11
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Vesselin Dimitrov Sep 27 '15 at 5:19
  • 3
    $\begingroup$ 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/… . $\endgroup$ – Vesselin Dimitrov Sep 27 '15 at 5:21
  • 1
    $\begingroup$ You could have a look to THEOREM 3.4 in Milne's Class Field Theory notes (p. 188). $\endgroup$ – Watson Dec 22 '16 at 8:52
19
$\begingroup$

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.

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

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.