6
$\begingroup$

By way of context: it is known that the prime number theorem $\pi(x) \sim x/\log x$ is (nontrivially) equivalent to the statement that $\zeta(s)$ does not vanish on the line $\Re s=1$.

I would like to have it clarified what are the analogous equivalent statements for primes in arithmetic progressions. In particular, I would love to have specific references to papers or books in which the following equivalences are proved (or at least stated explicitly):

  • Dirichlet's theorem, that there are infinitely many primes in any admissible arithmetic progression (mod $q$). is equivalent to the nonvanishing of $L(1,\chi)$ for all Dirichlet characters $\chi\pmod q$. [Edit: as has been pointed out, this nonvanishing is more likely equivalent to the equality of the Dirichlet densities of primes among the reduced residue classes (mod $q$). What analytic statement could be equivalent to the mere infinitude of primes in all such classes?]
  • The prime number theorem in arithmetic progressions $\pi(x;q,a)\sim x/(\phi(q)\log x)$ is equivalent to the statement that $L(s,\chi)$ does not vanish on the line $\Re s=1$ for all Dirichlet characters $\chi\pmod q$.

Of course, if these statements themselves are incorrect, I would like to be corrected as well as being pointed to the literature.

$\endgroup$
2
  • 3
    $\begingroup$ I am confused. Dirichlet's theorem, the PNT in arithmetic progressions, and the nonvanishing of $L(s,\chi)$ for $\Re(s)=1$ are three true statements. True statements are always equivalent. $\endgroup$ – GH from MO Apr 1 '20 at 7:07
  • 4
    $\begingroup$ One sense in which such statements are equivalent is in that analogous statements are known to be equivalent in greater generality, for instance for L-functions in the Selberg class. You can deduce equivalence for PNT and for PNT in APs this way. As for Dirichlet's Theorem, I doubt it's equivalent to nonvanishing at 1. Instead, I would expect nonvanishing to be equivalent to the result about Dirichlet density of primes in APs. $\endgroup$ – Wojowu Apr 1 '20 at 11:32
3
$\begingroup$

Regarding Dirichlet's Theorem, I think you need the quantitative version (in terms of Dirichlet density) for the equivalence to hold.

Suppose $$\sum_{\substack{p\equiv a(q)\\ p<x}} \frac1p = \frac{\chi_0(a)}{\phi(q)} \log\log x+O(1).$$

Then summing by parts gives for $s=1+\delta$ that, uniformly in $\delta$,

\begin{align*} \sum_p \chi(p)p^{-s} &= \sum_{a(q)} \chi(a)\sum_{p\equiv a(q)} p^{-s} \\ &= \sum_{a(q)} \chi(a) \sum_n \bigg(\frac1{\phi(q)}\log\log n +O(1)\bigg)\big(n^{-\delta}-(n+1)^{-\delta}\big) \\ &= \left(\sum_{a(q)} \chi(a)\right) \left (\sum_n \frac{\log\log n}{\phi(q)}\big(n^{-\delta}-(n+1)^{-\delta}\big) \right) \\ &\qquad{}+ \sum_{a(q)} \chi(a) \sum_n O(1)\big(n^{-\delta}-(n+1)^{-\delta}\big) \\ &\ll \sum_{n\geq 1} \big(n^{-\delta}-(n+1)^{-\delta}\big) \ll 1 \end{align*}

And since $\log L(s,\chi) = \sum_p \chi(p)p^{-s} + O(1)$ we see that $L(1,\chi)\neq 0$.

On the other hand if you only told me there were infinitely many primes in each AP, but with different asymptotics for different $a$ mod $q$, I don't think the conclusion would follow.

I have no idea if this is written up anywhere but I'll now add it to the homework in my course notes.

$\endgroup$
1
  • 1
    $\begingroup$ In (2.27) of Iwaniec and Kowalski's book, it is explained how the divergence of $\sum 1/p$ over just the one arithmetic progression $1 \mod{N}$ yields at once the non-vanishing of all $L(1,\chi)$ to the conductor $N$. $\endgroup$ – Vesselin Dimitrov Apr 13 '20 at 4:47
1
$\begingroup$
  • For $\pi(x;q,a)\sim x/(\varphi(q)\log x)$: I don't know of a source that states such results for progressions to modulus $q$ except on p. 40 of Iwaniec and Kowalski, and only for $q=1$. Davenport does not seem to say anything on the matter. My next guess would be to check Montgomery and Vaughan or Chandrasekharan. One direction follows from the explicit formula, and the other direction follows from applying the Weiner-Ikehara tauberian theorem to each $-L'/L(s,\chi)$ and combining them via the orthogonality relations.

  • For the infinitude of primes in progressions, we have (for $s>1$)

$\displaystyle \sum_{p\equiv a\pmod{q}}p^{-s} = \frac{1}{\varphi(q)}\log\frac{1}{s-1}+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod{q} \\ \chi\neq 1}}\bar{\chi}(a)\log L(s,\chi)+O_q(1)$

The boundedness of $\log L(s,\chi)$ for all $s>1$ is equivalent to the nonvanishing of $L(1,\chi)$ by means of Dirichlet's test for uniform convergence. Thus

$\displaystyle \sum_{\substack{p\equiv a\pmod{q}}}p^{-s}=\frac{1}{\varphi(q)}\log\frac{1}{s-1}+O_q(1)$ for all $s>1\iff L(1,\chi)\neq 0$ for all $\chi\pmod{q}$.

(See Chapter 1 of Davenport.) But this statement is stronger than merely the infinitude of primes in progressions.

$\endgroup$

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.