# Determine a Normal Basis for Galois Extension of $\mathbb{Q}$ with primitive pth root of unit (p prime)

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.

• Do you know a Q-basis of K? – eduard Jan 17 at 10:58
• 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... – DonAntonio Jan 17 at 11:17
• @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: $1 \leq k(\sigma) \leq p-1$) – Zorro_C Jan 17 at 14:30
• @eduard I think $\{1, \xi_p, \xi_p^2, ..., \xi_p^{p-1}\}$ is a Basis of K? I'm right? – Zorro_C Jan 17 at 14:30
• 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? – Zorro_C Jan 17 at 14:46