0

I'm trying to find a bijection between the quadratic subfields of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$, where $\zeta_n$ denotes an $n$-th primitive root of unity and the set of Kronecker symbols of conductor $d$ such that $d|n$.

By now, I've noticed that this bijection must have something to do with the cyclotomic characters

$$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\stackrel{\chi}{\longleftrightarrow} (\mathbb{Z}/n\mathbb{Z})$$

GSF
  • 924
  • 5
  • 15

1 Answers1

1

Not sure this is what you want, but there is what I would show :

  • A quadratic field is of the form $\mathbb{Q}[x]/( ax^2+bx+c )$ where $a,b,c \in \mathbb{Z}$ and $ax^2+bx+c$ is irreducible over $\mathbb{Q}$.

    $\Delta = b^2-4a, \alpha = \frac{-b+\sqrt{\Delta}}{2a},\beta = \frac{-b-\sqrt{\Delta}}{2a}$ and $ax^2+bx+c = a(x-\alpha)(x-\beta)$, so that $ax^2+bx+c$ irreducible iff $\alpha \not \in \mathbb{Q}$ i.e. $\Delta$ not a square.

    Clearly $\mathbb{Q}[x]/( \ x^2+ax+b \ ) \simeq \mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt{\Delta})$. Since $\Delta \in \mathbb{Z}$ you get that a quadratic field is of the form $ \mathbb{Q}(\sqrt{\pm d})$ where $d$ is squarefree.

    And (forgetting about the $\pm$) those are clearly in bijection with the real pritimive characters, since a real character modulo $d$ is of the form $\chi(n) = \left(\frac{n}{d}\right)$ and is primitive (not of the form $\chi'(n) 1_{gcd(n,d/d') = 1}$ for some other character $\chi'$ modulo $d' | d$) iff $d$ is squarefree.

  • Finally, one of $\sqrt{p},\sqrt{-p} \in \mathbb{Q}(\zeta_p)$ because for a prime $p$ and a Dirichlet character modulo $p$ : $X(k) = \sum_{n=0}^{p-1} \chi(n) e^{2i \pi nk/p} = \sum_{m=0}^{p-1} \chi(m/k) e^{2i \pi m/p} = \sum_{m=0}^{p-1} \overline{\chi(k)} \chi(m) e^{2i \pi m/p} = \overline{\chi(k)}X(1)$. Now $X(k)$ is the discrete Fourier transform of $\chi(n)$, so that $p \ |X(1)|^2 = \sum_{k=0}^{p-1} |X(k)|^2 = p \sum_{n=0}^{p-1} |\chi(n)|^2 = p^2$ i.e. $|X(1)| = \sqrt{p}$. And if $\chi(n) = \left(\frac{n}{p}\right)$ then $X(1)$ is purely real of purely imaginary (*), it follows that $X(1) = \pm \sqrt{\pm p}$, where $X(1) \in \mathbb{Q}(\zeta_p)$. Hence one of $\sqrt{p},\sqrt{-p} \in \mathbb{Q}(\zeta_n)$ whenever $p | n$.

    (*) $\left(\frac{p-n}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{n}{p}\right)$ so that $2 X(1) = \sum_{n=0}^{p-1}\left(\frac{n}{p}\right)e^{2i \pi n / p}+\left(\frac{p-n}{p}\right)e^{2i \pi (n-p) / p} = \sum_{n=0}^{p-1}\left(\frac{n}{p}\right)(e^{2i \pi n / p}+\left(\frac{-1}{p}\right)e^{-2i \pi n / p})$ that is purely real or purely imaginary depending on $\left(\frac{-1}{p}\right) =\pm 1$ i.e. $p \equiv 1$ or $3 \bmod 4$.

  • conclude by showing $\sqrt{\pm p} \not \in \mathbb{Q}(\zeta_n)$ whenever $gcd(p,n) = 1$, not sure how

reuns
  • 79,880