# Infinitely many primes don't split?

Milne's note says when $$L,K$$ are number fields and $$L/K$$ is an abelian extension, infinitely many primes don't split. I want to ask if it's right in general case and how to prove it?

Thank you!

• Overkill: Chebotarev's density theorem - en.wikipedia.org/wiki/Chebotarev%27s_density_theorem Commented Mar 13, 2020 at 6:25
• I believe existence of non-split primes in abelian extensions was historically the main stumbling block for providing fully algebraic proofs of class field theory (but don't quote me on that historical account). An algebraic proof now exists, I believe due to Chevalley, but is quite involved, see Proposition VII.4.6 of Milne's CFT. There is a simpler analytic proof too (which works in all Galois extensions), see Theoren VI.4.6 ibidem. Commented Mar 13, 2020 at 8:37
• You can use Chebotarev's density theorem as François said, or mimick the proof of Dirichlet's theorem in this particular case. All you need to know is that $L$ is the class field of some congruence class group $I_{\mathfrak m}/H$ of $K$, and that a prime of $K$ splits in $L$ if and only if it is in $H$. Commented Mar 13, 2020 at 9:27

Assume that all but finitely many primes split in $$L/K$$. Then the Dedekind zeta function $$\zeta_L(s)$$ agrees with $$\zeta_K(s)^{(L:K)}$$ apart from finitely many Euler factors. Comparing the order of the pole $$s=1$$, we conclude that $$(L:K)=1$$, that is, $$L=K$$.
• Why do you get a power of $\zeta_K$? Doesn't that need the stronger assumption that all but finitely many primes split completely in $L$? (I read "don't split" as "inert", but maybe I misread the question.) Commented Mar 13, 2020 at 18:09
• @R.vanDobbendeBruyn If you interpret "don't split" as "stay inert", then this is false if $L/K$ is not cyclic. Indeed, if $\mathfrak p$ stays inert (in particular is unramified), then $Gal(L/K)$ injects into the Galois group of the residue fields, which is cyclic. Commented Mar 13, 2020 at 19:30