Re: Q[sqr p]

William Elliot wrote:

This now brings to mind the multiplicate groups
Q*[sqr 2] = (Q[sqr 2] - {0}, *)
Q*[sqr p] = (Q[sqr p] - {0}, *)
Are they isomorphic and why not?


The arguments of Jyrki can be complemented by the the
following remarks: the ring R of algebraic integers
in Q[sqr p] is noetherian. Hence every element of R
can be written as a product of a unit of R with powers
of finitely many irreducibles. Choosing a representative
of each class of associated irreducibles and denoting
the set thus obtained by I this leads to a surjective
group homomorphism

R*x F(I)-->Q(sqr p)*

where R* is the group of units of R, F(I) is the free
abelian group with basis I.
This homomorphism is injective if and only if R is a UFD.

Note also that we get the follwoing nice result:
let K and L be the fraction fields of UFDs R and S
with a countably infinite number of non-associated primes.
Then the groups K* and L* are isomorphic if and only if
the groups of units R* and S* are isomorphic.

In the case of algebraic number fields the groups
R* are described by Dirichlet's unit theorem.

H
.

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