# Thread: Field generated by a complex number over the rationals.

1. ## Field generated by a complex number over the rationals.

Hi:
Let $\displaystyle \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q \text{ the field of rational numbers and } R \text{ the field of real numbers}$. I want to prove that
$\displaystyle Q(\zeta_{n})\cap R = Q(\zeta_{n} + \zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) \subseteq Q(\zeta_{n})\cap R \text{ trivially. Set }\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}$. To prove the opposite
inclusion, I develop $\displaystyle \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}$. This gives, by using global induction, that $\displaystyle cos 2\pi(2k/n) \epsilon Q(\eta_{n})$ for every
natural k. In like manner, $\displaystyle cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n})$.Now, it is easy to prove the inclusion, because $\displaystyle Q(\zeta_{n})\cap R={a_{0}+a_{1}cos2\pi/n+a_{2}cos2\pi(2/n)+...+a_{n-1}cos2\pi[(n-1)/n]}$. Question: is there
a more elegant or simpler proof? Any hint will be welcome. Thanks for reading.

2. Originally Posted by ENRIQUESTEFANINI
Hi:
Let $\displaystyle \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q \text{ the field of rational numbers and } R \text{ the field of real numbers}$. I want to prove that
$\displaystyle Q(\zeta_{n})\cap R = Q(\zeta_{n} + \zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) \subseteq Q(\zeta_{n})\cap R \text{ trivially. Set }\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}$. To prove the opposite
inclusion, I develop $\displaystyle \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}$. This gives, by using global induction, that $\displaystyle cos 2\pi(2k/n) \epsilon Q(\eta_{n})$ for every
natural k. In like manner, $\displaystyle cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n})$.Now, it is easy to prove the inclusion, because $\displaystyle Q(\zeta_{n})\cap R={a_{0}+a_{1}cos2\pi/n+a_{2}cos2\pi(2/n)+...+a_{n-1}cos2\pi[(n-1)/n]}$. Question: is there
a more elegant or simpler proof? Any hint will be welcome. Thanks for reading.

I'm not sure about shorter but I think simpler: as $\displaystyle \zeta_n^k+\zeta_n^{-k}=2\cos\frac{2\pi k}{n}$, we get that

$\displaystyle w=\sum\limits_{k=0}^na_k\zeta_n^k\in \mathbb{Q}(\zeta_n)\cap\mathbb{R} \Longrightarrow\,Im\left(\sum\limits_{k=0}^na_k\ze ta_n^k\right)= \sum\limits_{k=0}^na_k\sin\frac{2\pi k}{n}=0$, so then only the real part of w plays a role here, thus:

$\displaystyle w=Re\left(\sum\limits_{k=0}^na_k\zeta_n^k\right)=\ sum\limits_{k=0}^na_k\cos\frac{2\pi k}{n}=\frac{1}{2}\sum\limits_{k=0}^na_k(\zeta_n^k+ \zeta_n^{-k})\in \mathbb{Q}(\zeta_n+\zeta_n^{-1})$

Tonio

3. Originally Posted by ENRIQUESTEFANINI
Hi:
Let $\displaystyle \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q \text{ the field of rational numbers and } R \text{ the field of real numbers}$. I want to prove that
$\displaystyle Q(\zeta_{n})\cap R = Q(\zeta_{n} + \zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) \subseteq Q(\zeta_{n})\cap R \text{ trivially. Set }\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}$. To prove the opposite
inclusion, I develop $\displaystyle \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}$. This gives, by using global induction, that $\displaystyle cos 2\pi(2k/n) \epsilon Q(\eta_{n})$ for every
natural k. In like manner, $\displaystyle cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n})$.Now, it is easy to prove the inclusion, because $\displaystyle Q(\zeta_{n})\cap R={a_{0}+a_{1}cos2\pi/n+a_{2}cos2\pi(2/n)+...+a_{n-1}cos2\pi[(n-1)/n]}$. Question: is there
a more elegant or simpler proof? Any hint will be welcome. Thanks for reading.
$\displaystyle (*)$ for any $\displaystyle z \in \mathbb{C}$ and $\displaystyle j \in \mathbb{Z}$ we obviously have: $\displaystyle z^{j+1} + z^{-(j+1)}=(z^j + z^{-j})(z+z^{-1}) - (z^{j-1} + z^{-(j-1)}).$ thus $\displaystyle z^k + z^{-k} \in \mathbb{Q}(z+z^{-1}), \ \forall k \in \mathbb{Z}.$

now let $\displaystyle x = \sum_{k=0}^{n-1}r_k \zeta_n^k \in \mathbb{Q}(\zeta_n) \cap \mathbb{R},$ where $\displaystyle r_k \in \mathbb{Q}.$ then $\displaystyle x=\bar{x}=\sum_{k=0}^{n-1}r_k \zeta_n^{-k}$ and therefore $\displaystyle x=\frac{x + \bar{x}}{2}=\sum_{k=0}^{n-1} \frac{r_k}{2}(\zeta_n^k + \zeta_n^{-k}) \in \mathbb{Q}(\zeta_n + \zeta_n^{-1}),$ by $\displaystyle (*)$.

4. Originally Posted by NonCommAlg
$\displaystyle (*)$ for any $\displaystyle z \in \mathbb{C}$ and $\displaystyle j \in \mathbb{Z}$ we obviously have: $\displaystyle z^{j+1} + z^{-(j+1)}=(z^j + z^{-j})(z+z^{-1}) - (z^{j-1} + z^{-(j-1)}).$ thus $\displaystyle z^k + z^{-k} \in \mathbb{Q}(z+z^{-1}), \ \forall k \in \mathbb{Z}.$

now let $\displaystyle x = \sum_{k=0}^{n-1}r_k \zeta_n^k \in \mathbb{Q}(\zeta_n) \cap \mathbb{R},$ where $\displaystyle r_k \in \mathbb{Q}.$ then $\displaystyle x=\bar{x}=\sum_{k=0}^{n-1}r_k \zeta_n^{-k}$ and therefore $\displaystyle x=\frac{x + \bar{x}}{2}=\sum_{k=0}^{n-1} \frac{r_k}{2}(\zeta_n^k + \zeta_n^{-k}) \in \mathbb{Q}(\zeta_n + \zeta_n^{-1}),$ by $\displaystyle (*)$.