$\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^n \notin \Bbb{Q}(ζ_{11})$ for all positive integer $n$

Question

I want to prove $\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^n$ does not lie in $\Bbb{Q}(ζ_{11})$ for all positive integer $n$.

This problem arises from arithmetic geometry, but this problem itself is purely a field theoretic one.

For $n$ is even, this is equivalent to $\sqrt{5}\notin \Bbb{Q}(ζ_{11})$. But if $\sqrt{5}\in \Bbb{Q}(ζ_{11})$, prime ideal above $5$ will not be unratified, but from class field theory, prime ideal above $5$ does not ramify, this is a contradiction.

But I'm sticking with the case that $n$ is odd.

Answer 1

If $\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^n\in \Bbb{Q}(\zeta_{11})$ then $\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^{2n}=(-\frac{21 \ + \ 5\sqrt{5}}{2})^n \in \Bbb{Q}(\zeta_{11})$.

$Gal(\Bbb{Q}(\zeta_{11})/\Bbb{Q})$ is cyclic so $\Bbb{Q}(\zeta_{11})$ has a unique quadratic subfield, which is $\Bbb{Q}(\sqrt{-11})$.

So we get that $(21 \ + \ 5\sqrt{5})^n\in \Bbb{Q}(\zeta_{11})\cap \Bbb{Q}(\sqrt5)=\Bbb{Q}$.

Take the non-trivial automorphism $\sigma\in Gal(\Bbb{Q}(\sqrt5)/\Bbb{Q})$ such that $ \sigma(21 \ + \ 5\sqrt{5})=21 \ - \ 5\sqrt{5}$.

If $(21 + 5\sqrt{5})^n\in \Bbb{Q}$ then $(21 + 5\sqrt{5})^n=\sigma((21 + 5\sqrt{5})^n)=(21 - 5\sqrt{5})^n$ which is absurd as $|21 - 5\sqrt{5}|\ne |21 + 5\sqrt{5}|$.

Answer 2

Suppose for some $n$, $\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^n$ lies in $\Bbb{Q}(ζ_{11})$. Then $\left(i\sqrt {\frac{21 \ + \ 5\sqrt{5}}{2}}\right)^{2n}=\left( {\frac{333 \ + \ 105\sqrt{5}}{2}}\right)^{n}\in \Bbb{Q}(ζ_{11})$. So if we write $\left( {\frac{333 \ + \ 105\sqrt{5}}{2}}\right)^{n}=a+b \sqrt{5}$ for some $a,b \in \Bbb{Q}$, the last condition is equivalent to $\sqrt{5}\in \Bbb{Q}(ζ_{11})$.

Let's consider the prime ideal decomposition of $\sqrt{5}\Bbb{Z}[ζ_{11}]$ in $\Bbb{Q}(ζ_{11})$.

$\sqrt{5}\Bbb{Z}[ζ_{11}]=P_1^eP_2^e\cdots P_n^e$ ($P_1,P_2,...,P_n$ are prime ideals of $\Bbb{Z}[ζ_{11}]$ and $e$ is the ramification index), so $(5)= P_1^{2e}P_2^{2e}\cdots P_n^{2e}$. In particular, $(5)$ is not unramified.

On the other hand, from well known ramification property of cyclotomic fields, $(5)$ is unramified. This is a contradiction.