# Zeta functions

Here’s the rough idea: $E$ is a set of equations in a finite number of variables over the integers.  Set $a_p(E) = \#$ of solutions mod p.  The goal is to determine formulas that calculate $a_p(E)$.

Translating this to schemes, let $X$ be a scheme of finite type over $\mathbb{Z}$.  Then set $a_p(X) = \#X(\mathbb{F}_p)$ = cardinality of maps from $\mathbb{F}_p$ into X. For example, for $X = \mbox{Spec} A[x]/(x^2 - \alpha)$ for some $\alpha \in A$, then $a_p(X)$ counts the number of square roots for $\alpha \in \mathbb{F}_p$.

Here’s an explicit example.  $X = \mbox{Spec} \mathbb{Z}[x,y]/(y^2 - (x^3 - x))$.  To give a map from, for example, $\mathbb{F}_5$ into X is equivalent to give a ring map into the field which requires a choice of where to send x, such that $f(x) = x^3 -x$ has a square root in the field.  Now $f(x)$ for $x = 0, 1, 2, 3, 4$ is $0,0, 1, -1, 0$ and these latter numbers have $1,1,2,2,1$ square roots respectively.  Thus $a_5(X) = 7$.  As it happens $a_7(X) = 7$.

### Definitions and Easy properties

Generalizing a little bit, let $X$ be a scheme of finite type over a finite field $\mathbb{F}_q$.  Then define

$a_n(X/\mathbb{F}_q) = \# X/\mathbb{F}_q(\mathbb{F}_{q^n})$

I.e. the number of maps from $\mbox{Spec} \mathbb{F}_{q^n}$ into X in the category of $\mathbb{F}_q$ – schemes.

Here are some properties (all unions are meant to be disjoint).  If $X = X_1 \sqcup X_2$ then $a_n(X) = a_n(X_1) + a_n(X_2)$.  In what follows I write $\mathbb{A}^n, \mathbb{P}^n$ for $\mathbb{A}^n_{\mathbb{F}_q}, \mathbb{P}^n_{\mathbb{F}_q}$.  Its straightforward to see that $\mathbb{A}^m(\mathbb{F}_{q^n}) = (\mathbb{F}_{q^n})^m$ so $a_n(\mathbb{A}^m) = q^{nm}$.  Now using

$\mathbb{P}^m = \mathbb{A}^m \sqcup \mathbb{P}^{m-1} = \mathbb{A}^m \sqcup \mathbb{A}^{m-1} \sqcup \mathbb{P}^{m-2} = ...$

I find $a_n(\mathbb{P}^m) = \sum_{i = 0}^m a_n(\mathbb{A}^i) = \sum_i q^{ni} = \frac{z^{m+1}}{z-1}$, for $z = q^n$.

Some notation.  Let $F_n = \mathbb{F}_{q^n}$.  Another important example: $X = \mbox{Spec } F_e$ for fixed e.  Now the elements of $X(F_n)$ correspond to ring homomorphism

$F_e \xrightarrow{\phi} F_n$

Its easy to see the map must be injective.  If $a \ne 0$, then $a^{q^e - 1} = 1$, so $phi(a)^{q^e - 1} = 1$.  Further, $\phi(a)$ must now be a root of $x^{q^n -1 } - 1$.  Meaning $q^e - 1| q^n - 1$ which happens iff $e|n$.  In this case the number of such $\phi$ is equal to the order of the Galois group $G(F_e/F_1)$ which is $e$ because its a Galois extension.  Putting this all together I have

$a_n(F_e) = \begin{cases} 0 \mbox{ if } n/e \not \in \mathbb{Z}\\ e \mbox{ otherwise } \end{cases}$

From the previous post this is

$a_n(F_e) = \sum_{\eta \in \mu_e} \eta^n$

Where $\mu_e$ are the e-th roots of unity.

### The (local) zeta function

For $a_n := a_n(X/F_1)$ set

$Z(X/F_1, T) = \exp(\sum_n \frac{a_nT^n}{n}$

Also introduce the numbers

$b_d(X/F_1) = \# of \mbox{closed pts of X} s.t. \deg x = d$

$c_d(X/F_1) = \# of \eta = \sum_i n_i x_i s.t. \sum_i n_i \deg(x_i) = d$

Where $x_i$ are closed points and $n_i \ge 0$.

### A bunch of results

In the future I might add some of the proofs of these things.

1. $a_n = \sum_{d|n} d\cdot b_d$
2. $\sum_d c_dT^d = \prod_d (1 - T^d)^{-b_d}$
3. If $X/F_1$ is of finite type of dim = m, then there is a constant c s.t $a_n(X/F_1) \le c \cdot q^{mn}$ for all n.
4. If dim X is positive then $\{a_n(X/F_1)\}$ is unbounded.

Here’s a big theorem I wont prove

Thm: There exists a finite number of $\alpha_i , \beta_i \in \mathbb{C}$ s.t.

$a_n(X/F_1) = \sum_i (\beta_i^n - \alpha_i^n)$

In view of the last post, this says

$Z(X/F_1, T) = \prod_i \frac{1- \alpha_iT}{1 - \beta_i T}$

Further, if the hypothesis of #3 are satisfied then $Z(X/F_1,T)$ converges for $|T| < 1/q^m$

—————————-

The proof of #3 uses Noether Normalization and CRT; #4 uses the Nulstullenzats (spelling?) and the last statement of the theorem follows from #3 and nth root formula for radius of convergence.

### More Results

Thm: Let $X/F_1$ be a smooth geometrically connected projective curve of genus g.

1. $Z(X/F_1, T) = \frac{\prod_{i = 1}^{2g} (1 - \alpha_iT)}{(1 - T)(1 - qT)}$, $\alpha_i \in \mathbb{C}$
2. $\alpha_i \mapsto q/\alpha_i$ is a permutation of the $\alpha_i$ and $\prod_i \alpha_i = g$
3. $|\alpha_i| = \sqrt q$ for all i
4. The residue of $Z(T)$ at $T = 1$ is the order of a certain finite group having to do with $\mbox{Pic}_0(X)$ (I might actually explain this later).

Cor. $a_n = a_n(X/F_1) = 1 + q^n - \sum_{i = 1}^{2g} \alpha_i^n$ so that #3 also implies $|a_n - (1 + q^n)| \le 2g \cdot \sqrt{q}^n$.

One final thing.  There’s a degree homomorphism $Div (X) \to \mathbb{Z}$.  In general the image is of the form $e\mathbb{Z}$ for $e \ge 1$.  For $X$ a curve as the thm above Ogus proved that $e = 1$ as follows:

Lemma1: $Z(X/F_1, T) = \frac{f(T^e)}{(1 - q^eT^e)(1 - T^e)}$ where f is a poly and so the Z(T) has simple poles at $\eta/q$.

Lemma2: If $Z(X/F_1, T) = \prod_i \frac{1 - \alpha_iT}{1 - \beta_i T}$ and $X' = X \times_{F_1} F_r$ then $Z(X'/F_r, T) = \prod_i \frac{1 - \alpha_i^rT}{1 - \beta_i^rT}$.

Lemma2 is not so bad to prove, all the work is in understanding why $a_n(X'/F_r) = a_{rn}(X/F_1)$.

In any case, continuing to show e = 1, using lemma 1

$Z(X/F_1) = \frac{\prod_i (1 - \alpha_iT)}{(1 - T^e)(1-q^eT^e)}$

$= \frac{\prod_i (1 - \alpha_iT)}{\prod_\eta (1 - \eta T)(1-\eta qT)}$

So lemma 2 gives

$Z(X'/F_e) = \frac{\prod_i (1 - \alpha_i^e T)}{\prod_\eta (1 - \eta^e T)(1 - \eta^eq^eT)}$

$= \frac{\prod_i (1 - \alpha_i^e T)}{(1 -T)^e(1 - q^eT)^e}$

But by lemma 1 the zeta function has to have simple poles, so e = 1.