I want to prove that |$\mathbb{Q}(\omega):\mathbb{Q}(t)|=2$ with $\omega$ a primitive nth root of unity and $t=\omega +\omega ^{-1}$ So for any element $q \in \mathbb{Q}(\omega)$, I need to find a basis $f_1,f_2$ such that $q=af_1 + bf_2$ with $a,b \in \mathbb{Q}(t)$ and then prove it is a basis. Is this correct? and I can't find a suitable basis so any help is appreciated.

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    $\begingroup$ I think a slicker method would be via tower law, do you know it? $\endgroup$ – mdave16 Oct 27 '17 at 12:44
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    $\begingroup$ You do know that $\omega$ is a root of the polynomial $x^2-tx+1=(x-\omega)(x-\omega^{-1})$, don't you? $\endgroup$ – Jyrki Lahtonen Oct 27 '17 at 13:04
  • $\begingroup$ @mdave16 Yes I do, what would be the intermediary field? $\endgroup$ – Bradley Hill Oct 27 '17 at 13:29
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    $\begingroup$ Do you know the Fundamental Theorem of Galois Theory? $\mathbb{Q}(t)$ is the fixed field of complex conjugation $\tau$. $\tau$ generates a subgroup of order $2$, so its corresponding fixed field has index $2$. $\endgroup$ – André 3000 Oct 27 '17 at 17:04
  • $\begingroup$ @BradleyHill, I was thinking the intermediate group would be $\mathbb{Q}(t)$, as JykriLahtonen mentioned. $\endgroup$ – mdave16 Oct 27 '17 at 17:22

Notice that the polynomial $$p(x)=x^2-tx+1=(x-\omega)(x-\omega^{-1}) \in \mathbb{Q}(t)[x]$$ is irreducible in $\mathbb{Q}(t)$ (why?) and $x=\omega$ is one of it's roots, hence it's the minimal polynomial of $\omega$ over $\mathbb{Q}(t)$, therefore $[\mathbb{Q}(\omega):\mathbb{Q}(t)]=\partial(p)=2$.

  • $\begingroup$ I'm sorry but I'm quite new to this topic so is there a method to tell if a polynomial is irreducible? $\endgroup$ – Bradley Hill Oct 28 '17 at 13:36
  • $\begingroup$ Notice that the factorization of $p(x)$ is $(x-\omega)(x-\omega^{-1})$, and the factorization is unique. Hence the factors aren’t of the form $$a+bt=a+b(\omega+\omega^{-1}),$$ where $a,b \in \mathbb{Q}$. $\endgroup$ – Adrián Naranjo Oct 28 '17 at 17:52
  • $\begingroup$ That makes sense, thank you!! $\endgroup$ – Bradley Hill Oct 28 '17 at 18:43
  • $\begingroup$ You’re welcome :) $\endgroup$ – Adrián Naranjo Oct 28 '17 at 18:57

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