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I seem to recall that the prime number theorem (PNT) is equivalent to the fact that the Riemann zeta function $\zeta(s)$ is non-zero on all of $\text{Re}(s) = 1$ (see https://math.stackexchange.com/questions/1379583/why-is-zeta1it-neq-0-equivalent-to-the-prime-number-theorem or https://math.stackexchange.com/questions/706934/riemann-zeta-function-non-vanishing-on-the-line-mathrmre-z-1), where I suppose "equivalent" means that PNT and the non-vanishing of $\zeta(s)$ along $\text{Re}(s)=1$ can be both be proven in a relatively short manner from the other.

Is it the case that Dirichlet's theorem on primes in arithmetic progressions is also equivalent to the fact that all the Dirichlet $L$-functions $L(s,\chi)$ are non-zero at $s=1$, in any sense similar to the equivalence described above? Perhaps another way of phrasing the question is: is it "possible" that $L(s,\chi)$ could equal 0 (for some non-trivial character $\chi$) but still Dirichlet's theorem could hold?

It has come to my attention that this has already been discussed here Analytic equivalents for primes in arithmetic progressions. Apologies all, and thanks for your comments.

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    $\begingroup$ For Dirichlet's theorem on the infinitude of primes in arithmetic progressions, you only need that $L(1,\chi) \neq 0$ for each Dirichlet character $\chi$. For the prime number theorem in arithmetic progressions, you need $L(1 + it,\chi) \neq 0$ for all $t \in \mathbb{R}$. $\endgroup$ Sep 11, 2021 at 4:08
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    $\begingroup$ To clarify the previous comment, “Dirichlet’s theorem for primes in arithmetic progressions” means $\{p : p \equiv a \bmod m\}$ is infinite whenever $\gcd(a,m) = 1$, where $p$ runs over primes, and “the prime number theorem for arithmetic progressions” means $|\{p\leq x:p\equiv a \bmod m\} \sim (1/\varphi(m))x/\log x$ as $x\to\infty$ whenever $\gcd(a,m)=1$, where $p$ runs over primes. $\endgroup$
    – KConrad
    Sep 11, 2021 at 6:12
  • $\begingroup$ @PeterHumphries thank you and KConrad for your comments, but I did not mean "a version of Dirichlet's theorem for arithmetic progressions (DTAP) in the form of the PNT" (which would be the result you mentioned); I meant "a version of the equivalence between DTAP and the non-vanishing of the $L$-functions at $s=1$ in the form of the equivalence between PNT and the non-vanishing of $\zeta(s)$ along $\text{Re}(s)=1$". $\endgroup$
    – D.R.
    Sep 11, 2021 at 6:47
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    $\begingroup$ I don't think they are quite equivalent in any reasonable way. Nonvanishing of the L-functions implies not just infinitude of primes in APs, but also that their logarithmic/Dirichlet density is positive. One plausible way to make this into a formal statement is to say that for general L-functions (say in a Selberg class) nonvanishing at a point is equivalent to some density result. An analogous result holds for PNT statements, and I believe is due to Kaczorowski et al. $\endgroup$
    – Wojowu
    Sep 11, 2021 at 15:41
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    $\begingroup$ @Wojowu please see my answer below. There is no need to prove a sweeping claim about everything in the Selberg class to show a certain density version of Dirichlet's theorem is equivalent to nonvanishing of certain Dirichlet $L$-functions at $1$. Note that the equivalence is not with mere positivity of the Dirichlet density, but with the expected value of it (namely $1/\varphi(m)$). $\endgroup$
    – KConrad
    Sep 11, 2021 at 21:45

1 Answer 1

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Theorem: Fix a positive integer $m$. The following two conditions are equivalent:

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

(2) 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. When $m$ is $1$ or $2$, the condition (1) is vacuously true (there are no nontrivial Dirichlet characters mod $m$) and condition (2) is obvious, so from now on we can let $m \geq 3$.

The usual proof of Dirichlet's theorem shows (1) implies (2). It remains to show (2) implies (1).

For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except that $L(s,\mathbf 1_m)$ has a simple pole at $s = 1$, where $\mathbf 1_m$ denotes the trivial character mod $m$ (we have $L(s,\mathbf 1_m) = \zeta(s)\prod_{p \mid m} (1-1/p^s)$ and $\zeta(s)$ is analytic on ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$). Set $$ r_\chi := {\rm ord}_{s=1}(L(s,\chi)) $$ so $r_{\mathbf 1_m} = -1$ (the simple pole at 1) and $r_\chi \geq 0$ for all nontrivial $\chi$. Condition (1) is equivalent to $r_\chi = 0$ for all nontrivial $\chi$, so we want to show condition (2) implies the numbers $r_\chi$ vanish for all nontrivial $\chi$.

For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$, define $$ \log L(s,\chi) := \sum_{p,k} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}}. $$ This is a logarithm of $L(s,\chi)$ (meaning the exponential of that series is $L(s,\chi)$. The second series on the right is absolutely convergent for ${\rm Re}(s) > 1/2$, with absolute value bounded by $\sum_{p,k \geq 2} 1/(kp^{k\sigma})$, so for $s > 1$ we can say $$ \log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + O(1), $$ where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$. In the usual proof of Dirichlet's theorem, for each $a \in (\mathbf Z/m\mathbf Z)^\times$ and $s > 1$ we write $$ \sum_{p \equiv a \bmod m} \frac{1}{p^s} = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\left(\sum_p \frac{\chi(p)}{p^s}\right), $$ so $$ \sum_{p \equiv a \bmod m} \frac{1}{p^s} = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1), $$ where the $O$-constant on the right is an overall term (outside the sum).

Now let's bring in the order of vanishing $r_\chi$. For all $s$ near $1$, $L(s,\chi) = (s-1)^{r_\chi}f_\chi(s)$ where $f_\chi(s)$ is a holomorphic function in a neighborhood of $s = 1$ (in fact it is holomorphic on ${\rm Re}(s) > 0$, or even $\mathbf C$) and $f_\chi(1) \not= 0$. Therefore $f_\chi(s)$ has a logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi$), so for $s > 1$ $$ \log L(s,\chi) = r_\chi\log(s-1) + \ell_{f_\chi}(s) $$ where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. Thus $\log L(s,\chi) = r_\chi\log(s-1) + O(1)$ for $s$ near $1$ to the right, and plugging this into the formula for $\sum_{p \equiv a \bmod m} 1/p^s$ we can say $$ \sum_{p \equiv a \bmod m} \frac{1}{p^s} = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(r_\chi\log(s-1)) + O(1). $$ Let's extract the term for the trivial character mod $m$: since $r_{\mathbf 1_m} = -1$, $$ \sum_{p \equiv a \bmod m} \frac{1}{p^s} = -\frac{1}{\varphi(m)}\log(s-1) + \frac{1}{\varphi(m)}\left(\sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right)\log(s-1) + O(1). $$

In order to bring in a Dirichlet density, we want to divide both sides by $\sum_p 1/p^s$ for $s$ near $1$ to the right. For such $s$, $$ \log \zeta(s) = -\log(s-1) + O(1) $$ from the simple pole of $\zeta(s)$ at $s = 1$ and $$ \log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) $$ from the Euler product for $\zeta(s)$ when $s > 1$. Therefore $\sum_p 1/p^s = -\log(s-1) + O(1)$ as $s \to 1^+$, so $-\log(s-1) \sim \sum_p 1/p^s$ as $s \to 1^+$. Dividing through the last (big) formula above for $\sum_{p \equiv a \bmod m} 1/p^s$ by $\sum_p 1/p^s$ and letting $s \to 1^+$, we get $$ \frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} \to \frac{1}{\varphi(m)} - \frac{1}{\varphi(m)}\left(\sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right) $$ as $s \to 1^+$. So we have shown, without assuming condition (1) in the theorem, that for all $a \in (\mathbf Z/m\mathbf Z)^\times$ the set of primes $\{p \equiv a \bmod m\}$ has Dirichlet density $$ \frac{1}{\varphi(m)}\left(1 - \sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right). $$

Finally it is time to assume condition (2) in theorem, which implies $$ \sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi = 0 $$ for all $a \in (\mathbf Z/m\mathbf Z)^\times$. When $\chi = \mathbf 1_m$, $\overline{\chi}(a)r_\chi = 1(-1) = -1$, so condition (2) implies $$ \sum_{\chi} \overline{\chi}(a)r_\chi = -1 $$ for all $a \in (\mathbf Z/m\mathbf Z)^\times$, where the sum runs over all Dirichlet characters mod $m$ (including the trivial character). We want to show the above equation, for all $a$, implies $r_\chi = 0$ for all nontrivial $\chi \bmod m$.

Using vectors indexed by the Dirichlet characters mod $m$, let $\mathbf r_m = (r_\chi)_\chi$ 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 a 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 usual orthogonality of Dirichlet characters mod $m$. The equation $\sum_\chi \overline{\chi}(a)r_\chi = -1$ above says $\langle \mathbf r_m,\mathbf v_a\rangle = -1/\varphi(m)$ for all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so $$ \mathbf r_m = \sum_{a} \langle \mathbf r_m,\mathbf v_a\rangle\mathbf v_a = -\frac{1}{\varphi(m)}\sum_{a}\mathbf v_a. $$ For nontrivial Dirichlet characters $\chi \bmod m$, the $\chi$-component of $\sum_{a} \mathbf v_a$ is
$\sum_a \chi(a)$, which is $0$ (the $\mathbf 1_m$-component is $\varphi(m)$, but that's irrelevant). Since $\mathbf r_m$ has $\chi$-component $r_\chi := {\rm ord}_{s=1}L(s,\chi)$, we have $r_\chi = 0$ for all nontrivial $\chi$, so $L(1,\chi) \not= 0$ for all nontrivial $\chi$. QED

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    $\begingroup$ Nice answer. But I'd like to revisit Wojowu's remark about the Selberg class. I always feel some vague discomfort with claims that two theorems are equivalent. Not that I think that there's anything wrong with such a claim, but I'm always left wondering if there's some stronger sense of equivalence that can be proved. In the example at hand, I wonder if the equivalence can be proved for various kinds of fake integers? $\endgroup$ Sep 12, 2021 at 3:24
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    $\begingroup$ @TimothyChow do you have discomfort now with the statement of the Riemann Hypothesis (about nontrivial zeros lying on a line) being proved "to be equivalent" to each of a long list of other statements (a bound on $|\pi(x) - {\rm Li}(x)|$, a bound on the growth of $\sum_{n \leq x} \mu(n)$, a half-plane of convergence of $\sum_{n \geq 1} \mu(n)/n^s$, etc.), and if not would you have discomfort with such claims of equivalence in a hypothetical future where RH is proved? $\endgroup$
    – KConrad
    Sep 12, 2021 at 4:29
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    $\begingroup$ I don't have discomfort now with such statements. The discomfort would kick in after RH is proved. A claim that "RH is equivalent to X" is implicitly a claim that the equivalence is easier to prove than "Axioms for math imply RH." As long as nobody knows how to prove "Axioms for math imply RH," we have a more or less objective basis for claiming that "RH is equivalent to X" is easier. Once RH is proved, it becomes a judgment call. Admittedly, we expect that RH will be very hard to prove; if so, then the judgment call may be a no-brainer. But there is still a qualitative difference. $\endgroup$ Sep 12, 2021 at 11:49
  • $\begingroup$ But this is a bit of a digression. In the case at hand, it's not the distinction between a conjecture and a theorem that's at play. For the sake of argument, let's say that I had replied that I experienced discomfort with "RH is equivalent to X." Then the question still stands whether a stronger statement can be proved in the case at hand. $\endgroup$ Sep 12, 2021 at 11:54
  • $\begingroup$ Maybe a better example to have used than equivalences with RH would be equivalences for an elliptic curve over $\mathbf Q$ to be modular. See the 6 conditions in the middle of the third page of the paper proving the modularity theorem (ams.org/journals/jams/2001-14-04/S0894-0347-01-00370-8/…). The equivalence among those conditions was known before modularity could be proved outside of individual examples (i.e., pre-Wiles). [contd] $\endgroup$
    – KConrad
    Sep 12, 2021 at 16:24

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