Finding out an element of degree $p$ in

Question

Note that it is related but NOT duplicated from: Finding out the element of degree $p$ in $\Bbb Q(\zeta_{p^3})$

For any integer $n$, define $ζ_{3^n} = e^{\frac{2πi}{3^n}}$.

Then for the field $\Bbb Q(\zeta_{3^n})$? May I please ask how to find an element of degree $p$ over $\Bbb Q$? I am simply ask for an explicit example here, better with a general formula involves $n$. That is,a just "Claim and justify" answer.

I know that this question may be quite related to Galois theory. But I am asking for a solution that without using Galois theory, at least not to use it explicitly because I have not learnt it. However, I think the idea of Galois theory may help. Could someone please tell me how to deal with that? Thanks.

Answer 1

Well, dropping that element out of nowhere is easy enough, although as already mentioned in the last question, giving reason how to find it is difficult without Galois theory:

As MooS already mentioned, it suffices to solve the problem for $\zeta_9$. Here, $\zeta_9 + \zeta_9^{-1}$ will do the job.

For general $n$, we can then just take $$\zeta_9 := \zeta_{3^n}^{3^{n-2}}$$ and then the same argument holds. That might also work for $p \neq 3$, but proofs or generalizations mostly depend on results from Galois theory.

Answer 2

Let $\zeta$ be a primitive $9$-th root of unity. $a=\zeta+\zeta^{-1}$ is an element of degree $3$ over $\mathbb Q$. The minimal polyomial is $(x-(\zeta+\zeta^{-1}))(x-(\zeta^2+\zeta^{-2}))(x-(\zeta^4+\zeta^{-4}))=x^3-3x+1$.

Feel free to double-check that computation and that it is irreducible...

For another approach to get the minimal polynomial, just compute $1,a,a^2,a^3$ and find a non-trivial linear-combination. This is linear algebra and should be available to you.

Here is your element (and your minimal polynomial): https://www.wolframalpha.com/input/?i=2+cos((2+%CF%80)%2F9)