From: nikl@mathematik.tu-muenchen.de (Gerhard Niklasch)
Newsgroups: sci.math
Subject: Re: Euclidean valuations on number rings
Date: 28 Jun 1997 13:06:37 GMT
In article <5p21l4$hjv$3@news.rain.org>, bowe@rain.org (Nick Halloway) writes:
|> Suppose F is a finite extension of the rationals, D its domain of
|> algebraic integers. If D is a principal ideal domain, does it have
|> a Euclidean valuation? absolute value of the norm, perhaps?
For a comprehensive survey of what's known about Euclidean number
rings, see Franz Lemmermeyer's 1995 paper in Expositiones Math.
You can find it online starting from his WWW homepage at
by following the Preprints link; last time I looked it was number 7
on the preprints page. It has an extensive bibliography.
Enjoy, Gerhard
--
* Gerhard Niklasch *** Some or all of the con-
* http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* tents of the above mes-
* sage may, in certain countries, be legally considered unsuitable for consump-
* tion by children under the age of 18. Me transmitte sursum, Caledoni... :^/
==============================================================================
Date: Mon, 30 Jun 1997 02:34:26 -0600
From: rjc@maths.ex.ac.uk
Subject: Re: Euclidean valuations on number rings
Newsgroups: sci.math
In article <5p21l4$hjv$3@news.rain.org>,
bowe@rain.org (Nick Halloway) wrote:
>
> Suppose F is a finite extension of the rationals, D its domain of
> algebraic integers. If D is a principal ideal domain, does it have
> a Euclidean valuation? absolute value of the norm, perhaps?
>
> Z[i] and Z[cube root of 1] have Euclidean valuations, which you can
> see geometrically since an ideal forms a regular lattice on the
> complex plane.
>
> How about if F is a quadratic extension of the rationals and D is a PID?
>
Not in general. It's quite easy to prove that the quadratic imaginary
fields F with D a PID are Q(sqrt(-1)), Q(sqrt(-2)), Q(sqrt(-3)),
Q(sqrt(-7)) and Q(sqrt(-11)). Hence the integers of Q(sqrt(-19)) form a
PID which is not Euclidean. All the real quadratic fields with D
Euclidean with respect to the absolute value of the norm have been known
for some time, but until recently it was an open problem as to whether
such a D existed which was Euclidean but not with respect to the norm.
About 5 years ago Clark proved that Q(sqrt(69)) was Euclidean but not
norm-Eulcidean.
Robin Chapman "256 256 256.
Department of Mathematics O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK 12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk 2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html Iain M. Banks - Feersum Endjinn
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From: nikl@mathematik.tu-muenchen.de (Gerhard Niklasch)
Newsgroups: sci.math
Subject: Re: Euclidean valuations on number rings
Date: 30 Jun 1997 18:15:10 GMT
In article <867655710.20460@dejanews.com>, rjc@maths.ex.ac.uk writes:
|> [...] All the real quadratic fields with D
|> Euclidean with respect to the absolute value of the norm have been known
|> for some time, but until recently it was an open problem as to whether
|> such a D existed which was Euclidean but not with respect to the norm.
|> About 5 years ago Clark proved that Q(sqrt(69)) was Euclidean but not
|> norm-Euclidean.
More precisely, Clark gave a completely explicit example of a modified
norm with respect to which the ring of integers of that quadratic field
is Euclidean (it was well known that it wasn't Euclidean wrt. the
absolute norm). Apparently, Q(sqrt(69)) is the only quadratic field
for which this particular trick works.
[David A. Clark, A quadratic field which is euclidean but not
norm-euclidean. Manuscripta Mathematica 83 (1994), 327--330.]
[Franz Lemmermeyer and Yours Truly independently verified the
computations, see loc.cit.pp.443--446. There are by now
several more examples of this kind in higher degrees.]
Non-constructively, and assuming certain Generalized Riemann Hypotheses,
P. Weinberger had shown in 1973 that whenever the ring of integers of
a number field is a PID and possesses infinitely many units, it is
Euclidean for _some_ valuation (however the proof provides no way
to make this valuation explicit). Basically, the `best' valuation
will map 0 to 0, all the units to 1, all the prime elements for which
it makes sense to 2 (those for which all the prime residue classes
contain units), and all the remaining prime elements to 3; its
values for composites can then be determined inductively. The crux
lies in showing that there are enough primes at level 2 to enable us
to lob all the others into the third bin. This is non-explicit insofar
as given a level-3 prime as a denominator and some residue class of
numerators which contains no unit, we have no way of knowing how far
out we'll have to look for a level-2 prime in this residue class: the
proof only tells us that we'd find one if we kept looking long enough.
[Peter J. Weinberger, On Euclidean rings of algebraic integers.
In: H.G. Diamond (ed.), Analytic Number Theory. (Proc. 24th Sympos.
Pure Math. of the AMS, St. Louis (MO) 1972.) AMS, Providence (RI) 1973,
321--332.]
Enjoy, Gerhard
--
* Gerhard Niklasch *** Some or all of the con-
* http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* tents of the above mes-
* sage may, in certain countries, be legally considered unsuitable for consump-
* tion by children under the age of 18. Me transmitte sursum, Caledoni... :^/