# Analytic equivalents for primes in arithmetic progressions

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.

• 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. Apr 1, 2020 at 7:07
• 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. Apr 1, 2020 at 11:32

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

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.

• 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 $N$. Apr 13, 2020 at 4:47

Here are two equivalences.

Theorem 1. For each $$m \geq 1$$, the following are equivalent.

a) For all nontrivial Dirichlet characters $$\chi \bmod m$$, $$L(1,\chi) \not= 0$$.

b) For all $$a \in (\mathbf Z/m\mathbf Z)^\times$$, the set of primes $$p \equiv a \bmod m$$ has Dirichlet density $$1/\varphi(m)$$.

Proof of Theorem 1. We will will compute the Dirichlet density of $$\{p \equiv a \bmod m\}$$ without assuming (a) and then see why (a) and (b) are equivalent. The trivial Dirichlet character modulo $$m$$ will be written as $$\mathbf 1_m$$.

For each Dirichlet character $$\chi \bmod m$$, $$L(s,\chi)$$ is analytic for $${\rm Re}(s) > 0$$ except for $$L(s,{\mathbf 1}_m)$$ having a simple pole at $$s = 1$$. Set $$n(\chi) := {\rm ord}_{s=1}(L(s,\chi))$$ so $$n({\mathbf 1}_m) = -1$$ and $$n(\chi) \geq 0$$ for all nontrivial $$\chi$$.

For $${\rm Re}(s) > 1$$ and $$(a,m) = 1$$, $$\sum_{p \equiv a \bmod m} \frac{1}{p^s} = \frac{1}{\varphi(m)} \sum_{p}\sum_{\chi} \frac{\chi(p)\overline{\chi}(a)} {p^s} = \frac{1}{\varphi(m)} \sum_{\chi}\overline{\chi}(a)\left(\sum_{p} \frac{\chi(p)} {p^s}\right)$$ where the sum on the right run over all primes $$p$$ and all Dirichlet characters $$\chi \bmod m$$. For a Dirichlet character $$\chi \bmod m$$ and $${\rm Re}(s) > 1$$, $$\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + O(1),$$ where the $$O$$-constant is $$\sum_{p,k \geq 2} 1/(kp^k)$$, so $$\sum_{p \equiv a \bmod m} \frac{1}{p^s} = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1).$$

Now let's bring in the order of vanishing $$n(\chi)$$ above. For $$s$$ near $$1$$, $$L(s,\chi) = (s-1)^{n(\chi)}f_\chi(s)$$ where $$f_\chi(s)$$ is an analytic function in a neighborhood of $$s = 1$$ and $$f_\chi(1) \not= 0$$. Therefore $$f_\chi(s)$$ has an analytic logarithm around $$s = 1$$ (well-defined up to adding an integer multiple of $$2\pi i$$), so for $$s > 1$$, $$\log L(s,\chi) = n(\chi)\log(s-1) + \ell_{f_\chi}(s)$$, where $$\ell_{f_\chi}(s)$$ is a suitable logarithm of $$f_\chi(s)$$. Thus $$\log L(s,\chi) = n(\chi)\log(s-1) + O_\chi(1)$$ for $$s$$ near $$1$$ to the right, and plugging this into the above displayed formula, \begin{align} \sum_{p \equiv a \bmod m} \frac{1}{p^s} & = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(n(\chi)\log(s-1)+ O_\chi(1)) + O(1) \nonumber \\ & =\frac{1}{\varphi(m)}\left(\sum_{\chi} \overline{\chi}(a)n(\chi)\right)\log(s-1) + O_m(1). \end{align}

To compute a Dirichlet density, we want to divide both sides by $$\sum_p 1/p^s$$ for $$s$$ near $$1$$ to the right. For such $$s$$, $$\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) = -\log(s-1) + O(1).$$ Therefore $$\sum_p 1/p^s \sim -\log(s-1)$$ as $$s \to 1^+$$, so dividing through by $$\sum_p 1/p^s$$ and letting $$s \to 1^+$$ gives us $$$$\lim_{s \to 1^+} \frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} = \frac{1}{\varphi(m)}\left(-\sum_{\chi} \overline{\chi}(a)n(\chi)\right),$$$$ which expresses the Dirichlet density of $$\{p \equiv a \bmod m\}$$ in terms of the orders of vanishing $$n(\chi)$$ as $$\chi$$ runs over Dirichlet characters mod $$m$$.

If (a) is true then $$n(\chi) = 0$$ for all nontrivial $$\chi$$, so the right side of the above limit calculation is $$(1/\varphi(m))(-n({\mathbf 1}_m)) = 1/\varphi(m)$$, which is (b).

Conversely, if (b) is true then $$\sum_{\chi} \overline{\chi}(a)n(\chi) = -1$$ for all $$a \in (\mathbf Z/m\mathbf Z)^\times$$ by our limit calculation. Why does this imply $$n(\chi) = 0$$ for nontrivial $$\chi$$?

Using complex vectors indexed by all the Dirichlet characters mod $$m$$, let $${\mathbf n}_m = (n(\chi))_\chi$$ and $${\mathbf v}_a = (\chi(a))_\chi$$ for each $$a \in (\mathbf Z/m\mathbf Z)^\times$$. The space of all complex vectors $$\mathbf z = (z_\chi)_\chi$$ has dimension $$\varphi(m)$$ and it has the Hermitian inner product $$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum_{\chi} z_\chi\overline{w_\chi}$$ for which the vectors $${\mathbf v}_a$$ are an orthonormal basis by the orthogonality relations for Dirichlet characters mod $$m$$. The above displayed formula says $$\langle {\mathbf n}_m,{\mathbf v}_a\rangle = -1/\varphi(m)$$ for all $$a$$ in $$(\mathbf Z/m\mathbf Z)^\times$$, so $${\mathbf n}_m = \sum_{a} \langle {\mathbf n}_m,{\mathbf v}_a\rangle{\mathbf v}_a = -\frac{1}{\varphi(m)}\sum_{a}{\mathbf v}_a.$$ For each nontrivial $$\chi \bmod m$$, the $$\chi$$-component of $$\sum_{a} {\mathbf v}_a$$ is $$\sum_a \chi(a)$$, which is $$0$$. So the $$\chi$$-component of $${\mathbf n}_m$$, which is $$n(\chi)$$, is 0. That is (a).

QED Theorem 1. (I only realized after copying and pasting this that I had already copy and pasted it earlier as an answer to the MO question here.)

Theorem 2. For each $$m \geq 1$$, the following are equivalent.

a) For all Dirichlet characters $$\chi \bmod m$$, $$L(s,\chi) \not= 0$$ when $${\rm Re}(s) = 1$$.

b) $$\sum_{n \leq x} \chi(n)\Lambda(n) = o(x)$$ for nontrivial Dirichlet characters $$\chi \bmod m$$ and $$\sum_{n \leq x} \chi_{{\mathbf 1}_m}(n)\Lambda(n) \sim x$$,

c) For all $$a \in (\mathbf Z/m\mathbf Z)^\times$$, $$|\{p \leq x : p \equiv a \bmod m\}| \sim (1/\varphi(m))x/\log x$$.

Comparing the proof of Theorem 2 below to the sketch in the answer by 2734364041, we will also be using a Tauberian theroem (to prove (b) implies (c)), but we will not need an explicit formula.

Proof of Theorem 2.

We will show (a) is equivalent to (b) and (b) is equivalent to (c).

First we show (a) implies (b). Set $$\psi_\chi(x) = \sum_{n \leq x} \chi(n)\Lambda(n)$$ for all $$\chi$$, so (b) says $$\psi_\chi(x) = o(x)$$ for nontrivial $$\chi$$ and $$\psi_{{\mathbf 1}_m}(x) \sim x$$.

For $$\sigma > 1$$, $$-L'(s,\chi)/L(s,\chi) = \sum \chi(n)\Lambda(n)/n^s$$, for all Dirichlet characters $$\chi \bmod m$$, so $$\psi_\chi(x)$$ is a partial sum of coefficients of $$-L'(s,\chi)/L(s,\chi)$$. Since $$L(s,{\mathbf 1}_m) \not= 0$$ on $$\sigma = 1$$ by (a), $$-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$$ is holomorphic on $$\sigma \geq 1$$ except for a simple pole at $$s = 1$$ with residue 1 and it has nonnegative Dirichlet series coefficients with $$\psi_{{\mathbf 1}_m}(x) = O(x)$$. Therefore $$\psi_{{\mathbf 1}_m}(x) \sim x$$, which is part of (b), by Newman's Tauberian theorem. To get the rest of (b), namely $$\psi_\chi(x) = o(x)$$ for nontrivial $$\chi$$, we have $$-L'(s,\chi)/L(s,\chi)$$ being holomorphic on $$\sigma \geq 1$$ by (a) and its Dirichlet series coefficients satisfy $$|\chi(n)\Lambda(n)| \leq {\mathbf 1}_m(n)\Lambda(n)$$ for all $$n$$, so $$\psi_\chi(x) = o(x)$$ by a corollary of Newman's Tauberian theorem for $$-L'(s,\chi)/L(s,\chi)$$ using comparison Dirichlet series $$-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$$ .

Thus (a) implies (b).

To show (b) implies (a), we will use the following fact. For a function $$a(x)$$ on $$[1,\infty)$$ that is bounded and Riemann integrable on $$[1,T]$$ for all $$T \geq 1$$, so $$f(s) := \int_1^\infty (a(x)/x^s)dx/x$$ is absolutely convergent on $$\sigma > 1$$, if $$a(x) \to 0$$ as $$x \to \infty$$ and $$f$$ extends to a meromorphic function on $$\sigma = 1$$ then $$f$$ in fact is holomorphic on $$\sigma = 1$$. (This is used to show the condition $$\psi(x) \sim x$$ implies $$\zeta(s) \not= 0$$ on $$\sigma = 1$$ by using $$a(x) = \psi(x)/x - 1$$.) Because of the integral representations $$-\frac{L'(s,\chi)}{sL(s,\chi)} = \int_1^\infty \frac{\psi_\chi(x)}{x} \frac{dx}{x^s}$$ and $$-\frac{L'(s,{\mathbf 1}_m)}{sL(s,{\mathbf 1}_m)} - \frac{1}{s-1} = \int_1^\infty \left(\frac{\psi_{\mathbf 1}(x)}{x} -1\right)\frac{dx}{x^s},$$ for $${\rm Re}(s) > 1$$, where $$\chi$$ is nontrivial in the first equation. we can use the above fact when $$a(x) = \psi_\chi(x)/x$$ for nontrivial $$\chi$$ and $$a(x) = \psi_{{\mathbf 1}_m}(x)/x - 1$$ to conclude that $$L'(s,\chi)/L(s,\chi)$$ is holomorphic on $$\sigma = 1$$ for nontrivial $$\chi$$ and $$L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$$ is holomorphic on $$\sigma = 1$$ except for a simple pole at $$s = 1$$, so $$L(s,\chi)$$ is nonvanishing on $$\sigma = 1$$ and $$L(s,{\mathbf 1}_m)$$ is nonvanishing on $$\sigma = 1$$. Thus (b) implies (a).

That (b) implies (c) follows from the above integral representation of $$-L'(s,\chi)/L(s,\chi)$$ for all nontrivial $$\chi$$ by a standard method to prove (c).

Our last step is showing (c) implies (b). Set $$\pi(x;a \bmod m) = |\{p \leq x : p \equiv a \bmod m\}|$$ when $$(a,m) = 1$$ and $$\pi_\chi(x) = \sum_{p \leq x} \chi(p)$$, where $$\chi$$ is a Dirichlet character mod $$m$$. Write $$\chi$$ as a linear combination of delta-functions on $$(\mathbf Z/m\mathbf Z)^\times$$: $$\chi = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a$$. Then \begin{align*} \pi_\chi(x) & = \sum_{p \leq x} \chi(p) \\ & = \sum_{p \leq x} \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a(p) \\ & = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\left(\sum_{p \leq x} \delta_a(p)\right) \\ & = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\pi(x; a \bmod m), \end{align*} so $$\frac{\pi_\chi(x)}{x/\log x} = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\frac{\pi(x;a \bmod m)}{x/\log x}.$$ By (c), as $$x \to \infty$$ the right side tends to $$\sum_{a \in ({\mathbf Z}/m{\mathbf Z})^\times} \chi(a)/\varphi(m)$$, which is 0 if $$\chi$$ is nontrivial. Therefore when $$\chi$$ is nontrivial we have $$\pi_\chi(x) = o(x/\log x)$$, which implies $$\psi_\chi(x) = o(x)$$ by the same argument that $$\pi(x) \sim x/\log x$$ implies $$\psi(x) \sim x$$. To show (c) implies $$\psi_{{\mathbf 1}_m}(x) \sim x$$, sum the relation in (c) over all $$a$$ in $$(\mathbf Z/m\mathbf Z)^\times$$ to get $$\pi(x) \sim x/\log x$$, the Prime Number Theorem, which is equivalent to $$\psi(x) \sim x$$, so $$\psi_{{\mathbf 1}_m}(x) \sim x$$ since $$\psi_{{\mathbf 1}_m}(x) = \psi(x) + O_m(\log x)$$.

QED Theorem 2.

• 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.