show/hide this revision's text 2 Changed an incorrect minus to a correct plus.

We can actually do a good bit in the function field case, because the zeta function is of the form $$\zeta_F(s)={P(q^{-s})\over (1-q^{-s})(1-q^{1-s})}$$ where $P$ is a polynomial of degree equal to twice the genus of the underlying curve. When the genus of the curve is zero (e.g., $F=\mathbb F_q(t)$), $P(x)=1$. In this case, we can calculate the Laurent expansion of $\zeta_F(s)$ to be (using WolframAlpha to avoid thinking) $${q\over (s-1)(q-1)\log(q)}+{(q-3)q\over 2(q-1)^2}+O(s-1)$$ For the general case, we can multiply the above by the Laurent expansion for $P(q^{-s})$. For a genus-$g$ curve, the corresponding polynomial is $P(q^{-s})=1+a_1q^{-s}+\ldots+a_{2g}q^{-2gs}$. The Laurent expansion of $P(q^{-s})$ is $$\big(1+a_1 q^{-1}+\ldots+a_{2g}q^{-2g}\big)-(s-1)\log(q)\big(a_1 q^{-1}+2a_2q^{-2}\ldots+2g\cdot a_{2g}q^{-2g}\big)+O\big((s-1)^2\big)$$ Multiplying through, we get that the zero-th term in the Laurent expansion of $\zeta_F(s)$, where $F$ is the function field of a genus-$g$ curve, is $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})-{q\over P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$

show/hide this revision's text 1

We can actually do a good bit in the function field case, because the zeta function is of the form $$\zeta_F(s)={P(q^{-s})\over (1-q^{-s})(1-q^{1-s})}$$ where $P$ is a polynomial of degree equal to twice the genus of the underlying curve. When the genus of the curve is zero (e.g., $F=\mathbb F_q(t)$), $P(x)=1$. In this case, we can calculate the Laurent expansion of $\zeta_F(s)$ to be (using WolframAlpha to avoid thinking) $${q\over (s-1)(q-1)\log(q)}+{(q-3)q\over 2(q-1)^2}+O(s-1)$$ For the general case, we can multiply the above by the Laurent expansion for $P(q^{-s})$. For a genus-$g$ curve, the corresponding polynomial is $P(q^{-s})=1+a_1q^{-s}+\ldots+a_{2g}q^{-2gs}$. The Laurent expansion of $P(q^{-s})$ is $$\big(1+a_1 q^{-1}+\ldots+a_{2g}q^{-2g}\big)-(s-1)\log(q)\big(a_1 q^{-1}+2a_2q^{-2}\ldots+2g\cdot a_{2g}q^{-2g}\big)+O\big((s-1)^2\big)$$ Multiplying through, we get that the zero-th term in the Laurent expansion of $\zeta_F(s)$, where $F$ is the function field of a genus-$g$ curve, is $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})-{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$