finding number fields and a prime with desired $r$,$e$ and $f$. This is an exercise from Marcus's Number Fields, on page 116. Given $r$, $e$, and $f$ positive integers, find primes $p$ and $q$ such that $p$ splits (not necessarily completely) into $r$ distinct primes in $\mathbb Q(\zeta_q)$. I wonder if one can solve this without using the fact that there are primes in an arithmetic progression, because my solution assumed $q$ to be $\equiv 1 \mathrm{mod}\, ref$, and $p$ was also obtained similarly by requiring it to be congruent to some element of $(\mathbb Z/q \mathbb Z)^\times$.
 A: First, if $p = q$, then $p$ is totally ramified in $\mathbb{Q}(\zeta_q)$, so there is only one prime of $\mathbb{Q}(\zeta_p)$ above $p$. From now on, then, let us assume that $p$ and $q$ are distinct primes. In that case, $p$ is unramified in $\mathbb{Q}(\zeta_{q})$.
The splitting behavior of $p$ in $\mathbb{Q}(\zeta_{q})$ is governed by the factorization of the cyclotomic polynomial $\Phi_{q}(x)$ over $\mathbb{F}_p$: the primes of $\mathbb{Q}(\zeta_{q})$ above $p$ are in bijection with the irreducible factors of $\Phi_{q}(x)$ over $\mathbb{F}_p$. So, you want to know when $\Phi_{q}(x)$ splits into $r$ factors over $\mathbb{F}_p$. This in turn amounts to determining the degree of the field extension of $\mathbb{F}_p$ over which there is a nontrivial $q$th root of unity. In general, a finite field of order $p'$ has a nontrivial $q$th root of unity if and only if $p' \equiv 1$ (mod $q$). Thus $\Phi_{q}(x)$ splits completely over any such field.
Let $p'$ be the smallest power of $p$ such that $p' \equiv 1$ (mod $q$). If $p' = p^d$, then the irreducible factors of $\Phi_{q}(x)$ over $\mathbb{F}_p$ all have degree $d$, so there are $(q-1)/d$ such factors. Thus, the answer to your question is that this happens when $r = (q-1)/d$.
It might be possible to give a more convenient formula, but I am not sure about that. At any rate, you should easily be able to construct such $p$ and $q$ using the above.
