TITLE: From Zeta Functions to Curves
QUESTION [41 upvotes]: Let $C$ be a nonsingular projective curve of genus $g \geq 0$ over a finite field $\mathbb{F}_q$ with $q$ elements. From this curve, we define the zeta function
$$Z_{C/{\mathbb{F}}_q}(u) = \exp\left(\sum^{\infty}_{n = 1}{\frac{\# C(\mathbb{F}_{q^n})}{n} u^n}\right),$$
valid for all $|u| < q^{-1}$. This zeta function extends meromophicially to $\mathbb{C}$ via the equation
$$Z_{C / \mathbb{F}_q}(u) = \frac{P_{C / \mathbb{F}_q}(u)}{(1 - u) (1 - qu)}$$
for some polynomial with coefficients in $\mathbb{Z}$ that factorises as
$$P_{C/\mathbb{F}_q}(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$
with $|\gamma_j| = \sqrt{q}$ and $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$. This last point tells us that $Z_{C / \mathbb{F}_q}(u)$ has a functional equation and satisfies a version of the Riemann hypothesis.
What happens if we run this construction in reverse? What if we start with a set of numbers $\gamma_j$, $1 \leq j \leq 2g$, such that $|\gamma_j| = \sqrt{q}$, $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$, and such that the polynomial
$$P(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$
has coefficients in $\mathbb{Z}$? Is there a way of telling whether the function
$$\frac{P(u)}{(1 - u) (1 - qu)}$$
is the zeta function of some curve $C$? Furthermore, what is this curve exactly?
A simple case of this is if we look at the function
$$\frac{1 - au + qu^2}{(1 - u) (1 - qu)}$$
for some $a \in \mathbb{Z}$ with $|a| \leq 2 \sqrt{q}$. How do we determine whether this function is the zeta function $Z_{C / \mathbb{F}_q}(u)$ of an elliptic curve $C$ over $\mathbb{F}_q$? If it is indeed equal to $Z_{C / \mathbb{F}_q}(u)$, what is the Weierstrass equation for $C$ (assuming $\mathrm{char}(q) \geq 5$)?

REPLY [26 votes]: In your last paragraph, you suggest that you are particularly interested in the case of elliptic curves. This is much easier than the general case, which is addressed well by Torsten's answer.
If $q$ is prime, and $|a| \leq 2 \sqrt{q}$, then there is always an elliptic curve with $\zeta$-function $(1-au+qu^2)/(1-u)(1-qu)$. This is the one dimensional case of Honda's theorem. I'll sketch the proof. As you will see, it uses some very sophisticated methods, and it will be very hard to make it effective.
Sidenote: What happens when $q = p^k$ for $k>1$? Then there are $2 p^{k-1}-1$ ways to choose $a$ to be $0 \mod p$. When $a=0$, the elliptic curve must be supersingular. But there are only $\approx p/12$ supersingular elliptic curves over $\mathbb{F}_{p^k}$. So, once $2 p^{k-1}$ is much greater than $p/12$, there will be $a$'s which don't occur. To be honest, I am not clear what happens if you feed one of these $a$'s into Honda's theorem.
Proof Sketch: Let $R$ be the ring $\mathbb{Z}[\phi]/(\phi^2 - a \phi + q)$. $R$ is an order in an imaginary quadratic field (the inequality $a^2-4q<0$ is used to show that this is an imaginary extension.) Let $K$ be the the fraction field of $R$ and let $H$ be its class field. Let $\mathfrak{p}$ be the ideal $(\phi)$ in $\mathcal{O}_K$. Since $\mathfrak{p}$ is principal, it splits in $H$; let $\mathfrak{q}$ lie over $\mathfrak{p}$. So $\mathcal{O}_H/\mathfrak{q} \cong \mathbb{F}_p$.
Let $E$ be an elliptic curve with complex multiplication by $R$; then $E$ can be defined over $H$. ($E$ is only well defined up to a quadratic twist, at this point.) Take a model of $E$ over $\mathcal{O}_H$. Let $E_0$ be the fiber over $\mathfrak{q}$. Then $E_0$ is an elliptic curve over $\mathbb{F}_p$. One can show that the Frobenius acts on $E_0$ by $\pm \phi$ or $\pm \overline{\phi}$, where the bar is the automorphism of $R$ over $\mathbb{Z}$ coming from complex conjugation. By changing the quadratic twist, one can make sure the sign is $+$. Then the trace of the Frobenius is $\mathrm{Tr}(\phi)$, which is $a$, as desired.
Sorry for making this so advanced, I don't know an easier way. I think you can find most of the tools I am using in Silverman's Advanced topics in the arithmetic of elliptic curves.