Let $d \geq 1$ be an integer. Dirichlet's theorem on arithmetic progression implies that the arithmetic progression $a, a+d, a+2d, \ldots$ contains infinitely many primes if and only if $\gcd(a,d)=1$.


Suppose $K/\mathbb{Q}$ is a finite Galois extension. Cebotarev's density theorem implies that there are infinitely many primes that split completely in $K$.


Since there are finitely many $1 \leq a \leq d$ such that $\gcd(a,d)=1$, there exists at least some $1 \leq a\_0 \leq d$ with $\gcd(a\_0,d)=1$ such that the arithmetic progression $a\_0, a\_0+d, a\_0+2d, \ldots$ contains infinitely many primes that split completely in $K$.


My question is



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> Is it true that for all $1 \leq a \leq d$ with $\gcd(a,d)=1$, the arithmetic progression $a, a+d, a+2d, \ldots$ contains infinitely many primes that split completely in $K$?
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Infinitely many primes that split completely in an arithmetic progression
**Theorem**. *Let $K$ and $L$ be finite Galois extensions of $\mathbf Q$*. Set $F = K\cap L$.


(1) *If $F = \mathbf Q$, then for each conjugacy class $C$ in
${\rm Gal}(L/\mathbf Q)$ there are infinitely many primes that are unramified in $L$ with Frobebius conjugacy class $C$ and split completely in $K$*.


(2) *If $F \not= \mathbf Q$ then there is a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ such that no prime unramified in $L$ with Frobenius conjugacy class $C$ splits completely in $K$.*


(3) *A conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ is the Frobenius conjugacy class of some prime unramified in $L$ that splits completely in $K$ if and only if $C \subset {\rm Gal}(L/F)$, in which case $C$ is the Frobenius conjugacy class of infinitely many primes unramified in $L$ that split completely in $K$.*


(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ and $[L:\mathbf Q]$ are relatively prime and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta\_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta\_d)$ must divide $d$. Neither of these conditions is necessary.)


**Remark**. Using $L = \mathbf Q(\zeta\_d)$, we see the answer to the OP’s question if affirmative if and only if $K \cap \mathbf Q(\zeta\_d) = \mathbf Q$, and that even if $K \cap \mathbf Q(\zeta\_d) \not= \mathbf Q$ we can still describe exactly which elements of the group $(\mathbf Z/d\mathbf Z)^\times$, viewed as ${\rm Gal}(\mathbf Q(\zeta\_d)/\mathbf Q)$, contain a prime number that splits completely in $K$: it is the congruence classes mod $d$ that belong to ${\rm Gal}(\mathbf Q(\zeta\_d)/F)$, where $F = K \cap \mathbf Q(\zeta\_d)$.


*Proof*.
(1) We assume $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and
${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.


Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.


(2) We assume $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, and ${\rm Gal}(L/F)$ is a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/F)$ is a *proper* normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is *no* prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.


(3) We showed in the proof of (2) that if there is a prime unramified in $L$ that splits completely in $K$, then its Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ lies in ${\rm Gal}(L/F)$. Conversely, let $C$ be a conjugacy class of ${\rm Gal}(L/\mathbf Q)$ that lies in the normal subgroup ${\rm Gal}(L/F)$. Pick $\sigma \in C$, so $\sigma \in {\rm Gal}(L/F)$. By Galois theory the restriction mapping ${\rm Gal}(KL/K) \to {\rm Gal}(L/F)$ is an isomorphism, so we can lift $\sigma$ to an automorphism $\sigma'$ in ${\rm Gal}(KL/K)$. By Chebotarev there are (infinitely many) primes $p$ unramified in $KL$ whose Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is the conjugacy class of $\sigma'$. Let's show for such $p$ that (i) the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is $C$ and (ii) $p$ splits completely in $K$:


(i) since $\sigma'|\_{L} = \sigma$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is the conjugacy class of $\sigma$ in ${\rm Gal}(L/\mathbf Q)$, which is $C$,


(ii) since $\sigma'$ is trivial on $K$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(K/\mathbf Q)$ is trivial, so $p$ splits completely in $K$.


QED


