5
$\begingroup$

Here the conductor $N$ of an abelian extension $\Bbb Q \subset K$ is the smallest positive integer $N$ such that $K \subset \Bbb Q(\zeta_N)$.

Thanks to class field theory there is an equivalence between abelian extensions whose conductor divdes $N$ and subgroups of $(\Bbb Z / N\Bbb Z)^*$, and we have a natural isomorphism between the Galois group and the quotient of $(\Bbb Z / N\Bbb Z)^*$ by that subgroup.

It's fairly easy to check that the discriminant of quadratic extensions of $\Bbb Q$ coincide with the conductors of those extensions, because we know that if $N$ is an odd squarefree number congruent to $1$ mod $4$, the corresponding rings of integers are the $\Bbb Z[(1+\sqrt N)/2]$ (for the extension of conductor $|N|$) and $\Bbb Z[\sqrt {-N}]$ (for the extension of conductor $4|N|$).

When investigating for Something strange about $\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ and its friends I noticed that for every abelian cubic extension I looked at, its discriminant coincided with the square of its conductor (and we have a nice family of cubic polynomials with apparently the smallest possible discriminant for those extensions).

The situation for abelian cubic extensions doesn't seem as nice. We don't have (as far as I know) a nice complete description of the ring of integers for the extensions of conductor $N$ or $9N$.

The only thing I have is that if $H$ is a subgroup of index $3$ and $9$ doesn't divide $N$, then the family $(a,b,c) = (\sum_{g \in H} \zeta^{kg} ; k \in (\Bbb Z/N\Bbb Z)^*/H)$ gives a normal integral basis, but then we have to check that $(3abc-a^3-b^3-c^3)^2 = N^2$.

When $9$ divides $N$ we have to replace one of the elements with $1$ to get an integral basis so we have to check instead that $(ab+bc+ac-a^2-b^2-c^2)^2=N^2$ (which is also valid in the previous case, I think it's just what you get when you divide the other expression by $(a+b+c)^2$)

I have checked all possible index $3$ subgroups for various values of $N$ and so far the equality $(ab+bc+ac-a^2-b^2-c^2)^2=N^2$ has always been true. What gives ?

Letting $e_2 = ab+bc+ac$, since $a+b+c = \mu(N)$, we get $e_2 = \frac 13(\pm N+\mu(N)^2)$

As a side note, $abc$ seems to also have some nice properties regarding its factors, though it doesn't say anything about the discriminant of the extension.

$\endgroup$
7
$\begingroup$

The discriminant of an abelian extension $L/K$ is computed by the conductor-discriminant formula.

So if $K \subset \mathbb Q(\zeta_N)$ is a cubic extension of $\mathbb Q$ of conductor $N$, then $K$ is cut out by a subgroup $H$ of index $3$ in $(\mathbb Z/N)^{\times}$, i.e. Gal$(K/\mathbb Q) = (\mathbb Z/N)^{\times}/H$.

This quotient group has two non-trivial characters, each of conductor $N$ (since $N$ was chosen minimally), and one trivial character, whose conductor is $1$. Hence the discriminant of $K$ is equal to $N^2$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.