2
$\begingroup$

Let $N$ be a positive integer and $p$ a prime not dividing $N$. Let $K$ be a finite Galois extension of $\mathbb{Q}$ which is unramified outside the primes dividing $Np$. Let $G$ be the Galois group of $K$ over $\mathbb{Q}$. Then for any conjugacy class $C \subset G$, there is a prime number $q$ not dividing $Np$ such that $\text{Frob}_q = C$.

My question is the following: In the same situation as above, for any conjugate $C \subset G$ does there always exist a prime number $q$ not dividing $Np$ such that

(a) $\text{Frob}_q=C$;

(b) $q\not\equiv 1 \pmod p$?

$\endgroup$
5
  • $\begingroup$ If $O$ is the open set of all questions obviously not suitable for MO, this question is in $\overline{O}$. I conjecture (but cannot prove, although I've tried) that this is an exercise in some algebraic number theory textbook. This questions will probably get some close votes, it will probably also get some answers. $\endgroup$ Nov 7, 2017 at 17:21
  • 4
    $\begingroup$ The condition $q\not\equiv1\pmod p$ is equivalent (for $q\ne p$) to Frob${}_q$ lying in a union of conjugacy classes in the Galois group of the cyclotomic extension $\Bbb Q(e^{2\pi i/p})/\Bbb Q$. Therefore presumably your desired primes form a union of conjugacy classes in the extension $L/\Bbb Q$ where $L$ is the compositum of $K$ and $\Bbb Q(e^{2\pi i/p})/\Bbb Q$, hence can be detected by the Chebotarev density theorem. $\endgroup$ Nov 7, 2017 at 17:25
  • 3
    $\begingroup$ If $K={\mathbb Q}(e^{\frac{2\pi i}{Np}})$, and $C$ is the trivial conjugacy class in $G$, then clearly there is no such prime $q$, since ${\mathbb Q}(e^{\frac{2\pi i}{p}})$ is contained in $K$. The comment of Greg Martin is relevant to see what happens in general. $\endgroup$ Nov 7, 2017 at 18:21
  • $\begingroup$ I didn't rule out the obvious exception that $C=\{1 \}$. Sorry for confusion. $\endgroup$
    – user116950
    Nov 7, 2017 at 23:06
  • 1
    $\begingroup$ $q \not \equiv 1 \bmod p$ means $Frob_{q,\mathbb{Q}(\zeta_p)} \ne Id$. Since $\mathbb{Q}(\zeta_p)/(K \cap \mathbb{Q}(\zeta_p))$ is abelian, it suffices that $\zeta_p \not \in K$ or that some $\sigma \in C, \sigma|_{K \cap \mathbb{Q}(\zeta_p)} \ne Id$. Then apply Cebotarev to find $q$ with $Frob_q = \sigma$ $\endgroup$
    – reuns
    Nov 8, 2017 at 2:33

1 Answer 1

8
$\begingroup$

As noted in the comments, this is false in general. For instance, if $K=\mathbb Q(e^{2i \pi/p})$, then $G=(\mathbb Z/p\mathbb Z)^\ast$. If you take $C=\{1\}$, then condition (a) is that $q \equiv 1 \pmod{p}$ and condition (b) is that $q \not \equiv 1 \pmod{p}$ which makes finding such primes $q$ a desperate task.

There is however a general situation where the answer is positive: when the group $G$ has no quotient isomorphic to a quotient of $(\mathbb Z/p \mathbb Z)^\ast$, in particular, when the group $G$ is perfect (equal to its own derived subgroup). In this case, indeed, the extension $K(e^{2i \pi/p})/\mathbb Q$ has a Galois groups with subjects to both $G$ and $(\mathbb Z/p\mathbb Z)^\ast$, hence which is isomorphic by Goursat's lemma and our assumption to $G \times (\mathbb Z/p\mathbb Z)^\ast$, and by Cebotarev you can find primes $q$ such that $Frob_q$ is in the class $C \times 1$ in this group. (Of course, this is a very standard argument in the field)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.