic to ... fields, up to isomorphism: ...
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  • Re: Field isomorphism
    ... R/M is isomorphic to K? ... However not all maximal ideals of R have the form M_j. ... The idea of the proof is to show that R/M has a subfield isomorphic to ... fields, up to isomorphism: ...
    (sci.math)