Let p be a prime, $\xi_p \in \mathbb{C}$ a primitive p-th unit root and $K = \mathbb{Q}(\xi_p)$.

Give a normal basis for $K/\mathbb{Q}$.

I know, that a basis of $L/K$ (finite and galois) is called a normal basis if it has the form $\{\sigma(a)\}_{\sigma \in \text{Gal}(L/K)}$ for an $a \in L$. A procedure to determine such an element a if $|K| = \infty$ is the following:

  1. Determine the minimal polynomial $m_{\beta}(X)$ for a primitive Element $\beta$ ($L = K(\beta)$)
  2. Set $g_{\sigma}(X) := \prod\limits_{\substack{\tau \in G\\\tau \neq \sigma}} (X - \tau(\beta))$ and $A(X) := (g_{\tau^{-1}\sigma})_{\tau,\sigma \in G}$ with $G := \text{Gal}(L/K)$
  3. Search a $\gamma \in K$ with $det(A(\gamma)) \neq 0$
  4. Set $a := \frac{m_{\beta}(\gamma)}{\gamma - \beta}$

So far so good. I have only been able to achieve that $\Phi_p(X) = \frac{X^p - 1}{X-1} = X^{p-1} + . . . + X + 1$ is the minimal polynomial of $\xi_p$ over $\mathbb{Q}$ since p is prime. The elements of the Galois group $G:=\text{Gal}(K/\mathbb{Q})$ should affect the primitive root of unit $\beta$ like potencies. That means it exists an isomorphism \begin{align} k : G \longrightarrow (\mathbb{Z}/n\mathbb{Z})^x\\ \sigma \mapsto \kappa(\sigma) + n\mathbb{Z} \end{align} so that $\sigma(w) = w^{\kappa(\sigma)}$. [$(K)^x$ means the group of invertible units in K]

Now I'm unsure how to proceed. Thanks for help.

  • $\begingroup$ Do you know a Q-basis of K? $\endgroup$ – eduard Jan 17 at 10:58
  • $\begingroup$ Can you tell what is $\;\zeta_p^k\;$ , for $\;k\in\Bbb N\;$ ? Do you know what the effect of any automorphism of $\;K/\Bbb Q\;$ is over $\;\zeta_p\;$ ? Well, there you go... $\endgroup$ – DonAntonio Jan 17 at 11:17
  • $\begingroup$ @DonAntonio For all $1 \leq k \leq p-1$ is $\zeta_p^k \neq 1$ since $\zeta_p$ is primitive. And the effect of an automorphism over $\zeta_p$ should be as I already wrote: If $\sigma$ is an Automorphism of $K/\mathbb{Q}$, so $\sigma(\zeta_p) = \zeta_p^{k(\sigma)}$ with $k(\sigma) \in (\mathbb{Z}/p\mathbb{Z})^x$ (means: $1 \leq k(\sigma) \leq p-1$) $\endgroup$ – Zorro_C Jan 17 at 14:30
  • $\begingroup$ @eduard I think $\{1, \xi_p, \xi_p^2, ..., \xi_p^{p-1}\}$ is a Basis of K? I'm right? $\endgroup$ – Zorro_C Jan 17 at 14:30
  • 1
    $\begingroup$ Yeah, I just noticed as well. Think it must be $\{\xi_p, ..., \xi_p^{p-1}\}$? And that could be even the normal base, right? $\endgroup$ – Zorro_C Jan 17 at 14:46

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