Presentation: ANT Spring 2019

This is a presentation I gave for my Algebraic Number Theory class during the Spring 2019 semester. In it I prove the functional equation for the Riemann zeta function for global function fields of transcendence degree one. The written version here quite closely resembles the talk I gave, so the edits are slim to none. The argument presented here is a more detailed combination of section 2.0.3 here and the proof presented in Rosen’s Number Theory in Function Fields. An understanding of the Riemann-Roch theorem and its consequences, divisor classes, and finiteness of the class number are assumed. If you are unfamiliar with these ideas or need a refresher, see chapter 5 of Rosen’s text.

Presentation: ANT Spring 2019

Using the Riemann-Roch theorem we’re going to prove the functional equation for the $\zeta$-function associated function fields of transcendence degree one over $\mathbb{F}_{q}$. As a consequence we will show that the $\zeta$-function satisfies a rational function in $u = q^{-s}$.

Theorem (The Functional Equation): Let $K$ be a function field of transcendence degree one over $\mathbb{F}_{q}$, and let $\zeta_{K}$ be the corresponding $\zeta$-function. Define $\xi(s) = q^{(g-1)s}\zeta_{K}(s)$ where $g$ is the genus of $K$. Then

$\xi(s) = \xi(1-s)$.

proof: Let $\deg(A) = n$, so that if $NA$ is the norm of $A$, $NA^{-s} = q^{-ns}$. Summing over $n \ge 0$ and making the substitution $u = q^{-s}$ we may write

$\zeta_{K}(s) = \sum_{n \ge 0}b_{n}u^{n} := Z_{K}(u) \qquad \zeta_{K}(1-s) = \sum_{n \ge 0}b_{n}\left(\frac{1}{qu}\right)^{n} := Z_{K}\left(\frac{1}{qu}\right)$,

where we define $b_{n}$ to be the number of effective divisors $A \in \mathcal{D}_{K}$ of degree $n$ (a divisor $A$ is effective if $A \ge 0$). The functional equation now takes the form

$u^{1-g}Z_{K}(u) = \left(\frac{1}{qu}\right)^{1-g}Z_{K}\left(\frac{1}{qu}\right)$.

Now notice if we substitute $\frac{1}{qu}$ for $u$ on the left-hand side we get the right-hand side. Denoting the left-hand side as $F'(u)$, the theorem is equivalent to showing $F'(u) = F'(\frac{1}{qu})$. To make the computations easier to work with we’re going to show the equivalent statement $F(u) = F(\frac{1}{qu})$ where $F(u) = (q-1)F'(u)$. We will accomplish this by rewriting the sum on the left-hand side as a sum over divisor classes and using Riemann-Roch. Define $Cl_{K}^{+}$ to be the set of all divisor classes $\overline{A}$ such that $\deg(\overline{A}) \ge 0$ (we can do this because all the divisors in a class have the same degree) and let $Cl_{K}^{i}$ be the set of all divisor classes $\overline{A}$ with degree $0 \le \overline{A} \le i$. Then we have the following:

\begin{aligned}[t] F(u) &\overset{1}{=} (q-1)u^{1-g}\sum_{n \ge 0}b_{n}u^{n} \\ &\overset{2}{=} (q-1)u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{+}}\frac{q^{l(\overline{A})-1}}{q-1}u^{\deg(\overline{A})}\right) \\ &\overset{3}{=} u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{+}}q^{l(\overline{A})}u^{\deg(\overline{A})}-\sum_{\overline{A} \in Cl_{K}^{+}}u^{\deg(\overline{A})}\right) \\ &\overset{4}{=} u^{1-g}\left(\sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})}+\sum_{\{\overline{A} \mid \deg(\overline{A}) \ge 2g-1\}}q^{l(\overline{A})}u^{\deg(\overline{A})}-\sum_{\overline{A} \in Cl_{K}^{+}}u^{\deg(\overline{A})}\right) \\ &\overset{5}{=} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1}+h_{k}\left(\sum_{n = 2g-1}^{\infty}q^{n-g+1}u^{n-g+1}-\sum_{n = 0}^{\infty}u^{n-g+1}\right) \\ &\overset{6}{=} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1}+h_{k}\left(\frac{q^{g}u^{q}}{1-qu}-\frac{u^{1-g}}{1-u}\right) \end{aligned}

The first equality is by definition $F(u)$. To get the second, sum over effective divisors instead of degree and note that the number of effective divisors in $\overline{A}$ is exactly $\frac{q^{l(\overline{A})-1}}{q-1}$ where $l(\overline{A})$ is the dimension of $\overline{A}$, really any representative, as a vector space over $\mathbb{F}_{q}$. Use the $q-1$ term to clear denominators to get the third. Breaking up the first sum by degree $2g-2$ gives the fourth. The fifth follows since there are always $h_{K} = |Cl_{K}^{0}|$ classes of degree $n$ ($h_{K}$ is independent of $n$), and a corollary of Riemann-Roch which says that if $\deg(\overline{A}) \ge 2g-1$ then $l(\overline{A}) = n-g+1$. Reindexing the middle sum to start from $n = 0$ and noticing the two latter sums are infinite geometric series which converge because $u = q^{-s}$ with $\Re(s) > 1$ yields the sixth equality.

A direct substitution $u \mapsto \frac{1}{qu}$ will show that the second term remains unchanged. In fact, the two terms in the difference switch under this substitution. So, all that remains to show is the summation remains unchanged. Now let $\overline{C} \in Cl_{K}$ be the canonical class. By a corollary of Riemann-Roch, $\deg(\overline{C}) = 2g-2$ so that the map $\varphi:Cl_{K}^{2g-2} \to Cl_{K}^{2g-2}$ given by $\overline{A} \mapsto \overline{C}-\overline{A}$ is a well-defined involution. So in particular, is a bijection. Then

\begin{aligned} \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}u^{\deg(\overline{A})-g+1} &= \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{C}-\overline{A})}u^{\deg(\overline{C}-\overline{A})-g+1} \\ &= \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})-\deg(\overline{A})+g-1}u^{-\deg(\overline{A})+g-1} \end{aligned}

Since $\varphi$ is a bijection, the first equality holds. We get the second as follows. Use Riemann-Roch thrice. The first two times use it for $\overline{A}-\overline{C}$ and $\overline{A}$, and combine results. This gives the change in power for $u$. The change in power for $q$ is just Riemann-Roch again applied to $\overline{A}$.

On the other hand, if we make the substitution $u \mapsto \frac{1}{qu}$:

$\sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})}\left(\frac{1}{qu}\right)^{\deg(\overline{A})-g+1} = \sum_{\overline{A} \in Cl_{K}^{2g-2}}q^{l(\overline{A})-\deg(\overline{A})+g-1}u^{-\deg\overline{A}+g-1}$,

and we get the same result as above. This proves $F(u) = F(\frac{1}{qu})$ as desired.

Having proven the functional equation we now want to show the existence of a polynomial $L_{K}(u)$ of degree $2g$ such that

$\zeta_{K}(s) = \frac{L_{K}(u)}{(1-u)(1-qu)}$.

As before, take $\zeta_{K}(s) = Z_{K}(u)$. It’s a consequence of Riemann-Roch that if $n > 2g-2$, then $b_{n} = h_{K}\frac{q^{n-g+1}-1}{q-1}$. If we substitute this into $Z_{K}(u)$ and sum the geometric series for terms with $n > 2g-2$ we find

$Z_{K}(u) = \sum_{n = 0}^{2g-2}b_{n}u^{n}+\frac{h_{K}}{q-1}\left(\frac{q^{g}}{1-qu}-\frac{1}{1-u}\right)u^{2g-1}$.

If we combine the terms inside the parenthesis then their numerator divides $q-1$ since $q^{g}(1-u)-(1-qu) = (q^{g}-1)-qu(q^{g-1}-1)$ (we can manually check the case $g = 0$). Collecting all terms under the common denominator gives the result. We now wish to find the degree of $L_{K}(u)$. Start by noticing $L_{K}(0) = 1$ because $b_{0} = 1$. Since $u^{1-g}Z_{K}(u)$ is invariant under $u \mapsto \frac{1}{qu}$, computing $u^{1-g}Z_{K}(u)$ and using the substitution $u \mapsto \frac{1}{qu}$ easily shows $q^{-g}u^{-2g}L_{K}(u) = L_{K}(\frac{1}{qu})$. If we take the limit as $u \to \infty$, then $L_{K}(\frac{1}{qu}) \to 1$ and hence $q^{-g}u^{-2g}L_{K}(u) \to 1$. This implies $L_{K}(u)$ is of degree $2g$ and in fact its leading term is $q^{g}u^{2g}$. This rational expression for the $\zeta$-function has some other interesting properties such as giving an analytic continuation to all of $\mathbb{C}$ with simple poles at $s = 0$ and $s = 1$. Also, $L'_{K}(0) = a_{1}-1-q$ and $L_{K}(1) = h_{K}$ where $a_{1}$ is the number of primes of $K$ of degree one. These facts are easy to check and we omit their proof.

On that note we are finished, and thanks for reading!