Showing that the field generated by elements $e^{2\pi i/n} (n=1,2,...)$ is algebraic over $\mathbb{Q}$

Question

Let $K$ be the field generated by the elements $e^{2\pi i/n} (n=1,2,...)$. Show that $K$ is an algebraic extension of $\mathbb{Q}$, but that $[K:\mathbb{Q}]$ is not finite.

Suggestion: it may help to show that the minimum polynomial of $e^{2\pi i/p}$ for $p$ prime is $x^{p-1}+x^{p-2}+...+1$.

I have figured out that any term $\alpha\in K$ must be a linear combination of terms $c_ke^{2\pi i\frac{m_k}{n_k}}$ where $c_k\in\mathbb{Q}$ and $m_k,n_k\in\mathbb{Z}$. And that any term where $n_k$ is composite can be broken up into primes. And that for one term $e^{2\pi i/p}$ we can find a polynomial in $\mathbb{Q}$ where it is a root. But I can't see how to do it when you have linear combination of terms with different primes and different coefficients. Any help on this?

As far as non-finitness, I am guessing it comes from there been infinite primes, and therefore infinitely many linearly independent elements $e^{2\pi i/p}$.

Answer 1

$$\forall\,n\in\Bbb N\;,\;\;e^{2\pi i/n}\;\;\text{is algebraic, and since an extension by algebraic elements is algebraic...}$$

On the other hand, for any prime $\;p\;$, the minimal polynomial of $\;e^{2\pi i/p}\;$ over the rationals is $\;x^{p-1}+x^{p-2}+\ldots+x+1\;$ , so the extension must be at least of order $\;p\;$ for any prime $\;p\;$ ...

Answer 2

Each $e^{2\pi i/n}$ is algebraic as it is a root of $X^n-1$ (though this need not be the minimal polynomial). The general element of $K$ is alrady contained in some $\mathbb Q[e^{2\pi i/2},e^{2\pi i/3},\ldots ,e^{2\pi i/n}]$ and thus in $\mathbb Q[e^{2\pi i/n!}]$, i.e. each element of $K$ is algebraic.

Assume $d=[K:\mathbb Q]$ were finite. then$[\mathbb Q[e^{2\pi i/p}:\mathbb Q]=p-1>d$ for sufficiently big prime $p$, contardiction!