Saturday, 31 May 2008

fields - What is the prime spectrum of a Cauchy series ring?

In this answer I will treat the case in which |text| is not discrete.



I first claim that mathfrakm0 is not the restriction of any proper ideal in
kinfty. Indeed, choose xink such that 0<|x|<1. Then (xi)
is an element of mathfrakm0 which is invertible in kinfty (with
inverse equal to (xi), and so mathfrakm0 generates the unit ideal
of kinfty.



This doesn't contradict anything; the maximal ideals of kinfty pull-back
to prime ideals in mathcalC(k) which are simply not maximal (as often happens
with maps of rings).



Furthermore, this pull-back is injective.



To see this, we first introduce some notation; namely, we
let mathfrakmmathcalU denote the prime ideal of kinfty corresponding to
the non-principal ultra-filter mathcalU,and recall that mathfrakmmathcalU is defined as follows: an element (xi) lies in mathfrakmmathcalU if and only
if i,|,xi=0 lies in in mathcalU.



Now suppose that mathcalU1 and mathcalU2 are two distinct non-principal ultra-filters. Let A be a set lying in mathcalU1, but not in mathcalU2.
Then Ac, the complement of A, lies in mathcalU2.
Choose xink such 0<|x|<1, and let xi=xi if iinA and
xi=0 if inotinA. Then (xi) is an element of mathcalC(k),
in fact of mathfrakm0, and it lies in mathfrakmmathcalU2
but not in mathfrakmmathcalU1.



Thus mathfrakmmathcalU1 and mathfrakmmathcalU2 have distinct pull-backs.



So the map
Spec kinftyrightarrow Spec mathcalC(k)
is injective
and dominant (since it comes from an injective map of rings), but is not surjective.
Choosing the valuation |text| allows us to add to Spec kinfty (which is the
Stone-Cech compactification of mathbbZ+) an extra point dominating all the
other points at infinity (i.e. all the non-principal ultrafilters), because the valuation now gives
us a definitive way to compute limits (provided we begin with a Cauchy sequence).

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