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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 $K$?

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    $\begingroup$ 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. $\endgroup$ Jun 30 at 4:07
  • $\begingroup$ @StanleyYaoXiao Thank you very much for pointing this out! $\endgroup$
    – Xiao Xiao
    Jun 30 at 18:39

1 Answer 1

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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

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  • $\begingroup$ 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. $\endgroup$
    – Xiao Xiao
    Jun 30 at 18:39
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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Jun 30 at 20:59
  • $\begingroup$ 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 $F = K \cap \mathbb{Q}(\zeta_d)$? $\endgroup$
    – Xiao Xiao
    Jul 1 at 4:37
  • $\begingroup$ Yes. Do you see how it is consistent with the counterexample Stanley wrote about in his comment? $\endgroup$
    – KConrad
    Jul 1 at 4:50
  • $\begingroup$ 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 $(\mathbb{Z}/4\mathbb{Z})^{\times} = \{[1],[3]\}$. The identity element is $[1]$. 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 $[a]=[1]$. Did I get this? $\endgroup$
    – Xiao Xiao
    Jul 1 at 5:01

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