# Finding the Normal Basis of Cyclotomic field

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?

## 1 Answer

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.

• 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. – matt Jan 17 at 23:48
• 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 $a_i\in\Bbb Q$. – Lord Shark the Unknown Jan 18 at 4:49
• 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 $\Phi_p(X)$. – Jyrki Lahtonen Jan 18 at 21:52