# Lecture 14: The Hasse-Weil Zeta Function Recall the that *Hasse-Weil zeta function* of a one-variable function field $K/\FF_q$ over a finite ground field is defined in the following way: let $A_n = A_n(K)$ be the number of effective divisors of degree $n$. We have proved that $A_n$ is finite, and for $n>2g-2$ we have a formula \[ Z(t) = \sum_{n=0}^\infty A_n t^n = \sum_{D\in \Div^+(K)} t^{\deg(D)} \in \ZZ[[t]] ,\] which is a formal power series with integer coefficients. :::{.remark} Recall that we have proved that it is a rational function of $t$, and in particular when $g=0, \delta = 1$ [^delta_def_reminder] we get \[ Z(t) = {1 \over (1-qt)(1-t)} .\] We got another expression which isn't fantastic: it involves this $\delta$, which we'll work toward proving is equal to 1. When $g>1$, we broke the zeta function into two pieces $Z(t) = F(t) + G(t)$. For divisors of sufficiently high degree, Riemann-Roch tells you what the dimension of the Riemann-Roch space is, and $G(t)$ explains the part coming from divisors of large degree. We obtained a formula previously for $F(t)$ and $G(t)$, and once we show $\delta=1$ the formula for $G$ will simplify. For $F(t)$, we specifically had \[ F(t) = {1\over q-1} \sum_{0\leq \deg(c) \leq 2g-2} q^{\ell(c) t^{\deg(c)}} ,\] where the sum is over divisor classes and $\ell$ is the dimension of linear system corresponding to a divisor. But this isn't a great formula: what are these classes, dhow many are in each degree, and what is the dimension of the Riemann-Roch space? [^delta_def_reminder]: The *index* of the function field, least positive degree of a divisor. ::: :::{.remark} This is analogous to the Dedekind zeta function of a number field $K$, in which case \[ \zeta_K(s) = \sum_{T\in \ell(\ZZ_k)}^\bullet \abs{\ZZ_k/I}^{-s} ,\] which will be covered in a separate lecture on *Serre zeta functions*. ::: :::{.theorem title="F.K. Schmidt"} For all $K/\FF_q$, we have $\delta = I(K) = 1$ where $I$ is the index. ::: This will follow from the associated, but it much weaker. However, this is one of the facts we'd like to establish to use to *prove* the Riemann hypothesis. :::{.remark} Pete studied this in 2004 and found that every $I\in \ZZ^+$ arises as the index of a genus one function field $K/\QQ$. ::: Notation: for $n\in\ZZ^+$, let $\mu_n$ denote the $n$th roots of unity in $\CC$. :::{.lemma title="?"} For $m, r\in \ZZ^+$, set $d \da \gcd(m, r)$. Then \[ \qty{1-t^{mr/d}}^d = \prod_{\xi\in \mu_r} 1 - (\xi t)^m .\] ::: :::{.proof title="?"} In $\CC[x]$, we have \[ (X^{r/1} - 1)^d = \prod_{\xi\in \mu_r}(X - \xi^m) ,\] where both sides are monic polynomials whose roots include the $(r/d)$th roots of unity, each with multiplicity $d$. On the LHS, the distinct roots are the $r/d$th roots of unity, then raising to the $d$th power gives them multiplicity $d$. On the RHS, this is an exercise in cyclic groups: consider the $n$th power map on $\ZZ/r\ZZ$ and compute its image and kernel. As $\xi$ ranges over $r$th roots of unity, $\xi^m$ ranges over all $r/d$th roots of unity, each occurring with multiplicity $d$. Substituting $X= t^{-m}$ and multiplying both sides by $t^r$ yields the original result.[^special_case] [^special_case]: Special case: set $m=r$, so $d=r$, then the RHS is $r$ copies of 1. ::: ## Comparing Zeta Functions After Extending Scalars Next up, we want to compare the zeta function $Z(t)$ for a function field over $\FF_q$ to the zeta function obtained when extending scalars to $\QQ^r$. :::{.proposition title="Factorization identity for the zeta function"} Let $K/\FF_q$ be a function field, $r\in \ZZ^+$, and take the compositum $K_r$ of $K$ and $\FF_q^r$ viewed as a function field over $\FF_q^r$. Let $Z(t)$ be the zeta function of $K/\FF_q$ and $Z_r(t)$ the zeta function of $K_r/\FF_q^r$. Then \[ Z_r(t^r) = \prod_{\xi \in \mu_r} Z(\xi t) .\] ::: :::{.proof title="?"} We have an Euler product formula \[ Z(t) = \prod_{p\in \Sigma(K/\FF_q)} (1 - t^{\deg(p)})^{-1} .\] where the sum is over places of the function field.[^proving_euler_prod_rmk] [^proving_euler_prod_rmk]: Proving this Euler product formula might show up in a separate lecture, but it is not any more difficult than proving it for the Riemann zeta function. :::{.exercise} Why is this product expansion true? Write as a geometric series with ratio $t^{\deg(p)}$. Here just expand each summand to get \[ Z(t) = \prod_p \sum_{j=1}^\infty t^{j\deg(p)} .\] Multiplying this out and collecting terms is in effect multiplying out the prime divisors to get effective divisors. ::: We now use the result about splitting that was stated (but not proved): :::{.claim} If $p\in \Sigma_m(K/\FF_q)$ is a degree $n$ place and $r\in \ZZ^+$, then there exist precisely \[ d\da \gcd(m, r) \] places $p^r$ of $K_r$ lying over $p$, where each place $p^r$ has degree $m/d$. ::: In order to compare $Z_r(t)$ to $Z(t)$, we collect the $p'$ into ones that have the same fiber. We then can range over all $p$ first, then over all $p'$ in the fiber above $p$, yielding \[ Z_r(t^r) = \prod_{p\in \Sigma(K_{/\FF_q})} \prod_{p'/p} {1 \over 1 - t^{r\deg(p')}} .\] Using the Euler product identity, we have for $p\in \Sigma_m(K_{/\FF_q})$ and $d\da \gcd(m, r)$ we can express the innermost product as \[ \prod_{p'/p} {1 \over 1 - t^{r\deg(p')}} = (1 - t^{rm/d})^{-d} = \prod_{\xi\in \mu_r} (1- (\xi t)^m)^{-1} ,\] where we've used the fact that we know there are exactly $d$ places and each contributes the same degree in the first expression. By using $-d$ in the previous lemma, we get the last term. Combining all of this yields \[ Z_r(t^r) = \prod_{\xi \in \mu_r} \prod_{p\in \Sigma(K_{/\FF_q})} (1- (\xi t)^{\deg p})^{-1} = \prod_{\xi \in \mu_r} Z(\xi t) .\] ::: :::{.remark} Similar to taking an abelian extension of number fields and noting that the Dedekind zeta function factors into a finite product: the original zeta function, and in general, Hecke $L$ functions. If you do this for an abelian number field over $\QQ$, then the Dedekind zeta function of the upstairs number field will be a finite product where one of the terms in the Riemann zeta function and the others are Dirichlet $L$ functions associated to certain Dirichlet characters. So this is some (perhaps simpler) version of that. ::: ## Proof That $\delta = 1$ We can finally prove Schmidt's theorem that $\delta = 1$: :::{.proof title="$\delta = 1$"} Take a $\delta$th root of unity $\xi \in \mu_\delta$. Then for all places $p \in \Sigma(K_{/\FF_q})$, $\delta$ divides $\deg p$ by definition since it is a gcd, and so we have \[ Z(\xi t) = \prod_{p\in \Sigma(K_{/\FF_q})} (q - (\xi t)^{\deg p} )^{-1} = \prod_{p\in \Sigma_{K_{\FF_q}}} {1 \over 1 - t^{\deg p}} = Z(t) ,\] using the fact that $\xi^{\deg p} = 1$. We're now in a situation where we can apply the previous proposition, which gives the following identity for the zeta function over the degree $\delta$ extension: \[ Z_{\delta}(t^\delta) = \prod_{\xi \in \mu_\delta} Z(\xi t) = Z(t)^\delta .\] Our previous formulas show that any zeta function for a 1-variable function field over a finite field has a simple pole at $t=1$, and since $\ord_{t-1}(t^\delta) = 0$, we get \[ -1 = \ord_{t-1} Z_\delta(t^\delta) = \ord_{t-1} Z(t)^\delta) = -\delta ,\] where for the first equality we're using the fact that the $(t-1)\dash$adic valuation of $Z_\delta(t^\delta)$ is one, and for the RHS, the ordinary zeta function has a simple pole at $t=1$ and since we have a valuation, raising something to the $\delta$th power is just $\delta$ times the original valuation. ::: There is some modest representation theory (character theory) that shows up when looking at zeta functions of abelian extensions. :::{.remark} We can also conclude that every genus zero function field $\kfq$ is isomorphic to $\FF_q(t)$ and thus rational, since such a function field rational iff it has index one. Why? By Riemann-Roch, index one implies existence of a divisor of degree one, and taking a genus zero curve says that every divisor of nonnegative degree is linearly equivalent to an effective divisor. Thus if you have a divisor of degree one, you have an effective divisor of degree one, which makes the function field a degree one extension of a rational function field. ::: :::{.exercise title="?"} Let $K = \FF_q(t)$, then show that $g=0, \delta = 1$, and \[ Z(t) = {1 \over (1-qt)(1-t)} .\] > Hint: go back to complicated formulas and substitute $\delta=1$ to simplify things. ::: Thus for rationality of the zeta function, we can get rid of the $\delta$ cluttering up formulas. ## The Functional Equation Going back to the plan, we wanted to show 1. Rationality: $Z(t) \in \QQ(t)$ and thus $Z(t) = P(t) / Q(t)$, 2. Understand the degrees of $P$ and $Q$ in terms of the genus, and 3. Ask about the roots of $P(t)$ to understand the analog of the Riemann Hypothesis for Dedekind zeta functions We'll want to establish a functional equation, as is the usual yoga for zeta functions, since it helps establish a meromorphic continuation to $\CC$. The algebraic significance of the functional equation is that it aids in understand several equivalent packets of data: - The number of effective divisors of a given degree, - The number of places of a given degree, - The number of rational points over each finite degree extension of the base field. :::{.theorem title="Functional Equation"} Let $\kfq$ be a function field of genus $g$, then \[ Z(t) = q^{g-1} t^{2g-2} Z\qty{1\over qt} .\] ::: :::{.proof title="?"} For $g=0$, we know that \[ Z(t) = {1 \over (1-t)(1-qt)} ,\] and plugging in ${1\over qt}$ is a straightforward calculation. So assume $g\geq 1$. The idea was that we wrote $Z(t) = F(t) + G(t)$. The $F(t)$ piece came from summing over divisor classes of degree between $0$ and $2g-2$ and recording the dimension of the associated linear system. The tricky piece $G(t)$ came from summing an infinite geometric series to get a more innocuous closed-form expression of a rational function. So the strategy here is to separately establish the functional equation for each of $F$ and $G$ separately. How to do this: for $g=0$, there was no $F(t)$ piece. If we have a closed form it's just a computational check. For $F(t)$, we'll use our greatest weapon and dearest ally, the Riemann-Roch theorem. This will provide the extra symmetry we need. We essentially already applied Riemann-Roch to $G(t)$ to get the closed-form expression, but we haven't applied it to the small degree divisors. This doesn't tell you what the dimension is, but rather gives you a duality result: ti gives the dimension in terms of the dimension of a complementary divisor. Take a canonical divisor $\mathcal{K} \in \div(K)$, so $\deg \mathcal{K} = 2g-2$. As $C$ runs through all divisor classes of $\mathcal{K}$ of degree $d$ with $0\leq d \leq 2g-2$, so does the complementary divisor $\mathcal{K}-C$. We can thus write \[ (q-1) F(t) &= \sum_{0 \leq \deg C \leq 2g-2 } q^{\ell(C)} t^{\deg(C)} \\ (q-1)G(t) &= h \qty{ {q^g t^{2g-1} \over 1-qt} - {1 \over 1-t} } .\] We can thus compute \[ (q-1) F\qty{1\over qt} &= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(C)} \qty{1\over {qt} }^{\deg C} \\ &= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(\mathcal{K} - C)} \qty{1\over {qt} }^{2g-2-\deg C} ,\] where in the second step we've exchanged $C$ for $\mathcal{K}- C$ and noted that $\deg(\mathcal{K}-C) = 2g-2-\deg(C)$. We now do the calculation another way, \[ (q-1) F(t) &= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(C)} t^{\deg C} \\ &= q^{g-1} t^{2g-1} \sum_{0\leq \deg C \leq 2g-2} q^{\deg(C) - (2g-2) + \ell(\mathcal{K}-C)} t^{\deg(C) - (2g-2)} && \text{by Riemann-Roch} \\ &= q^{g-1} t^{2g-2} \sum_{0 \leq \deg C \leq 2g-2} q^{\ell(\mathcal{K} - C)} \qty{1\over qt}^{\deg(\mathcal{K} - C)} \\ &= q^{g-1} t^{2g-2} (q-1) F\qty{1\over qt} .\] where we've used Riemann-Roch to find that $\ell(C) = \ell(\mathcal{K}-C) + \deg(C) - g + 1$. Cancelling the common factor of $(q-1)$ establishes the functional equation for $F(T)$. Now using the fact that $\delta = 1$, we have \[ (q-1)G(t) = h \qty{ {q^g t^{2g-1} \over 1-qt} - {1\over 1-t} } ,\] and thus \[ (q-1) q^{g-1} t^{2g-2} G\qty{1\over qt} &= hq^{g-1} t^{2g-2} \qty{q^g \qty{1\over qt}^{2g-1} - {1\over 1 - q \qty{1\over qt}} - {1\over 1 - {1\over qt}} } \\ &= h\qty{ {-1\over 1-t} + {q^g t^{2g-1} \over 1-qt}} \\ &= (q-1) G(t) ,\] which establishes the functional equation for $G(t)$. ::: ## The $L$ Polynomial :::{.definition title="The $L$ Polynomial"} \[ L(t) \da (1-t) (1-qt) Z(t) \in \ZZ[t] .\] ::: This clears the denominators in $Z(t)$, so this is now a polynomial of degree at most $2g$. We can thus rewrite \[ Z(t) = {L(t) \over (1-t)(1-qt)} = {a_{2g} t^{2g} + \cdots + a_1 t + a_0 \over (1-t)(1-qt)} .\] Note that if we know $L(t)$, then we know $Z(t)$, and in particular we would like to know what the coefficients $a_j$ are. We'll be able to determine $a_0 = 1$ in all cases, as well as $a_{2g}$ in all cases pretty easily. So it looks like it only remains to compute $a_1, \cdots, a_{2g-1}$, but the functional equation will give a "mirror" relation between pairs of coefficients. The upshot is that the functional equation shows that we only need to know $a_1, \cdots, a_g$ to completely determine $Z(t)$. If $g=1$, just one coefficient suffices. It turns out that $a_1$ will be $q+1$ minus the number of degree one places. :::{.question} \envlist - What are the constraints on these quantities? - Can we write the zeta function in a nice way? - Exactly what do we need to compute to determine it? ::: It will turn out that computing the number of rational points over $\FF_{q}, \FF_{q^2}, \cdots, \FF_{q^g}$ will be possible. For example, for a hyperelliptic curve, we'll have an explicit defining equation and can make an explicit point count, and you only need $g$ of them.