2
$\begingroup$

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?
$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.