Basis of primitive nth roots of unity when n is squarefree?

• Dec 1st 2011, 01:34 PM
AshleyLin
Basis of primitive nth roots of unity when n is squarefree?
I found an unanswered question on Math.SE concerning a footnote in an expository paper by Keith Conrad. It states that in general, the primitive $\displaystyle n$th roots of unity in the $\displaystyle n$th cyclotomic field form a normal basis over $\displaystyle \mathbf{Q}$ if and only if $\displaystyle n$ is squarefree.

The forward direction is not difficult to show. However, if $\displaystyle n$ is squarefree, then how can one show that the primitive $\displaystyle n$th roots of unity form a basis for the cyclotomic extension over $\displaystyle \mathbb{Q}$?
• Dec 2nd 2011, 12:37 AM
NonCommAlg
Re: Basis of primitive nth roots of unity when n is squarefree?
Quote:

Originally Posted by AshleyLin
I found an unanswered question on Math.SE concerning a footnote in an expository paper by Keith Conrad. It states that in general, the primitive $\displaystyle n$th roots of unity in the $\displaystyle n$th cyclotomic field form a normal basis over $\displaystyle \mathbf{Q}$ if and only if $\displaystyle n$ is squarefree.

The forward direction is not difficult to show. However, if $\displaystyle n$ is squarefree, then how can one show that the primitive $\displaystyle n$th roots of unity form a basis for the cyclotomic extension over $\displaystyle \mathbb{Q}$?

do it by induction over $\displaystyle n$. first assume that $\displaystyle n$ is a prime and solve the problem. then write $\displaystyle n = kp,$ where $\displaystyle p$ is a prime and $\displaystyle k < n$ is an integer coprime with $\displaystyle p,$ and apply your induction hypothesis.
• Dec 2nd 2011, 08:23 AM
AshleyLin
Re: Basis of primitive nth roots of unity when n is squarefree?
Thanks NonCommAlg. I understand how it follows when $\displaystyle n$ is prime, since the powers of the adjoined element in a simple extension form a basis. But when $\displaystyle n=kp$, $\displaystyle k$ is squarefree, so by the induction hypothesis, the primitive $\displaystyle k$th roots of unity form a basis for $\displaystyle \mathbb{Q}(\zeta_k)/\mathbb{Q}$. By the result of simple extensions, the primitive $\displaystyle p$th roots of unity form a $\displaystyle \mathbb{Q}(\zeta_k)$ basis of $\displaystyle \mathbb{Q}(\zeta_k,\zeta_p)$ over $\displaystyle \mathbb{Q}(\zeta_k)$, so letting the basis run over products of primitive roots of $\displaystyle \zeta_p$ and $\displaystyle \zeta_k$ gives a basis of primitive $\displaystyle n$th roots of $\displaystyle \mathbb{Q}(\zeta_k,\zeta_p)$ over $\displaystyle \mathbb{Q}$? Is that the correct way to go?