Let $K$ be an Abelian cubic number field of conductor $n$. I want to prove that $\Delta_K=n^2$.

An answer is given in Are the discriminant of abelian cubic extensions of $\Bbb Q$ equal to the square of their conductor?, using the conductor-discriminant formula. However I'm interested in a more direct proof that only uses Kronecker-Weber and its relatively straightforward consequences (like a prime divides the conductor iff it ramifies, and it divides it twice iff it is wildly ramified).

This is how far I've gotten:

Write $K=\mathbb{Q}(\alpha)$ for some $\alpha\in\mathcal{O}_K$, and let $f\in\mathbb{Z}[X]$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Then $\Delta(f)=\Delta(\mathbb{Z}[\alpha])=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\cdot\Delta_K$. As $K/\mathbb{Q}$ is Galois, we know that $\Delta(f)$ is a square, which implies that $\mathrm{sign}(\Delta_K)=1$. Thus to prove that $\Delta_K=n^2$ it suffices to show that

\begin{equation} \quad\quad\quad\quad \quad\quad\quad\quad\quad\mathrm{ord}_p(\Delta_K)=2\cdot\mathrm{ord}_p(n)\quad\text{ for every prime }p.\quad\quad\quad\quad \quad\quad\quad\quad\quad(1) \end{equation}

This is clear for all primes $p$ that are unramified in $K$. If a prime $p$ ramifies in $K$ then it ramifies totally as $K/\mathbb{Q}$ is Galois of prime degree, say as $p\mathcal{O}_K=\mathfrak{p}^3$. If $p\neq 3$ then the ramification is tame, and hence $\mathrm{ord}_{\mathfrak{p}}(\mathfrak{D}_K)=2$, which yields $\mathrm{ord}_p(\Delta_K)=2$ as $\mathfrak{D}_K$ has ideal norm $\Delta_K$ and $\mathfrak{p}$ is the only prime of $K$ lying over $p$. Since also $\mathrm{ord}_p(n)=1$ we see that $(1)$ holds for all tamely ramified primes in $K$.

It remains to cover the case where $p=3$ ramifies in $K$, and this is were I'm stuck. In the special case where only $3$ ramifies in $K$, we must have that $n=3^k$ for some $k\geq 2$. As $\mathrm{Gal}(\mathbb{Q}(\zeta_{3^k})/\mathbb{Q})$ is cyclic and has order divisible by $3$ it contains a unique cubic subfield, which thus must be $K$. As $\mathbb{Q}(\zeta_9)$ already contains this cubic subfield we see that $k=2$, and that we must have $K=\mathbb{Q}(\zeta_9+\zeta_9^{-1})$. We have tame ramification of $\mathfrak{p}$ in $\mathbb{Q}(\zeta_9)$, say with factorisation $\mathfrak{p}\mathbb{Z}[\zeta_9]=\mathfrak{B}^2$. This implies that $\mathfrak{D}_{\mathbb{Q}(\zeta_9)/K}=\mathfrak{B}$, and if $\mathfrak{D}_{K/\mathbb{Q}}=\mathfrak{B}^m$ we have

$$ \mathfrak{B}^9=\mathfrak{D}_{\mathbb{Q}(\zeta_9)}=\mathfrak{D}_{\mathbb{Q}(\zeta_9)/K}\cdot(\mathfrak{D}_{K/\mathbb{Q}}\mathbb{Z}[\zeta_9])=\mathfrak{B}^{1+2m}, $$ which implies that $m=4$ and hence that $\Delta_K=3^4=81$. Thus we find $\Delta_K=n^2$ in this special case.

However I don't see how I can treat the case where for example $K$ ramifies at $3$ and some other prime $p$. Trying to construct such a $K$ out of the Abelian cubic field of discriminant $81$ and one of $p^2$ seems like a weird thing to do, in the quadratic case you can 'do this' by constructing $\mathbb{Q}(\sqrt{-5})$ out of $\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\zeta_5)$ and $\mathbb{Q}(i)$ using $\sqrt{-5}=i\sqrt{5}$, but in the cubic case we don't have such explicit generators and I don't see how to proceed.

Any help would be highly appreciated!

  • $\begingroup$ If $K$ is an abelian extension with discriminant $D = D_1D_2$, where $D_1$ and $D_2$ are coprime, then there are abelian extensions $K_1$ and $K_2$ with discriminants $D_1$, $D_2$, respectively, and $K$ is contained in the compositum $K_1K_2$ (only over the rationals; see Hilbert's proof of Kronecker-Weber and his technique of abelian crossing). $\endgroup$ – franz lemmermeyer Sep 10 '17 at 12:33

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