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](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions?rq=1). Apologies all, and thanks for your comments.



How essential is the vanishing of the Dirichlet $L$-functions to Dirichlet's theorem on primes in arithmetic progressions?
**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


