Let $p$ be a prime, and $p-1=fd$. There is exactly one unique subfield of $\mathbb{Q}(\zeta_p)$ which is of degree $d$. The defining polynomial $P_d(x)$ for this subfield has the root:
$$\sum_{r=1}^{f} (\zeta_p^{g^{rd}}) = \zeta_p^{g^{d}} + \zeta_p^{g^{2d}} + \zeta_p^{g^{3d}}... + \zeta_p^{g^{(f-2)d}} + \zeta_p^{g^{(f-1)d}} + \zeta_p^{g^{fd}}$$
where $g$ is a primitive root $\pmod p$.
When $d=3$, it is well known that $P_3(x) = x^3+x^2+cx+c_2x$ where
$c=-(p-1)/3$ and $c_2=(p(L+3)-1)/27$
The integer $L=1\pmod 3$ is given by the solution of $4p = L^2+27M^2$.
The discriminant of $P_3(x)$ is $p^2M^2$.
Are there any known results when $d=7$?
If $p=1\pmod 7$, then the subfield of $\mathbb{Q}(\zeta_p)$ of degree $7$ is given by:
Let $P_7(x): x^7 + x^6 + cx^5 + c_2x^4 + c_3x^3 + c_4x^2 + c_5x + c_6$
One finds that $c=-3(p-1)/7$, but the question of interest is how do $c_2, c_3, c_4, c_5,$ and $c_6$ depend on $p$? Any links to articles or Diophantine equations necessary to find how the remaining coefficients depend on $p$ would be helpful. Thanks in advance!
We can use the Fundamental Theorem of Galois Theory. Note that the Galois Group Gal$(\mathbb{Q}(\zeta_p))$ is cyclic of order $p-1$. Therefore it has exactly one subgroup $G$ of index $d$ for each divisor $d$ of $p-1$. By the Galois Correspondence there is exactly one intermediate field $E\leq\mathbb{Q}(\zeta_p)$ such that $[E:\mathbb{Q}]=d$.