Re: Q[sqr p]

In article <Ixn5KE.E96@xxxxxx> "Peter L. Montgomery" <Peter-Lawrence.Montgomery@xxxxxx> writes:
In article <Pine.BSI.4.58.0604121909350.11304@xxxxxxxxxxxxxxxxx>
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> writes:
On Wed, 12 Apr 2006, *** T. Winter wrote:
....
> > ...
> > > Ok, then Q^*[sqr 2] and Q^*[sqr 3] aren't isomorphic because
> > > the later is a UFD while the former has non-unique factorization
> > > 2 = (sqr 2)^2 = (2 - sqr 2)(2 + sqr 2)
....
So sqrt(2) and (2 - sqrt(2)) and (2 + sqrt(2)) are associated, and the
above is not an example of non-unique factorisation.

That I get. What I don't get is why the multiplicative groups
Q*[sqr 2] and Q*[sqr 3] aren't isomorphic.

They are isomorphic, see below. The conclusion that they are not
isomorphic is by you based on the wrong argument that for sqrt(2)
there was not a UFD.

We need to argue that each multiplicative
group is a direct product of infinite cyclic groups,
times a group with roots of unity. The only roots
of unity are +- 1 (whereas Q^([sqr -3]) has other
primitive roots.

This is wrong, it is about sqrt(2) and sqrt(3), not sqrt(-2) and sqrt(-3).
Both Q*[sqrt(2)] and Q*[sqrt(3)] have roots of unity different from +-1,
and both have primitive such roots. Also both are UFD and have infinitely
many irreducibles. For each take for each set of associated irreducibles
one representative, and also one representative primitive root of unity.
Then in both cases each number can be uniquely represented as:
u^k . i_1^k_1 . i_2^k_2 . ... , u_n^k^n
for some n, here u is the chosen primitive root of unity, i_x are the
chosen irreducibles, k is an integer, and k_x are positive integers.
This clearly defines the multiplicative structure, and also shows the
isomorphism (take some arbitrary mapping of the irreducibles from one
to the other).
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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