Polynomial of degree $7$ with cyclic Galois group So I want to find a degree $7$ polynomial whose Galois group is cyclic of order $7$. I know that to do this, I need to find a polynomial which is irreducible mod every prime (not dividing the discriminant). But this is where I'm stuck, since I don't know how to prove that any particular polynomial is irreducible mod every prime. Could I have some hints?
 A: Let $\zeta = \exp(2\pi i/29)$ be a primitive 29th root of unity.
Then the field $\mathbb Q(\zeta)/\mathbb Q$ has Galois group 
$$
G = (\mathbb Z/29\mathbb Z)^\times 
\simeq \mathbb Z/28 \mathbb Z 
\simeq \mathbb Z/4 \mathbb Z \times \mathbb Z/7 \mathbb Z .
$$
Since 2 is a primitive root mod 29 (a generator of $(\mathbb Z/29\mathbb Z)^\times$), this Galois group is generated by the automorphism $\zeta \mapsto \zeta^2$.
Furthermore, an element of order 4 can by found by the seventh power of this automorphism, namely $\zeta \mapsto \zeta^{2^7} = \zeta^{12}$.
Let $H$ be the subgroup of $G$ generated by $\zeta \mapsto \zeta^{12}$, which is cyclic of order 4.
Then $G/H \simeq \mathbb Z/7\mathbb Z$ is the Galois group of the fixed field $K := \mathbb Q(\zeta)^H$ over $\mathbb Q$.
Now all we have to do is find the minimal polynomial of a generator for $K/\mathbb Q$.
Since $\zeta$ generates $\mathbb Q(\zeta)/\mathbb Q$, the traces $\operatorname{Tr}_H(\zeta^i)$ generate $K/\mathbb Q$.
Let
$$
\xi 
= \operatorname{Tr}_H(\zeta) 
= \sum_{\sigma \in H} \sigma(\zeta)
= \zeta + \zeta^{12} + \zeta^{28} + \zeta^{17} .
$$
By solving a large system of linear equations involving the powers of $\xi$ (or using a computer to do so), we find that $\xi$ has minimal polynomial
$$
  f(x) = x^7 + x^6 - 12x^5 - 7x^4 + 28x^3 + 14x^2 - 9x + 1 .
$$
Remark: I chose 29 because it is the smallest prime which is 1 mod 7.
Your comment above about needing the polynomial to be reducible mod every prime except those dividing the discriminant is not true. The only primes dividing the discriminant of $f$ are 17 and 29, but $f$ splits completely mod 59 or 233, for example. This is not a coincidence. These primes are both 1 mod 29, and so they split completely in $\mathbb Q(\zeta)$, and hence also in $K$
