0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think a slicker method would be via tower law, do you know it? $\endgroup$ – mdave16 Oct 27 '17 at 12:44
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $[\mathbb{Q}(\omega):\mathbb{Q}(t)]=\partial(p)=2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.