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rupertmccallum@xxxxxxxxx wrote:
> jstevh@xxxxxxx wrote:
> > Turns out you can use some numbers that are special in that they are
> > coprime to each other with respect to some prime:
> >
> > 5 + (sqrt(37) - sqrt(13))/2 and (sqrt(37) - sqrt(13))/2 are coprime to
> > each other with respect to 3 which can be seen by subtracting one from
> > the other, but each has its own factors in common with 3.
> >
If d_1 and d_2 are squarefree integers, not equal to 1, and both
congruent to 1 modulo 4, then the ring of integers in
Q(sqrt(d_1),sqrt(d_2)) is Z[(1+sqrt(d_1))/2, (1+sqrt(d_2))/2] and the
discriminant is (d_1)^2.(d_2)^2. This is a special case of a more
general result of which I will post a proof later on.
Given an algebraic integer z which generates a number field as a field,
the discriminant of the minimum polynomial will be k^2 times the
discriminant of the field for some integer k. If k is not divisible by
a prime p, then to obtain the prime factorization of the ideal
generated by p in the ring of integers in the field, take the
factorization of the minimum polynomial for z into irreducibles over
the field Z_p. To each irreducible factor of degree f will correspond a
prime ideal in the factorization of norm p^f, and the exponent e to
which it occurs will equal the exponent to which the ideal occurs in
the factorization. This theorem is proved by Dedekind in "Theory of
Algebraic Integers". I was hoping to use the theorem to obtain an
explicit factorization of the ideal generated by 3 in
Z[(1+sqrt(13))/2,(1+sqrt(37))/2], so that I could get explicit prime
factorizations for the ideals generated by the numbers you were talking
about, and thereby test whether your claims were true or false. If they
turned out to be false, then hopefully knowing what the prime
factorizations were would enable you to see where your argument went
wrong.
However, frustratingly it turns out you can't use this theorem to get
the prime factorization of the ideal generated by 3. If
z=a((1+sqrt(13))/2)+b((1+sqrt(37))/2)+c((1+sqrt(13))/2)((1+sqrt(37))/2),
then the discriminant of the minimum polynomial of z is
13^2.37^2.(a^2+ac-9c^2)^2.(b^2+bc-9c^2)^2.(13a^2-37b^2)^2, and
unfortunately this is always divisible by 3. Dedekind said there were
cases like this where you couldn't use the theorem. This is the first
one I have seen. So now I will have to find out how you obtain prime
factorizations in a case like that. I'll get back to you.
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