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Let $\zeta_n$ be a $n$-th primitive root of unity. How to prove that $[\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n+\zeta_{n}^{-1})]=2$ ?

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  • $\begingroup$ Use the FTGT: it's the fixed field of complex conjugation, which has order 2. $\endgroup$ – Adam Hughes Aug 13 '16 at 15:28
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This is only true for $n>2$. Since $\zeta_n$ is a root of the polynomial $$X^2 -(\zeta_n+\zeta_n^{-1})X + 1 = (X-\zeta_n)(X-\zeta_n^{-1})\in \mathbb Q(\zeta_n+\zeta_n^{-1})[X], $$ it follows that $[\mathbb Q(\zeta_n): \mathbb Q(\zeta_n+\zeta_n^{-1})]\le 2$. Now, we have $\mathbb Q(\zeta_n+\zeta_n^{-1})\subseteq \mathbb R$ and $\zeta_n\notin \mathbb R$. Together it follows that $\mathbb Q(\zeta_n)\neq \mathbb Q(\zeta_n+\zeta_n^{-1})$ and hence the degree must be 2.

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    $\begingroup$ It would be better to ban any analysis from the arguments. For instance, in @user218931's answer,introduce, instead of R, the automorphism f of order 2 sending zeta to zeta^-1. By Galois, our cyclotomic field has degree 2 over the fixed field of f. But f obviously fixes Q(zeta + zeta^-1), and we are done. It is all the more surprising that an object constructed in a purely algebraic way contains cos (2 pi/n), which is transcendental. An interesting related problem is to determine Q(sin (2 pi/n)). For simplicity, take n = an odd prime; but still a certain amount of number theory is needed. $\endgroup$ – nguyen quang do Aug 11 '16 at 15:09
  • $\begingroup$ Edit : contains the value cos (2 pi/n) of a transcendental function $\endgroup$ – nguyen quang do Aug 11 '16 at 16:04
  • $\begingroup$ But is it clear that there is an automorphism given by $\zeta\mapsto \zeta^{-1}$? Is there a way to show it without using that the $n$-th cyclotomic polynomial is irreducible? Because showing its irreducibility is quite involved and not trivial at all. $\endgroup$ – Claudius Aug 12 '16 at 3:13
  • $\begingroup$ Here zeta is a primitive element of the extension over Q, hence a Q-automorphism can be defined just by sending zeta to another root of its minimal polynomial, or more generally to another primitive element (and zeta^-1 is obviously such). $\endgroup$ – nguyen quang do Aug 12 '16 at 12:45
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    $\begingroup$ Why are $\zeta$ and $\zeta^{-1}$ roots of the same minimal polynomial? Your last statement is obviously false, since $\zeta+1$ is another primitive element of $\mathbb Q(\zeta)$, but $\zeta\mapsto \zeta+1$ does not define an automorphism. $\endgroup$ – Claudius Aug 12 '16 at 14:17

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