# Lecture 16 ## Weil Bounds Last time: we finished a discussion of the Hasse-Weil zeta function over a finite ground field $\fq$. :::{.exercise title="?"} Let $K/\fq$ be a function field of genus 1. a. Show \[ Z(t) = { 1-at + qt^2 \over (1-t)(1-qt)} ,\] where $a = q+1 - \abs{\Sigma_1(K/\FF_q)}$. b. Let $L(t) \da (1-\alpha_1 t)(1-\alpha_2 t)$. Show that $a = \alpha_1 + \alpha_2$, and that for all $r\in \ZZ^{+}$, \[ \abs{\Sigma_1 (K_r / \FF_{q^r}} = q^r +1 - \alpha_1^r - q^r/\alpha_1^r .\] Thus for elliptic curves $E/\FF_q$, knowing $\abs{E(\FF_q)}$ determines $\abs{E(\FF_{q^r})}$ for all $r\in \ZZ^+$. c. Suppose $a^r = 0$ and show \[ r \text{ odd} &\implies \abs{\Sigma_1 (K_r/\FF_{q^r}) } = q^r + 1 \\ r \equiv 2\mod 4 &\implies \abs{\Sigma_1 (K_r/\FF_{q^r}) } = (q^{r/2} + 1)^2 \\ r \equiv 0\mod 4 &\implies \abs{\Sigma_1 (K_r/\FF_{q^r}) } = (q^{r/2} - 1)^2 \\ .\] ::: :::{.theorem title="?"} Let $K/\FF_q$ be a function field of genus $g$ with $L\dash$polynomial $L(t) = \prod_{i=1}^g (1 - \alpha_i t)$. Then $\abs \alpha_i = \sqrt{q}$ for all $i$. ::: :::{.remark} In order to prove this, Weil had to develop foundations for algebraic geometry in positive characteristic. His original proof used intersection theory on algebraic surfaces. ::: :::{.corollary title="Weil Bounds"} If $K/\FF_q$ is a function field of genus $g$, then \[ \abs{ \abs{\Sigma_1 (K/\FF_q)} - (q+1) } \leq 2g \sqrt{ q} .\] ::: :::{.remark} This says that the number of $\FF_q$ points is approximately $q+1$< where the error is controlled by the genus. ::: :::{.proof title="?"} We know \[ \abs{\Sigma_1(K/\FF_q)} = q + 1 - \sum \alpha_i .\] Thus \[ \abs{ \Sigma_1(K/\FF_q) - (q+1) } = \abs{ \sum \alpha_i } \leq \sum_{i=1}^{2g} \abs{\alpha_i} =2g\sqrt q .\] ::: :::{.corollary title="?"} For $r\gg 0$, $N_r \da \abs{\Sigma_1(K_r/ \FF_{q^r})} \geq 1$. ::: :::{.proof title="?"} The Weil bounds yield \[ N_r \geq q^r + 1 - 2g q^{r/2} \geq q^{r/2} \qty{q^{r/2} - 2g} \converges{r\to \infty} \infty .\] This can be alternatively phrased as \[ N_r = q^r + O_g(q^{r/2}) .\] Note that we've used the fact that making any separable extension of a function field will preserve the genus, and so $g$ is fixed. ::: :::{.remark} So for $r$ large enough, there is an $\FF_{q^r}\dash$rational point, and $N_r \to \infty$ exponentially fast in $r$. ::: :::{.remark} As a consequence, if $\omega\in \Sigma_1(K_r/ \FF_{q^r})$, let $v\da \omega\intersect K$. We saw that $\deg(v) = d$ which divides $r$, so we can form the divisor $r/d[p] \in \Div^r K$ for any place $p$ below $v$. So any degree 1 place yields a degree $r$ divisor, which shows that $\Div^r K \neq \emptyset$ for all $r$ large enough and thus $\delta=1$ (which is Schmidt's theorem). ::: :::{.exercise title="?"} \envlist a. Show that there is a constant $C$ depending only on $g$ such that if $r\geq C$ then for all $n\geq 2$, $N_{nr} > N_r$. b. Use the following fact to show that there exists a constant $D$ depending on $g$ such that for all $d\geq D$, $\Sigma_d(K/\FF_q) \neq \emptyset$: \[ N_r = \sum_{d\divides r} d \abs{ \Sigma_d(K/\FF_q) } .\] ::: :::{.remark} Note that this is stronger than Schmidt's theorem: it implies that not only do you have a divisor of degree $d$, but also a place of degree $d$. ::: :::{.exercise title="?"} \envlist a. Use the Weil bounds to show that when $g=0$, $\abs{\Sigma_1(K/\FF_q)} = q+1$. Deduce that every genus zero function field is rational. b. Use the Weil bounds to show that if $g=1$, then $\abs{\Sigma_1(K/\FF_q)} \geq \qty{\sqrt q - 1}^2 > 0$, and thus every genus 1 function field over $\FF_q$ is elliptic.[^def_ell_ff] [^def_ell_ff]: By definition, a genus 1 function field with a degree 1 place, which can be used as the origin for the group structure. ::: :::{.corollary title="Serre Bounds"} Let $K/\FF_q$ be a function field of genus $f$, then \[ \abs{ \Sigma_1(K/\FF_q) - (q + 1) } \leq g \floor{2 \sqrt q} .\] ::: :::{.remark} We write \[ M_q(g) &\da \text{The maximal } \abs{\Sigma(K/\FF_q)} \text{ as $K$ ranges over genus $g$ function fields} \\ m_q(g) &\da \text{The minimal } \abs{\Sigma(K/\FF_q)} \text{ as $K$ ranges over genus $g$ function fields} \\ A(q) &\da \limsup_{g\to \infty} {M_q(g) \over g} .\] This is essentially the best constant that can be put in front of $g$ in the bounds. With these definitions, we have \[ A(q) \leq 2\sqrt q && \text{by Weil bounds }\\ A(q) \leq \floor{2\sqrt q} && \text{by Serre bounds }\\ .\] Note that we can do much better, e.g. $A(2) \leq \sqrt{2} - 1$, but it is not known if this is sharp. ::: :::{.proof title="of Serre's bound"} We may assume $g \geq 1$, and write $L(t) = \prod_{i=1}^{2g} (1-\alpha_i t)$, where the $\alpha_i$ may be ordered such that $\alpha_i \alpha_{g+-} = g$ for all $i$. By the Riemann hypothesis, we have $\abs {\alpha_i} = \sqrt q$, and so for each $i$ we have $\bar{\alpha_i} = q/\alpha_i = \alpha_{g+i}$. We now pair in the following way: set \[ \gamma_i &\da \qty{ \alpha_i + \bar{\alpha_i} } + \floor{2\sqrt q} + 1 \\ \delta &\da -\qty{ \alpha_i + \bar{\alpha_i} } + \floor{2\sqrt q} + 1 \\ .\] These are real algebraic integers and by the Riemann hypothesis they are positive. Since $L^{\perp}(t) \da \prod t-\alpha_i \in \ZZ[t]$, take any complex embedding \[ \sigma: \QQ\qty{\ts{\alpha_i}} \injects \CC ,\] which preserves the $\alpha_i$. If $\sigma(\alpha_i) = \alpha_j$, then we have \[ \sigma(\bar{\alpha_i}) = \sigma\qty{q/\alpha_i} = q/\sigma(\alpha_i) = \bar{\sigma(\alpha_i)} = \alpha_j ,\] and thus $\sigma$ preserves the multisets of the $\gamma_i$ and $\delta_i$. Now set $\gamma\da \prod \gamma_i$ and $\delta \da \prod \delta_i$, making these both positive real integers that are fixed by every embedding $\sigma$, and thus $\gamma, \delta\in \ZZ^+$. We can now apply the AM-GM inequality: \[ {1\over g} \sum_{i=1}^g \gamma_i &\geq \qty{ \prod_{i=1}^g \gamma_i}^{1\over g} = \gamma^{1/g} \geq 1 ,\] and thus \[ g \leq \sum \gamma_i = \sum_{\alpha_i + \bar{\alpha_i}} + g\floor{2\sqrt q} + g = \sum \alpha_i + g\floor{2\sqrt q} + g ,\] and we can conclude that $-\sum \alpha_i \leq g\floor{2\sqrt g}$. Repeating the argument with the $\delta_i$ yields $\sum \alpha_i \leq g \floor{2\sqrt q}$, meaning that $\abs{\sum \alpha_i } \leq g \floor{2\sqrt q}$. Thus \[ \abs{ \abs{\Sigma_1(K/\FF_q)} - (q+1) } = \abs{\sum \alpha_i} \leq g \floor{2\sqrt q} .\] ::: :::{.remark} An application to class numbers: since $g = L(1)$, by the Riemann hypothesis we have \[ \sqrt{q} - 1 \leq \abs{\alpha_i - 1} \leq \sqrt{q} + 1 \] and thus \[ \qty{\sqrt q - 1}^{2g} \leq h \leq \qty{\sqrt q + 1}^{2g} .\] There is a slight improvement: \[ \qty{\sqrt q - 1}^{2g} \leq \qty{q+1 - \floor{2\sqrt q} }^g \leq h \leq \floor{\qty{\sqrt q + 1}^2}^g = \qty{q+1 + \floor{2\sqrt q}}^g .\] ::: :::{.corollary title="?"} \envlist a. For all $q\geq 5$, we have \[ h\ geq \qty{\sqrt 5 - 1}^{2g} \geq 1.236^{2g} .\] b. The class number grows exponentially in the genus not just for each fixed $q$ but uniformly over all $q\geq 5$. In particular, if $g\geq 1$, this forces $h> 1$. ::: What exactly is happening at small $q$, such as $q\in \ts{2,3,4}$? :::{.theorem title="?"} For $K/\FF_q$ of genus $g\geq 1$, \[ h \geq \qty{q-1 \over 2} \qty{q^{2g} + 1 - 2gq^g \over g\qty{q^{g+1} - 1}} .\] ::: :::{.exercise title="?"} \envlist a. Show that \[ h\geq \qty{q-1 \over 2}\qty{{ q^{g-1} \over g} - {2\over q}} ,\] which still grows exponentially in $g$. b. Show that for any $1 2g-2$. Let $Q\in \Sigma_1(K_{2g} / \FF_{q^{2g}})$ be a degree one place and restrict to $K$ to obtain $P \da Q\intersect K$ which has degree $\ell$ dividing $2g$. Then $\qty{2g \over \deg(P)} P$ has degree $2g$, and this yields a map \[ \Sigma_1(K_{2g} / \FF_{q^{2g}}) &\to \text{Effective degree $2g$ divisors} \\ Q &\mapsto {2g \over \deg P} P .\] This is not necessarily surjective, but how far is it from being injective? The fibers have size at most $2g$ since we have a degree $2g$ extension of Dedekind domains. Thus \[ \abs{\Sigma_1 (K_{2g} / \FF_{q^{2g}}) } \leq A_{2g} ,\] and substituting the known value of $A_{2g}$ and rearranging yields \[ h &\geq {1\over 2g} \qty{q-1 \over q^{g+1} - 1} N_{2g} \\ &\geq \qty{1\over 2g} \qty{q-1 \over q^{g+1} - 1} \qty{ q^{2g} + 1 - 2g q^g} && \text{by the Weil bounds} \\ &= \qty{q-1 \over 2} \qty {q^{2g} + 1 - 2gq^g \over g\qty{q^{g+1} - q}} .\] :::