So let $p$ be a prime number and $\zeta_p$ the p-th roots of unity. I want to proof that $ B = \{ \zeta_p, \zeta_{p}^{2}, \dots, \zeta_{p}^{p-1} \}$ is the normal basis of $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ (which I hope is true ...).

First of all, I know that $B$ is a $\mathbb{Q}$-basis of $\mathbb{Q}(\zeta_p)$ since

  1. the elements of $B$ are $\mathbb{Q}$-linearly independent, because $\zeta_p$ is a $p$-th primitive roots of unity,
  2. and $B$ has exactly $p-1$ elements and the degree of $\mathbb{Q}(\zeta_p)$ over $\mathbb{Q}$ is $p-1$ too, because the Galois group of $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^\times$ which has $p-1$ elements.

Now, I have trouble proving that $B$ is indeed a normal basis. According to the definition of a normal basis, if we let $\sigma_i \in \text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$, I would have to find an $a \in \mathbb{Q}(\zeta_p)$ such that

$$ \{\sigma_1(a),\sigma_2(a), \dots, \sigma_{p-1}(a)\} $$

forms a $\mathbb{Q}$-basis of $\mathbb{Q}(\zeta_p)$.

Since $\text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ is cyclic, there is a generating automorphism which I will just call $\sigma$. Then, just reformulating the definition, we would have

$$ \{\sigma(a),\sigma^{2}(a), \dots, \sigma^{p-1}(a)\} \text{.} $$

Now, I have troubles determining this element $a$.

Right now, I believe that $a = \zeta_p$ in which case, we would have

$$\{\sigma(\zeta_p),\sigma^{2}(\zeta_p), \dots, \sigma^{p-1}(\zeta_p)\}\text{.}$$

So, my questions are

  1. Am I right with my assumption to set $a = \zeta_p$?
  2. And if so, how do I proceed with my proof in order to show that $ B = \{ \zeta_p, \zeta_{p}^{2}, \dots, \zeta_{p}^{p-1} \}$ is a normal basis?

The $\zeta_p^j$ ($1\le j\le p-1$) are the zeros of the $p$-th cyclotomic polynomial $\Phi_p(X)=X^{p-1}+X^{p-2}+\cdots+X+1$, which is well-known to be irreducible over $\Bbb Q$. Thus the Galois group of $\Bbb Q(\zeta_p)$ acts transitively on the zeros of $\Phi_p(X)$. Thus there is a Galois group element $\sigma_j$ with $\sigma_j(\zeta_p)=\zeta_p^j$. This is unique: its action on $\zeta_p$ determines its action on all of $\Bbb Q(\zeta_p)$. So $B=\{\sigma_1(\zeta_p),\sigma_2(\zeta_p),\cdots,\sigma_{p-1}(\zeta_p)\}$ really is a normal basis.

  • $\begingroup$ Thank you for your answer! However, I have some troubles following you. When you say the Galois group acts transitively, you mean that $\sigma(\zeta_p)$ is also a root of $\Phi_p(X)$, right? And, if possible, could you elaborate on the part where you say its action on $\zeta_p$ determines its action on all of $\mathbb{Q}(\zeta_p)$? Thank you. $\endgroup$ – matt Jan 17 at 23:48
  • $\begingroup$ I mean that $\sigma(\zeta)$ is also a zero of $\Phi_p$, and all such zeroes arise. If $\sigma$ takes $\zeta$ to $\zeta'$, then it takes $a_0+a_1\zeta+a_2\zeta^2+\cdots$ to $a_0+a_1\zeta'+a_2\zeta'^2+\cdots$ where the $a_i\in\Bbb Q$. $\endgroup$ – Lord Shark the Unknown Jan 18 at 4:49
  • $\begingroup$ Did you guys deal with the linear independence of this set of conjugates? Anyway, if it were linearly dependent this would imply the existence of a polynomial (i) of degree $\le p-1$, (ii) with rational coefficients, (iii) zero constant term, and (iv) $\zeta_p$ as a zero, contradicting irreducibility of $\Phi_p(X)$. $\endgroup$ – Jyrki Lahtonen Jan 18 at 21:52

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