Let $a$ and $b$ be two relatively prime positive integers. Denote by $S$ the set of all positive primes of the form $a+bn$ (where $n$ is an integer, possibly zero or negative). Let $S'$ be any set of positive primes such that $T=S\,\triangle\, S'$ is finite. Does there necessarily exist a finite Galois extension $K/\mathbb{Q}$ and a conjugacy class $C\subset \mathrm{Gal}(K/\mathbb{Q})$ such that $S'$ is exactly the set of primes unramified in $K$ whose Frobenius conjugacy class is equal to $C$?


If $T=\emptyset$ we can take $K$ to be a cyclotomic field. Can one at least solve the problem when $|T|=1$?



Sets of primes with a given Frobenius conjugacy class
The answer is no for $a=1$ and $b\geq 3$ arbitrary. Indeed, let $S$ be the set of (positive) primes congruent to $1$ modulo $b$. Let $S'$ be any set of (positive) primes such that $T=S\,\triangle\, S'$ is finite. Assume that $K/\mathbb{Q}$ is a finite Galois extension, and $C\subset \mathrm{Gal}(K/\mathbb{Q})$ is a conjugacy class, with the required property. Then I claim that $S'=S$.


To see this, let $L/\mathbb{Q}$ be a finite Galois extension that contains both $K$ and the $b$-th cyclotomic field $\mathbb{Q}(\zeta\_b)$. There are infinitely many primes $p$ that split in $L$. If $p$ is sufficiently large, then $p\in S'$, which forces $C=\langle\mathrm{Frob}\_p\rangle=\{1\}$ by the initial assumptions. Invoking again the initial assumptions, we see that $S'$ is the set of primes that split in $K$. Therefore, a well-known theorem (last Corollary in Section VIII.5 of Weil's Basic number theory) shows that $K=\mathbb{Q}(\zeta\_b)$, and hence $S'=S$.


