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From: Jude on 23 Jun 2010 04:06
Hi~

I have a question in algebraic number theory.

It is known that the ideal class group of a number field is finitely
generated abelian group and so
it is isomorphic to the direct product of finite cyclic groups.

Then, for each cyclic component of ideal class group, is it possible
to choose a prime ideal representing the ideal class of generaor of
the cyclic component?
From: Arturo Magidin on 23 Jun 2010 14:01
On Jun 23, 3:06 am, Jude <classnu...(a)gmail.com> wrote:
> Hi~
>
> I have a question in algebraic number theory.
>
> It is known that the ideal class group of a number field is finitely
> generated abelian group and so
> it is isomorphic to the direct product of finite cyclic groups.

This is still incorrectly stated. The "so" is unjustified. The reason
you know the ideal class group is isomorphic to a direct product of
finite cyclic groups is not because it is merely finitely generated
and abelian, but because it is *finite* and abelian. In the function
field case, the ideal class group is also finitely generated and
abelian, but need not be finite and in that case you can get direct
summands that are *infinite* cyclic groups. (Also, it is incorrect to
say "the" direct product here; *a* direct product is more accurate).


>
> Then, for each cyclic component of ideal class group, is it possible
> to choose a prime ideal representing the ideal class of generaor of
> the cyclic component?

I don't have a counterexample at hand, but I don't believe this will
be the case always. You can always find a set of prime ideals whose
classes generate the ideal class group, and of course you can always
find a basis in which a generator of the largest cyclic factor will be
one of these distinguished generators, but in general a set of
generators for a finitely generated abelian group need not contain a
basis; e.g., if you had an ideal class group isomorphic to Z_2 x Z_4,
generated by two prime ideals of order 4 (you can certainly find a
generating set for the group made up of two elements of order 4).
That said, as I mentioned I don't have a counterexample, so maybe this
can finessed somehow.

--
Arturo Magidin
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