Finding out the element of degree $p$ in $\Bbb Q(\zeta_{p^3})$ For a prime $p$, define $ζ_{p^3} = e^{\frac{2πi}{p^3}}$.
Then for the field $\Bbb Q(\zeta_{p^3})$? 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 $p$.
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.
P.S. Also I am interested in the general case: for any integer which is not less then $2$, consider  $\Bbb Q(\zeta_{p^n})$, how can I see if I have an element with degree $p$ over $\Bbb Q$ here?
If it is really hard to do it without the usage of the Galois theory, may I please just ask for a "Claim and justify" answer which simply point out which element is of order $p$ and prove that its order is $p$. Many thanks!
 A: I know this is not the answer expected, as it uses Galois theory, but it is a little long for a comment.
Finding an element of order $p$ in the field $K = \mathbb{Q}(\zeta_{p^3})$ is equivalent to finding a sub-extension, say $E$, of $K$ over $\mathbb{Q}$. To find it, one needs first to look at its Galois group (if want to avoid it, you can work also with embeddings from $K$ to $\mathbb{C}$), which in this case is isomorphic to the multiplicative group $(\mathbb{Z}/p^3\mathbb{Z})^{\times}$, which I'll denote by $G_p$. 
(ASIDE: A class represented by $a \in \mathbb{Z}/p^3\mathbb{Z}$ is mapped to the automorphism of $K$ determined by sending $\zeta_{p^3}$ to $\zeta_{p^3}^a$.) 
For the prime $p=2$, this group is of order 4, and indeed is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. 
(ASIDE: this group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{r-2}\mathbb{Z}$ when 3 is replaced by $r$.)
This is basically why the case $p=2$ is easy. 
For $p>2$, $G_p$ is cyclic, generated by the product of image, say $a$ of the generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ (you need to know that this group is cyclic) in $(\mathbb{Z}/p^3\mathbb{Z})^{\times}$ with the class of $1+p$. 
(ASIDE: You can show easily that the class of $1+p$ generates a sub-group of order $p^2$ in $G_p$.)
Hence the subgroup say $H_p$ generated by the product $a(1+p)^p$ is of size $p(p-1)$, hence is of index $p$. Now, fundamental theorem of Galois theory tells you that the elements that are fixed by $H_p$, say $E$, is a sub-field of K of degree $[G_p:H_p]=p$.
As we work in characteristic 0, this field can be written as $\mathbb{Q}(\alpha)$, for some $\alpha \in \mathbb{Q}(\zeta_{p^3})$. Then, of course, this $\alpha$ is of degree $p$. This partly answers your question. Of course, it is not easy to find such an element explicitly.
One should check by hand that a sum of the element $\zeta_{p^3}$ over all classes  in $G_p/H_p$ should work, I guess.
