-1
$\begingroup$

Let $a,m$ an integers s.t $(a,m)=1$. Let $K$ a quadratic field, I would like to calculate the natural density of the set

$$\{p \;\; \text{rational prime}\; /\; p\;\text{inert in}\; K,\; p\equiv a\pmod m\}$$

I think that is equal to $1/2\phi(m)$, but I couldn't prove that.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ This is obviously false. Think about $p \equiv 3 \pmod{4}$ in $\mathbb Q(i)$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 25, 2017 at 19:29
  • 3
    $\begingroup$ The general idea is to say that both conditions correspond to the Frobenius of the prime p lying in certain conjugacy classes in $Gal(K/Q)$ and $Gal(Q(\mu_m)/Q)$. If these two fields are disjoint, then the Galois group of the compositum is the product of these two groups and the density is multiplicative. Otherwise, $K$ will be contained in the cyclotomic field and the density will be either $1/\phi(m)$ or $0$ depending on the image of $a$ in the map $Gal(Q(\mu_m)/Q) \to Gal(K/Q)$ being the non identity element or the identity element respectively. $\endgroup$
    – Asvin
    Commented Sep 25, 2017 at 19:36
  • $\begingroup$ @R.vanDobbendeBruyn, Thanks for your comment. $\endgroup$
    – square-free
    Commented Sep 25, 2017 at 19:42
  • $\begingroup$ @Asvin, Thanks a lot for your comment. $\endgroup$
    – square-free
    Commented Sep 25, 2017 at 19:43
  • 1
    $\begingroup$ Look up Chebotarev density (sometimes spelled differently). $\endgroup$
    – Kimball
    Commented Sep 25, 2017 at 21:21

1 Answer 1

Reset to default
5
$\begingroup$

This is not true in general. Take $K = \mathbb{Q}(i), a =3$ and $m = 4$. Then the density is just $1/2$ in this case. This is because a prime $p$ is inert in $K$ if and only if $p \equiv 3 \bmod 4$. So the congruence condition implies already that $p$ is inert.

In general, saying that a prime is inert in $K$ can be written in terms of congruence conditions modulo $|d_K|$, where $d_K$ is the discriminant of $K$. So if $\gcd(d_K,m) = 1$, then the density is indeed $1/2\varphi(m)$. But otherwise you can get different densities occuring.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .