December 23rd
Today I learned the correct context for the zeta function of the ring of integers in a function field. Fix a global field $K$ with a ring of integers $\mathcal O_K.$ The zeta function of interest, then, is\[\zeta_{\mathcal O_K}(s)=\sum_{I\subseteq\mathcal O_K}\frac1{\op{Norm}(I)^s},\]where the sum ranges over nonzero ideal of $\mathcal O_K.$ Here we define $\op{Norm}(I)$ to be $\#\mathcal O_K/\mf p$ for primes and then extend multiplicatively to all ideals by unique prime factorization. This of course matches with how we defined ideal norms in number fields, and the zeta functions also match.
However, the $\zeta$ function for function fields turns out to be a lot nicer. Let $K=\FF_q(t)$ so that $\mathcal O_K=\FF_q[t],$ and abbreviate $\zeta_q:=\zeta_{\mathcal O_K}.$ What makes $\zeta_q$ so nice is that $\mathcal O_K$ is always a principal ideal domain and has finitely many units, which of course is rarely true for number fields.
We show $\mathcal O_K$ is a principal ideal domain for completeness and because it's the core of the argument. Fix any nonzero ideal $I\subseteq\mathcal O_K.$ It's nonzero, so it contains some nonzero polynomial, so it contains some nonzero polynomial of least degree, say $f(t).$ We claim $I=(f(t)).$ Certainly $(f(t))\subseteq I$ because $f(t)\in I.$ In the other direction, if $p(t)\in I,$ then apply Euclidean division (we're in a field) so that\[p(t)=q(t)f(t)+r(t).\]Here $r(t)=p(t)-q(t)f(t)\in I$ is either $0$ or is nonzero with strictly smaller degree than $f.$ The latter is impossible by the minimality of $f,$ so $r(t)=0,$ forcing $p(t)\in(f(t)),$ as desired.
The above tells us that we can represent any nonzero ideal $I$ as $(f)$ for some polynomial in $\mathcal O_K\setminus\{0\}.$ To account for over-counting, we have to deal with units. Note we can make this more precise by forcing $f$ monic by just multiplying whatever $f$ we chose by the inverse of its leading coefficient. Then if $I$ is represented by $(f)$ and $(g),$ then they both have the same degree (by construction) are both monic (without loss of generality), so when we write\[f=ug\]for some polynomial $u\in\mathcal O_K^\times,$ we have to have $\deg u=0$ and $u=1$ to make the leading coefficients match. Therefore $f=g,$ so nonzero ideals are in fact uniquely represented by monic polynomials.
Before touching $\zeta_q,$ we still have to talk about ideal norms. Suppose we have a prime factorization into irreducibles\[f=\prod_{k=1}^r\pi_k.\]All polynomials here are monic to account for leading coefficients. By equating generators,\[\op{Norm}((f))=\op{Norm}\left(\prod_{k=1}^r(\pi_k)\right)=\prod_{k=1}^r\op{Norm}((\pi_k)).\]As usual, $\pi$ being irreducible is equivalent to $(\pi)$ being prime (this is a principal ideal domain), so $\op{Norm}((\pi))=\#\mathcal O_K/(\pi).$ But this is\[\#\frac{\FF_q[t]}{(\pi(t))}\cong\#\FF_{q^{\deg\pi}}=q^{\deg\pi}\]by the construction of finite fields. So we see\[\op{Norm}((f))=\prod_{k=1}^rq^{\deg\pi_k}=q^{\deg\prod\pi_k}=q^{\deg f}.\]So norms are also well-behaved in function fields.
Now we can destroy $\zeta_q.$ Taking hints from the class number formula, we sum ideals by their norm. (If $\zeta_q$ converges, then it absolutely converges.) We know all norms have the form $q^d.$ The number of ideals of norm $q^d$ is the number of monic polynomials of degree $d,$ which is also $q^d$ because there are $d$ remaining coefficients in $\FF_q$ to determine. It follows\[\zeta_q(s)=\sum_{I\subseteq\mathcal O_K}\frac1{\op{Norm}(I)^s}=\sum_{d=0}^\infty\frac{q^d}{q^{ds}}=\sum_{d=0}^\infty\left(q^{1-s}\right)^d=\frac1{1-q^{1-s}},\]where we have simplified using the geometric series formula. This converges provided that $\left|q^{1-s}\right|=q^{1-\op{Re}(s)} \lt 1,$ which happens if and only if $\op{Re}(s) \gt 1.$
However, we now see the function $\frac1{1-q^{1-s}}$ actually provides an analytic continuation of $\zeta_q(s)$ to all of $\CC,$ defined everywhere except at the pole at $s=1.$ We could take the derivative to prove being meromorphic, but we don't bother. Blindly using L'H\^ospital's Rule lets us say the residue of this pole is\[\lim_{s\to1}\frac{s-1}{1-q^{1-s}}=\lim_{s\to1}\frac1{-(-s\log(q))q^{1-s}}=\frac1{\log q}.\]Notably, this does not align with what we said during the class number formula that the residue of this pole should be a constant $\rho$ such that\[\#\{I\subseteq\mathcal O_K:\op{Norm}(I)\le t\}=\rho t+o(t).\]As for why, I think the fact that the ideal norms are so sparsely distributed makes the error term in our estimations die. I'm not sure right now, but I think $\frac1{\log q}$ more or less gives an "average'' value of $\rho$ as $t$ varies.
We even have a Riemann hypothesis because this analytic continuation actually has no zeroes at all, so all of its zeroes in any critical strip lie vacuously on whatever critical line of interest.