# Are the discriminant of abelian cubic extensions of $\Bbb Q$ equal to the square of their conductor?

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.

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 $N^2$.