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

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

• A closely related other thread. It offers a different route with Gauss sums used in place of items (b) and (c). Nov 17 '19 at 19:31

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.
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 $$L_2 = \mathbb{Q}(\sqrt{-n})$$.