Re: Field isomorphism

in article <f14pil$g2n$1@xxxxxxxxxxxx>, james dolan <jdolan@xxxxxxxxx>
wrote:

|in article <1177932999.220299.155860@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
|rupert <rupertmccallum@xxxxxxxxx> wrote:
|
||On Apr 30, 9:30 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:
||> On Apr 30, 9:09 pm, Jose Capco
||> <cliomse...@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 (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?) ?
||>
||> > Sincerely,
||> > Jose Capco
||>
||> Assuming any sequence (k_1, k_2, ...) is allowed to be in K^N, I
||> don't think this is true because you have a subring of K^N
||> isomorphic to a simple transcendental extension of K,
||
||Sorry, this is not as obvious as I thought it was.
||
||
||> consider an ideal which is maximal with respect to the property of
||> not containing any element of this subring.
|
|
|can't you contrive for r/m to be non-archimedean even when k is
|archimedean? isn't that pretty much how non-standard analysis works?

depending on whether the term "archimedean" works in the way that i
thought that i remembered it does, perhaps it's a good idea to take k
to be a field where the order structure is nicely (that is
first-order) definable in terms of the field structure, such as k =
the real numbers. but i think that the idea basically works.


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