Primitive element for each subfield of a cyclotomic extension Given an odd prime number $p$, a natural number $r$, and a $p^r-th$ primitive root $\zeta=\zeta_p$, I have to find an explicit expression of a primitive element for each subfield of $\mathbb{Q}(\zeta)$.
By now, I know that $Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$ is cyclic of order $\varphi(p^r)=p^{r-1}(p-1)$, and I know how to calculate all primitive elements for each extension thanks to symmetric polynomials and when the Gauss periods are not zero. That is, when
$$\eta_i(n,g,\zeta)=\sum_{j=1}^{d-1}\zeta_{i+jn}$$
where $g$ is a generator of the multiplicative group $U(p^r)$, $d=\frac{p^{r-1}(p-1)}{n}$, and $\zeta_i=\zeta^{g^i}$, is not zero. 
Therefore, my question is:
How does it affect to the subfield lattice the fact that the Gaussian periods are zero and when and why does that happen  exactly? What can I do in such cases to find a primitive element?
 A: We may answer this question using the following result, which links totally ramified primes with Eisenstein polynomials:
Theorem. Let $ K $ be a number field where the rational prime $ p $ is totally ramified, with $ p = \mathfrak p^n $. For any $ x \in \mathcal O_K $ that is divisible by $ \mathfrak p $ exactly once, the characteristic polynomial of $ x $ is Eisenstein at $ p $.
You may find a proof of this in this text of Keith Conrad.
In particular, this implies that $ x $ is a primitive element, since the characteristic polynomial has degree $ n = [K : \mathbf Q] $, and by virtue of being Eisenstein at $ p $ it is irreducible over $ \mathbf Q $. Now, let $ K = \mathbf Q(\zeta) $ where $ \zeta $ is a primitive $ q $th root of unity, where $ q = p^n $ for a prime $ p $. By writing
$$ \Phi_q(X) = \frac{X^{p^n} - 1}{X^{p^{n-1}} - 1} = \sum_{k=1}^{p-1} X^{kp^{n-1}} $$
and evaluating at $ X = 1 $, we find that $ N_{K/\mathbf Q}(1 - \zeta) = p $, and the prime $ p $ totally ramifies as $ (p) = (1 - \zeta)^{\varphi(q)} $. On the other hand, for any subfield $ F $, the norm $ N_{K/F}(1 - \zeta) $ lies in $ F $, and it is a product of $ [K:F] $ elements that are associated to $ 1 - \zeta $. Therefore, 
$$ (N_{K/F}(1 - \zeta)) = (1 - \zeta)^{[K:F]} $$
as ideals, and raising both sides to the $ [F : \mathbf Q] $th power gives that the prime $ p $ totally ramifies in the subfield $ F $ as $ p = ((N_{K/F}(1 - \zeta))^{[F : \mathbf Q]} $. Now, $ x = N_{K/F}(1 - \zeta) $ satisfies the conditions of the theorem, and therefore $ F = \mathbf Q(x) $.
