$\mathbb{Q}(\sqrt{n})$ is Contained in the Cyclotomic Field of the $4n$'th Primitive Root of Unity.

Question

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 $\mathbb{Q}(\zeta')$.

Answer 1

Do the following steps. Let $K=\Bbb{Q}(\zeta_{4n})$.

• Let $p$ be an odd prime factor of $n$. It follows that $\zeta_p$ is a power of $\zeta_{4n}$. Therefore $\Bbb{Q}(\zeta_p)\subseteq K$.
• By item (b) we know that the discriminant of $\Bbb{Q}(\zeta_p)$ is $\pm p^p$. By item (c) this implies $\Bbb{Q}(\zeta_p)$ contains $\sqrt{\pm p}$ for an appropriate choice of sign. Therefore $K$ contains either $\sqrt{p}$ or $\sqrt{-p}$.
• Because $4\mid 4n$, $i$ is a power of $\zeta_{4n}$. Therefore $i\in K$. Therefore $\sqrt{p}\in K$ for every odd prime factor $p$ of $n$.
• If $n$ is odd, we are done.
• If $2\mid n$, then $8\mid 4n$. Therefore $(1+i)/\sqrt2=\zeta_8\in K$. Consequently $\sqrt2 \in K$. Therefore $\sqrt n\in K$ also when $n$ is even.

Answer 2

Let $K = \mathbb{Q}(\zeta_{4n})$ be the $4n$-th cyclotomic field with $n$ odd. Each prime $p$ that splits in $K$ is congruent to $1\pmod {4n}$. You are asking to prove the field $L = \mathbb{Q}(\sqrt{n})$ is a subfield of $K$. Since $n$ is assumed positive, $L$ is a real field (not imaginary). Each prime $p$ (completely) splitting is congruent to $1\pmod n$ if $n = 1\pmod 4$ and $1\pmod {4n}$ if $n = 3\pmod 4$ (concerning those $p=1\pmod n$ only). Thus, we have proved the case with $n= 3\pmod 4$.

When $n = 1\pmod 4$, only primes $p=1\pmod 4$ will split in both $L$ and $L_2 = \mathbb{Q}(\sqrt{-n})$.