Today I Learned

May 12th

Today I learned an alternative presentation of the Dirichlet form of the Chebotarev application from a few days ago, from here . Namely, we show the following.

We do this by showing the following stronger statement.

Indeed, this statement would let us say that the numerator of\[\lim_{s\to1^+}\left(\frac{\sum_p\frac{\omega_f(p)}{p^s}}{\sum_p\frac1{p^s}}-1\right)=\lim_{s\to1^+}\frac{\sum_p\frac{\omega_f(p)-1}{p^s}}{\sum_p\frac1{p^s}}\]is in fact finite as $s\to1^+,$ implying that the total limit vanishes. (Note that for given $s \gt 1,$ the sums in the numerator all absolutely converge because $\omega_f(p)\le\deg f$ is $O(1),$ so there is no need to worry about the rearrangements.) This gives the theorem. $\blacksquare$

It remains to show the stronger lemma; we focus on the $\omega_f(p)$ part of the sum because it is the part we don't understand. Like with our last proof, fix $K:=\QQ(\alpha)$ of degree where $\alpha$ is a root of $f(x).$ The main idea is to note that, for all but finitely many $p,$ we can associate the prime factorisation of $(p)$ in $\mathcal O_K$ with the factorization of $f(x)\pmod p.$

In particular, for all but finitely many primes, we may ignore ramification and say that the number of linear factors in $f(x)\pmod p$ is the number of degree-$1$ primes above $(p).$ But now we say that this means, for sufficiently large $x$ covering these cases,\[\sum_{p\le x}\frac{\omega_f(p)}p=\sum_{\substack{\op N(\mf p)\le x\\f(\mf p/p)=1}}\frac1{\op N(\mf p)}+C_{DK},\]where $C_{DK}$ comes from the finitely many primes from Dedekind-Kummer or ramification. However, degree-$1$ primes are really the only primes of substance for our measure, for\[\sum_p\sum_{f(\mf p/p)\ge2}\frac1{\op N(\mf p)}\le\sum_p\frac{\deg f}{p^2} \lt \deg f\sum_{n=1}^\infty\frac1{n^2} \lt \infty.\]In particular, we can write\[\sum_p\sum_{f(\mf p/p)\ge2}\frac1{\op N(\mf p)}-\sum_p\sum_{\substack{\op N(\mf p)\le x\\f(\mf p/p)\ge2}}\frac1{\op N(\mf p)}=\sum_p\sum_{\substack{\op N(\mf p) \gt x\\f(\mf p/p)\ge2}}\frac1{\op N(\mf p)},\]and we can bound this sum as\[\sum_p\sum_{\substack{\op N(\mf p) \gt x\\f(\mf p/p)\ge2}}\frac1{\op N(\mf p)}\le\sum_{p\ge x}\frac{\deg f}{p^2} \lt \deg f\sum_{n \gt x}\frac1{n^2} \lt \deg f\sum_{n \gt x}\left(\frac1{n-1}-\frac1n\right),\]which is $O(1/x)$ by telescoping. We said all of this to say that\[\sum_{\substack{\op N(\mf p)\le x\\f(\mf p/p)=1}}\frac1{\op N(\mf p)}=\sum_{\op N(\mf p)\le x}\frac1{\op N(\mf p)}+C_f+O(1/x),\]where $C_f:=\sum_p\sum_{f(\mf p/p)\ge2}\frac1{\op N(\mf p)}$ comes from the primes of degree larger than $1.$ Putting this back together,\[\sum_{p\le x}\frac{\omega_f(p)}p=\sum_{\op N(\mf p)\le x}\frac1{\op N(\mf p)}+C_f+C_{DK}+O(1/x).\]It remains to bound that sum reasonably.

For this we pick up Merten's second theorem in the number field case.

We are not going to prove this rigorously here (because I don't know its proof), but I will provide a reference here . From the proof I have skimmed, it looks like the Prime number theorem for number fields makes quick work of this. $\blacksquare$

Anyways, from this it follows\[\sum_{p\le x}\frac{\omega_f(p)}p=\log\log x+C_K+C_f+C_{DK}+o(1),\]where we have sadly had to increase the error term. On the other hand, Merten's over $K=\QQ$ tells us\[\sum_{p\le x}\frac1p=\log\log x+C_\QQ+o(1)\]as well. Finally taking the difference, we see that\[\sum_{p\le x}\frac{\omega_f(p)-1}p=C_K+C_f+C_{DK}-C_\QQ+o(1).\]This is exactly the convergence we required. $\blacksquare$