Prove that $\Bbb Q(ζ_n)∩\Bbb R = \Bbb Q(ζ_n + ζ^{−1}_{n})$

Question

Let $n ≥ 1$ be an integer. Let $ζ_n = e^{2πi/n}$ be a primitive $n$-th root of unity. I need to prove that $\Bbb Q(ζ_n)∩\Bbb R =\Bbb Q(ζ_n + ζ^{−1}_{n})$.

It is clear that by the Euler's formula, once we add $ζ_n + ζ^{−1}_{n}$ we can cancel the imaginary part and so it is clear that $\Bbb Q(ζ_n + ζ^{−1}_{n})\subseteq \Bbb Q(ζ_n)∩\Bbb R$. But I wonder how can I formally prove that the reverse of inclusion also holds. Could someone please help? Any hint would also be appreciate.

Answer 1

You know this is in all the textbooks on cyclotomic fields...

Let $\sigma$ be the complex conjugation operator. Then $\sigma$ acts on $\Bbb Q(\zeta)$ (where $\zeta=\zeta_n$) and $$\Bbb Q(\zeta)\cap\Bbb R=\{\xi\in\Bbb Q(\zeta):\sigma(\xi) =\xi\}=\{\eta+\sigma(\eta):\eta\in\Bbb Q(\zeta)\}.$$ The elements of $\Bbb Q(\zeta)$ are the $\Bbb Q$-spans of the $\zeta^k$. Therefore the elements of $\Bbb Q(\zeta)\cap\Bbb R$ are the $\Bbb Q$-spans of the $\zeta^k+\sigma(\zeta^k)=\zeta^k+\zeta^{-k}$. The numbers $a_k =\zeta^k+\zeta^{-k}$ satisfy the recurrence $a_{k+1}=(\zeta+\zeta^{-1})a_k-a_{k-1}$ and an easy induction gives $a_k\in\Bbb Z[\zeta+\zeta^{-1}]$.

Answer 2

For $n=1$, the claim is immediate.

Fix a positive integer $n > 1$.

\begin{align*} \text{Let}\;\;\omega &= \zeta_n\\[4pt] s &= \omega + \omega^{-1}\\[12pt] K &= \mathbb{Q}(\omega)\\[4pt] F &= K \cap \mathbb{R}\\[4pt] E &= \mathbb{Q}(s)\\[4pt] \end{align*}

We know that $s$ is real, and $\omega$ is non-real, hence we have the chain of fields

$$E \subseteq F \subset K$$

with $F$ a proper subfield of $K$.

Since $\omega$ and $\omega^{-1}$ have sum $s$ and product $1$, $\omega$ is a root of the quadratic polynomial $f = x^2 - sx + 1$, which has coefficients in $E$, hence $[K:E] \le 2$.

Since $F$ is a proper subfield of $K$, we must have $[K:E]>1$, hence $[K:E]=2$.

It follows that $[F:E]=1$, so $E=F$, as was to be shown.

Answer 3

$\mathbb{Q}(\zeta_n + \zeta_n^{-1}) \subset \mathbb{Q}(\zeta_n) \cap \mathbb{R}$, since $\zeta_n + \zeta_n^{-1} \in \mathbb{Q}(\zeta_n), \mathbb{R}$.

Furthermore, $\mathbb{Q}(\zeta_n) \cap \mathbb{R}$ is a field and so $\mathbb{Q}(\zeta_n + \zeta_n^{-1})$ is a subfield given by the containment above.

Note that $[\mathbb{Q}(\zeta_n): \mathbb{Q}] = \phi(n)$, where $\phi$ is the Euler Phi.

Using the tower law $$[\mathbb{Q}(\zeta_n): \mathbb{Q}] = [\mathbb{Q}(\zeta_n): \mathbb{Q}(\zeta_n + \zeta_n^{-1})][\mathbb{Q}(\zeta_n + \zeta_n^{-1}): \mathbb{Q}]$$

But $[\mathbb{Q}(\zeta_n): \mathbb{Q}(\zeta_n + \zeta_n^{-1})] = 2$ since $(\zeta_n)^2 + A(\zeta_n) + B = 0$ for $A = -(\zeta_n + \zeta_n^{-1})$, $B = 1$.

Let $k =[\mathbb{Q}(\zeta_n + \zeta_n^{-1}): \mathbb{Q}]$.

Using the tower law once again, invoking the subfield argument from before $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = [\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n) \cap \mathbb{R}] [\mathbb{Q}(\zeta_n) \cap \mathbb{R}: \mathbb{Q}(\zeta_n + \zeta_n^{-1})][\mathbb{Q}(\zeta_n + \zeta_n^{-1}): \mathbb{Q}]=:mnk$$

Combining the two tower laws gives us $$\phi(n) = 2 k = m n k$$

and so $mn = 2$, so either $m,n = 1$.

But $[\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n) \cap \mathbb{R}] \neq 1$ since $\mathbb{Q}(\zeta_n)\subset \mathbb{C}$ and $\mathbb{Q}(\zeta_n) \cap \mathbb{R} \subset \mathbb{R}$.

Therefore $$[\mathbb{Q}(\zeta_n) \cap \mathbb{R}: \mathbb{Q}(\zeta_n + \zeta_n^{-1})] =1$$ which gives us an isomorphism.

Since we have the containment $\mathbb{Q}(\zeta_n + \zeta_n^{-1}) \subset \mathbb{Q}(\zeta_n)$, and the isomorphism, it gives us equality.$$\mathbb{Q}(\zeta_n + \zeta_n^{-1}) = \mathbb{Q}(\zeta_n)$$

I'll leave you to fill in the remaining details, good luck with your exam on Monday!