Elementary proof of a special case of Chebotarev's density theorem

A special case of Cheboratev's density theorem states that, for $K/\mathbb{Q}$ a Galois number field of degree $n$, then the rational primes that split completely in $K$ have density $1/n$.

Is there an elementary proof of this fact?

Actually I'm asking this only to apply it to $K=\mathbb{Q}(\zeta_n)$ and get that the density is nonzero, and there is an elementary proof of this. However I'm also interested in the more general result.

• Should this be called Chebotarev's theorem? It is after all a rather more basic 19th century theorem, known to Kronecker and Frobenius (with the logarithmic notion of density), and also the Galois $K/\mathbb{Q}$ case (with the natural notion of density) of Landau's prime ideal theorem. Chebotarev's contribution treating the non-identity conjugacy classes in the Galois group was the difficult new step that he solved at the time, leading Artin to the proof of his conjectured abelian reciprocity law. – Vesselin Dimitrov Sep 27 '15 at 5:19
• With the logarithmic notion of density being understood, this case has fully elementary proofs by Chebyshev ideas. I have sketched them here: mathoverflow.net/questions/25794/… and here: mathoverflow.net/questions/208604/… . – Vesselin Dimitrov Sep 27 '15 at 5:21
• You could have a look to THEOREM 3.4 in Milne's Class Field Theory notes (p. 188). – Watson Dec 22 '16 at 8:52

There does exist an elementary proof, and is given in many books on algebraic number theory. I think it is in Lang's book. I reproduce the proof below.

Consider $\zeta _K(s)=\prod _{\mathfrak p}(1-\frac{1}{(N\mathfrak p)^{s}})^{-1}$ where $\mathfrak p$ runs over all prime ideals in the ring of integers in $K$. Therefore,

$$\log (\zeta _K(s))=\sum _{\mathfrak p}\sum _ {m\geq 1} \frac{1}{m(N\mathfrak p)^{ms}}$$

Since $\zeta _K(s)=\frac{1}{s-1}(a_0+a_1(s-1)+\cdots)$ with $a_0\neq 0$, it follows that $\frac{\log (\zeta _K(s))}{\log \frac{1}{s-1}}$ tends to $1$ as $s$ tends to $1$ from the right. On the other hand, in this limit, only the term $m=1$ and $\mathfrak p$ of degree one over $\mathbb Q$ need be considered. Hence we get $$1= \lim _{s \rightarrow 1} \frac{\sum _{\mathfrak p} \frac{1}{(N\mathfrak p) ^s}}{\log \frac{1}{s-1}}.$$ Now, degree $1$ primes $\mathfrak p$ lie over rational primes $p$ which split completely in $K$, and over each $p$, there are $n=\deg(K/{\mathbb Q})$ primes $\mathfrak p$ of degree $1$. Hence we get

$$1=\lim _{s\rightarrow 1} n\left( \frac{\sum _p \frac{1}{p^s}}{\log \frac{1}{s-1}}\right)$$ where the sum is over primes $p$ which split completely. This gives what you want.

• I've added \ to log, lim and deg in the TeXt, I hope you don't mind :) – M.G. Sep 20 '15 at 12:38
• Thank you. Of course I don't mind, and it will help me in future too! – Venkataramana Sep 20 '15 at 12:38
• Why does this proof only work when K/Q is Galois? What would be the main difficulty in extending it? – Campello Oct 26 '15 at 16:57
• If $K/\mathbb Q$ is not Galoisian, then, degree one primes $\mathfrak p$ in $K$ do not correspond to primes in $\mathbb Q$ which split completely in $K$. So the above count does not work – Venkataramana Oct 27 '15 at 0:03
• But then you can consider the closure of K :) – Campello Oct 27 '15 at 9:20