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

learning_math
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1 Answers1

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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.

Claudius
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    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. – nguyen quang do Aug 11 '16 at 15:09
  • Edit : contains the value cos (2 pi/n) of a transcendental function – nguyen quang do Aug 11 '16 at 16:04
  • 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. – Claudius Aug 12 '16 at 03:13
  • 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). – nguyen quang do Aug 12 '16 at 12:45
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    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. – Claudius Aug 12 '16 at 14:17
  • You are right, I was too hasty. My first statement is blatantly wrong (I thought of polynomial maps, and forgot the minimal polynomials !). As for the first, it misses a key ingredient - as you pointed out - which is the irreducibility of the n-th cyclotomic polynomial, or equivalently, the property that all the primitive n-th roots of 1 are mutually conjugate over Q. The usual proof is not entirely algebraic, in the sense that it uses a bit of arithmetic mod p. Nevertheless, it does not appeal to any analysis. – nguyen quang do Aug 13 '16 at 05:41