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Math Help - Field generated by a complex number over the rationals.

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

    Hi:
    Let \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q <br />
\text{ the field of rational numbers and } R \text{ the <br />
field of real numbers}. I want to prove that
    Q(\zeta_{n})\cap R = Q(\zeta_{n} + <br />
\zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) <br />
\subseteq Q(\zeta_{n})\cap R \text{ trivially. Set <br />
}\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}. To prove the opposite
    inclusion, I develop \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}. This gives, by using global induction, that cos 2\pi(2k/n) \epsilon Q(\eta_{n}) for every
    natural k. In like manner, cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n}).Now, it is easy to prove the inclusion, because 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.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:
    Let \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q <br />
\text{ the field of rational numbers and } R \text{ the <br />
field of real numbers}. I want to prove that
    Q(\zeta_{n})\cap R = Q(\zeta_{n} + <br />
\zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) <br />
\subseteq Q(\zeta_{n})\cap R \text{ trivially. Set <br />
}\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}. To prove the opposite
    inclusion, I develop \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}. This gives, by using global induction, that cos 2\pi(2k/n) \epsilon Q(\eta_{n}) for every
    natural k. In like manner, cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n}).Now, it is easy to prove the inclusion, because 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 \zeta_n^k+\zeta_n^{-k}=2\cos\frac{2\pi k}{n}, we get that

    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:

    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
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  3. #3
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:
    Let \zeta_{n} \text{ be }e^{2\pi i/n} \text{, } Q <br />
\text{ the field of rational numbers and } R \text{ the <br />
field of real numbers}. I want to prove that
    Q(\zeta_{n})\cap R = Q(\zeta_{n} + <br />
\zeta_{n}^{-1})\text{. }Q(\zeta_{n} + \zeta_{n}^{-1}) <br />
\subseteq Q(\zeta_{n})\cap R \text{ trivially. Set <br />
}\eta_{n}=\zeta_{n} + \zeta_{n}^{-1}. To prove the opposite
    inclusion, I develop \eta_{n}^{2k}\text{ and }\eta_{n}^{2(k+1)}. This gives, by using global induction, that cos 2\pi(2k/n) \epsilon Q(\eta_{n}) for every
    natural k. In like manner, cos 2\pi[(2k+1)/n]\epsilon Q(\eta_{n}).Now, it is easy to prove the inclusion, because 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.
    (*) for any z \in \mathbb{C} and j \in \mathbb{Z} we obviously have: z^{j+1} + z^{-(j+1)}=(z^j + z^{-j})(z+z^{-1}) - (z^{j-1} + z^{-(j-1)}). thus z^k + z^{-k} \in \mathbb{Q}(z+z^{-1}), \ \forall k \in \mathbb{Z}.

    now let x = \sum_{k=0}^{n-1}r_k \zeta_n^k \in \mathbb{Q}(\zeta_n) \cap \mathbb{R}, where r_k \in \mathbb{Q}. then x=\bar{x}=\sum_{k=0}^{n-1}r_k \zeta_n^{-k} and therefore 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 (*).
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    (*) for any z \in \mathbb{C} and j \in \mathbb{Z} we obviously have: z^{j+1} + z^{-(j+1)}=(z^j + z^{-j})(z+z^{-1}) - (z^{j-1} + z^{-(j-1)}). thus z^k + z^{-k} \in \mathbb{Q}(z+z^{-1}), \ \forall k \in \mathbb{Z}.

    now let x = \sum_{k=0}^{n-1}r_k \zeta_n^k \in \mathbb{Q}(\zeta_n) \cap \mathbb{R}, where r_k \in \mathbb{Q}. then x=\bar{x}=\sum_{k=0}^{n-1}r_k \zeta_n^{-k} and therefore 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 (*).
    Thank you very much for your useful and didactic answer. Regards,
    Enrique.
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