From Silverman and Tate, Rational Points on Elliptic Curves. Exercise 6.1
Let $\zeta'$ be the $4n$'th primitive root of unity. Use (c) to prove that $\mathbb{Q}(\sqrt{n})$ is contained in the cyclotomic field of the $4n$'th primitive root of unity.
From previous exercises I know that
(b) If $f(X) = X^n-1$. Then $ \text{Disc}(f) =(-1)^{(n-1)(n-2)/2}n^n.$
(c) Let $\zeta$ be a primitive $n$'th root of unit. Then the cyclotomic field $\mathbb{Q}(\zeta)$ contains $\sqrt{\text{Disc}(f)}.$
We need to show that $\sqrt{n} \in \mathbb{Q}(\zeta')$.
Since $\zeta'$ is a $4n$'th primitive root of unity, it is a solution to $X^{4n} = 1$. (so it is a root of $f(X) = X^{4n} -1$).
By (b), $ \text{Disc}(f) =(-1)^{(4n-1)(4n-2)/2}(4n)^{(4n)}.$
By (c), $\sqrt{\text{Disc}} \in \mathbb{Q}(\zeta)$. Since fields are closed under inverses and multiplication by integers, then $(n)^{(4n)}\in \mathbb{Q}(\zeta)$.
I don't see how to get from $\mathbb{Q}(\zeta)$ to $\mathbb{Q}(\zeta')$.
Best Answer
Do the following steps. Let $K=\Bbb{Q}(\zeta_{4n})$.