# Infinitely many primes that split completely in an arithmetic progression

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

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 $$K$$?

• One has to be a bit careful formulating this question. Take for example the Gaussian field $K = \mathbb{Q}(i)$. An odd prime $p$ splits in $K$ if and only if $p \equiv 1 \pmod{4}$. Now take $a = 3$ and $d = 4$. Then there is no prime $p$ in this arithmetic progression which splits in $K$. On the other hand, if $d$ is coprime to the conductor of $K$ and $K$ is abelian, then the answer is surely yes because the splitting behaviour of primes in such a field is controlled by congruence conditions. In non-abelian extensions this is probably still true, but I don't have a proof on hand. Jun 30 at 4:07
• @StanleyYaoXiao Thank you very much for pointing this out! Jun 30 at 18:39

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

• Thank you very much for the thorough response! If I understand your argument correctly, both Part (1) and (2) are corollaries of Part (3)? I do appreciate the exposition, which makes it easier to understand the obstruction. Jun 30 at 18:39
• Sure, but it is easier to see the ideas at work in (1) and (2) first. Also, (1) and (2) are just what I wrote up first as a reply to the question you asked. Only later did I formulate (3) to unify (1) and (2) into a single result. Jun 30 at 20:59
• Just to be sure that I understand the Remark in your answer. A congruence class $[a] \in (\mathbb{Z}/d\mathbb{Z})^{\times}$ contains infinitely many primes $p$ that split completely in $K$ if and only if $[a] \in \text{Gal}(\mathbb{Q}(\zeta_d)/F) \subset \text{Gal}(\mathbb{Q}(\zeta_d)/\mathbb{Q})\cong (\mathbb{Z}/d\mathbb{Z})^{\times}$ where $F = K \cap \mathbb{Q}(\zeta_d)$? Jul 1 at 4:37
• Yes. Do you see how it is consistent with the counterexample Stanley wrote about in his comment? Jul 1 at 4:50
• In that case, we have $K = L = \mathbb{Q}(i) = \mathbb{Q}(\zeta_4)$ so $F = \mathbb{Q}(i)$ and $\text{Gal}(\mathbb{Q}(\zeta_4)/F)$ consists of the identity element of $(\mathbb{Z}/4\mathbb{Z})^{\times} = \{,\}$. The identity element is $$. So a congruence class $[a] \in (\mathbb{Z}/4\mathbb{Z})^{\times}$ contains infinitely many primes $p$ that split completely in $\mathbb{Q}(i)$ if and only if $[a]=$. Did I get this? Jul 1 at 5:01