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As mentioned by @MichaelZieve in his comment re Quadratic residue, Chebotarev's density theorem was preceded by an allegedly much easier theorem of Frobenius (Mike Zieve is certainly not the only one to mention that the Frobenius theorem is much easier than Chebotarev) -- the difference between the two theorems is that Frobenius talks of conjugacy in $S_n,$ while Chebotarev of conjugacy in the Galois group itself. Does anyone know what the "easier" argument is? Is there a good reference?

EDIT If you look at the theorem in Janusz's book it proves the theorem for the Dirichlet density. Does proving it for the natural density require the full Chebotarev machine?

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I just changed the link, since it pointed neither to the question nor the answer you mention, but to an other answer. –  quid Jul 7 '13 at 19:40
    
@quid: thanks..! –  Igor Rivin Jul 7 '13 at 21:20
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@IgorRivin: To pass from Dirichlet density to natural density, you just need powerful enough tauberian theorems. I would bet that Newman's "easy tauberian theorem" (see Zagier's AMM paper) is enough for Frobenius. –  ACL Mar 6 at 1:41
    
@ACL So are you saying (for example) that to get from Dirichlet's theorem on primes in progressions to a natural density statement you just need a Tauberian argument? I thought (just as Speyer does, apparently) that you needed more analytic information on the $L$-functions involved. –  Igor Rivin Mar 6 at 12:50
    
I say exactly that. I wrote the details in a mid-graduate course on number theory (4th year univ. in France). See the text on math.u-psud.fr/~chambert/enseignement/2007-08/h4/coursh4.pdf –  ACL Mar 6 at 12:59
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2 Answers

up vote 10 down vote accepted

You can find Frobenius's theorem (and proof) on p.134 of Janusz, Algebraic Number Theory, 1973, and many other places. It really is easier. It was known long before Chebotarev's theorem, and Chebotarev had to come up with new ideas to prove his theorem (which in turn helped Artin prove his reciprocity law).

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Thanks! I am not sure about "many other places", but this certainly works.... –  Igor Rivin Jul 8 '13 at 0:28
    
@Igor: I'm not aware of any modern book besides Janusz which contains Frobenius's density theorem. It's hard to believe there could be no others, but I've looked in maybe 40 other books on algebraic number theory and can't recall seeing it elsewhere. –  Michael Zieve Jul 10 '13 at 13:04
    
@Michael, yes, Google search is similarly unproductive (you would think that someone would teach it in a course). Weird, since for computing Galois groups that's the only thing people use. –  Igor Rivin Jul 10 '13 at 14:09
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Regarding the edit: I think natural density should still be easier for Frobenius. Let $G$ be $\mathrm{Gal}(F/\mathbb{Q})$ and let $c$ be a class function $G \to \mathbb{C}$.

Cebatarov with natural density is the statement $$ \sum_{p \leq N} c(\mathrm{Frob}(p)) \sim \left( \sum_{g \in G} c(g) \right) \cdot \pi(N) $$ Frobenius is the special case of a class function where $c(g)=c(h)$ whenever $g$ and $h$ generate the same cyclic subgroup.

In both cases, the set of such class functions is a vector space, so it is enough to prove this for a spanning set of class functions. In the Cebatarov case, the spanning set is inductions of one dimensional characters of subgroups $H$ of $G$. In the Frobenius case, it is enough to induct only trivial characters.

Let $H$ be a subgroup of $G$, let $L$ be the fixed field and let $\chi: H \to \mathbb{C}^{\ast}$ be a character. Let $K \subset H$ be the kernel of $\chi$ and let $M$ be the fixed field of $K$, so $M/L$ is an abelian extension.

Then I believe (this is the part I am not sure of) that Cebatarov for $\mathrm{Ind}_H^G \chi$ should reduce to showing that $L(M/L, \chi)(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except for a simple pole at $s=1$ for $\chi$ trivial. (For Dirichlet density, we only need to show this at $s=1$.) Frobenius amounts proving this only in the case that $\chi$ is trivial. In this case, we are trying to show that $\zeta_L(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except a simple pole at $s=1$.

So the key question is, are the easiest proofs that $\zeta_L(s)$ has no zeroes or poles on the critical line easier than the corresponding proofs for $L(M/L, \chi)(s)$? I think the answer is yes. The proofs that I am aware involve two main parts

(1) Show that the function in question has no poles (other than the stated one at $s=1$) on $\mathrm{Re}(s)>0$.

(2) Write down some clever product of various $L$-functions which is manifestly non-vanishing, and conclude that none of the factors can vanish. This argument requires (1) to have been already done, to rule out the possibility that a zero and pole cancel.

Now, I don't know whether (2) gets significantly easier when $\chi$ is trivial. But (1) definitely does! The fact that $\zeta_H(s)$ extends to $\{ \mathrm{Re}(s)>0,\ s \neq 1 \}$ follows from the same sort of computations that lead to the class number formula for number fields. For $L$ functions, the way I know how to do it requires first proving that the character $\chi$ on $\mathcal{O}_L$ is periodic, so that we can group together sums over a complete period and guarantee that they converge. This is equivalent to Artin reciprocity, which as far as I know is as hard as all of class field theory.

Chebatarov's original proof (see the last section of Lenstra and Stevenhagen) avoids this. To my limited understanding of this, Cebatarov first proves the result when $M$ is a cyclotomic extension of $L$, where Artin reciprocity is easy, and then some how cleverly reduces to this case. But, even so, it seems to me that it must be easier when we work with just trivial characters.

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