I know following theorem (and its proof):

Let $K\subset L \subset M$ be number fields, $[L:K] = n, [M:L]=m$, and let $\{\alpha_1,\ldots,\alpha_n\}$ and $\{\beta_1,\ldots,\beta_m\}$ be bases for $L$ over $K$ and $M$ over $L$, respectively. Then we have $$\text{disc}_K^M(\alpha_1\beta_1,\ldots, \alpha_n\beta_m) = \left(\text{disc}_K^L\left(\alpha_1,\ldots,\alpha_n\right)\right)^m N_K^L\left(\text{disc}_L^M\left(\beta_1,\ldots, \beta_m\right)\right)$$ where, for example, $\text{disc}_{K}^L(\alpha_1, \ldots, \alpha_n) = \det(\sigma_i(\alpha_j))^2=\det(T_{K}^L(\alpha_i\alpha_j))$ where the embeddings $\sigma_i$ of $L$ in $\mathbb{C}$ fix $K$ point-wise; $N_K^L$ is the norm function defined for a number field $L$ which fixes $K$.

Using this I have to prove

Let $p$ be an odd prime number, then $\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}(\xi_p) = p^{(p-3)/{2}}$, where $\xi_p = \zeta_p +\zeta_p^{-1}$ and $\zeta_p = e^{2\pi i /p}$.

I have observed that

  1. $\mathbb{Q}\subset \mathbb{Q}[\xi_p] \subset \mathbb{Q}[\zeta_p]$
  2. $[\mathbb{Q}[\xi_p]:\mathbb{Q}] = \frac{\varphi(p)}{2}$
  3. $[\mathbb{Q}[\zeta_p]:\mathbb{Q}[\xi_p]]=2$
  4. $\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}(\zeta_p) = \pm p^{p-2}$, where + sign holds iff $p\equiv 1,2 \pmod 4$.
  5. $\{1,\zeta_p,\xi_p,\xi_p\zeta_p, \ldots, \xi_p^{n-1}, \xi_p^{n-1}\zeta_p\}$ is an integral basis of $\mathbb{Q}[\zeta_p]$.

[Source of question: ex. 35, Ch 2, "Number Fields" by D. A. Marcus.]

  • $\begingroup$ Do you know integral bases for these extensions? $\endgroup$ Jul 19, 2016 at 9:10
  • $\begingroup$ @JyrkiLahtonen I know that the integral basis for $\mathbb{Q}[\xi_p]/\mathbb{Q}$ is $\{1,\xi_m, \ldots, \xi_m^{n-1}\}$ where $n=(p-1)/2$. But I'm not sure about the basis of $\mathbb{Q}[\zeta_p]/\mathbb{Q}[\xi_p]$, I think it's $\{1,\sqrt{-1}\}$. $\endgroup$ Jul 19, 2016 at 9:43
  • $\begingroup$ I would think that $\{1,\zeta_p\}$ is more likely. Why should $\sqrt{-1}$ even be an element of $\Bbb{Q}(\zeta_p)$? $\endgroup$ Jul 19, 2016 at 9:45
  • $\begingroup$ @JyrkiLahtonen Yes, you are right. So, I can conclude that: $$\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}(1,\zeta_p,\xi_p,\xi_p\zeta_p, \ldots, \xi_p^{n-1}, \xi_p^{n-1}\zeta_p) = \left(\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}\left(1,\xi_p,\ldots,\xi_p^{n-1}\right)\right)^2 N_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}\left(\text{disc}_{\mathbb{Q}[\xi_p]}^{\mathbb{Q}[\zeta_p]}\left(1,\zeta_p\right)\right)$$ $\endgroup$ Jul 19, 2016 at 9:50
  • $\begingroup$ @JyrkiLahtonen Also in related problem I proved that: If $n=\frac{\varphi(m)}{2}$, then $\{\xi_m^i\zeta_m^j : i = 0,1,\ldots, n-1; j = 0,1\}$ is an integral basis for $\mathbb{Z}[\zeta_m]$ $\endgroup$ Jul 19, 2016 at 9:52

1 Answer 1


Based on hints by Jyrki Lahtonen, here is the answer:

We know that, the integral basis for $\mathbb{Q}[\xi_p]/\mathbb{Q}$ is $\{1,\xi_p, \ldots, \xi_p^{n-1}\}$ and the integral basis of $\mathbb{Q}[\zeta_p]/\mathbb{Q}[\xi_p]$ is $\{1,\zeta_p\}$.

We will use above result to calculate discriminant: $$\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}(1,\zeta_p,\xi_p,\xi_p \zeta_p, \ldots, \xi_p^{n-1}, \xi_p^{n-1}\zeta_p) = \left(\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}\left(1,\xi_p,\ldots,\xi_p^{n-1}\right)\right)^2 N_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}\left(\text{disc}_{\mathbb{Q}[\xi_p]}^{\mathbb{Q}[\zeta_p]}\left(1,\zeta_p\right)\right)$$

Keeping in mind invariance of discriminant of a ring of integers and using the observations stated in question we get (note that the $\pm$ signs cancel out on both sides)

$$\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}(\zeta_p) = \left(\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}(\xi_p)\right)^2 N_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}\left(N_{\mathbb{Q}[\xi_p]}^{\mathbb{Q}[\zeta_p]}\left(f'(\zeta_p)\right)\right)$$ where $f(x) = x^2-\xi_p x+1$, is the minimal polynomial for $\mathbb{Q}[\zeta_p]$ over $\mathbb{Q}[\xi_p]$. Using transitivity property of norm we can re-write it as

$$\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}(\zeta_p) = \left(\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}(\xi_p)\right)^2 N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(f'(\zeta_p)\right)$$

Note that $\xi_p = \zeta_p + \zeta_p^{-1}$, therefore

$$N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(f'(\zeta_p)\right) = N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(2\zeta_p-\xi_p\right)= N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(\zeta_p-\zeta_p^{-1}\right)=\frac{N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(\zeta_p-1\right)N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(\zeta_p+1\right)}{N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_p]}\left(\zeta_p\right)}$$

Observe that $\Phi_p(x-1)$ is the minimal polynomial for $\zeta_p+1$ where $\Phi_p(x) = 1+x+\ldots+x^{p-1}$, therefore $N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_{p}]} (1+\zeta_p)$ is equal to constant term in $\Phi_p(x-1)= 1+(x-1)+\ldots+(x-1)^{p-1}$, since $p-1$ is even we will have $\frac{p-1}{2}$ times $-1$ and $\frac{p-1}{2}+1$ times $+1$, thus leaving $+1$ as constant term. Hence $N_{\mathbb{Q}}^{\mathbb{Q}[\zeta_{p}]} (1+\zeta_p) = 1$. We can use similar argument to calculate $N(1-\zeta_p)=p$. Using these we get (note that we already cancelled out sign arising due to $p$)

$$\text{disc}_{\mathbb{Q}}^{\mathbb{Q}[\xi_p]}(\xi_p) = \text{disc}(\xi_p) = \pm \sqrt{\frac{p^{p-2}}{p}} = \pm p^{\frac{p-3}{2}}$$

But + sign must hold since, $\mathbb{Q}[\xi_p]$ contains $\sqrt{\text{disc}(\xi_p)}$ (algebraic closure property of number field).


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