Let (E) be an elliptic curve $y^2=x^3x1$ over $\mathbb{F}_3$, $a_0=1$, $a_n$ is the number of positive divisor of degree $n\geq 1$. $a_1$ in this case is the number of points of E, i.e., $a_1=1$ only a point $\infty$. The Zeta function $Z(t)=\sum_{n=0}^{\infty}a_n t^n$ is computable $$Z(t)=\dfrac{13t+3t^2}{(1t)(13t)}=1+t+4t^2+13t^3+...$$ My question is: What are positive divisors of degree 2 on Elliptic Curve $y^2=x^3x1$ over $\mathbb{F}_3$? Is a positive divisor of degree 2 a sum of 2 points of E? If it is true, I can't explain $a_2=4$. Thank you for your helps.

3$\begingroup$ You also have divisors coming from $\mathbb{F}_9$points, which should give you the remaining divisors counted by $a_2.$ $\endgroup$– dhyDec 25, 2017 at 3:57

$\begingroup$ @dhy: can you explain more? I don't understand which divisors coming from $\mathbb{F}_9$points? Could you give me an example or reference? Thanks. $\endgroup$– ZolaEllipticDec 25, 2017 at 4:25

1$\begingroup$ An effective divisor of degree 2 either comes from a sum of two points of degree $1$ over your base field or from one point of degree $2$ over your base field. Similarly an effective divisor of degree $3$ on your curve would either be a sum of $3$ $\mathbb{F}_3$points, a sum of a $\mathbb{F}_9$point and a $\mathbb{F}_3$point, or a single $\mathbb{F}_{27}$point. $\endgroup$– dhyDec 25, 2017 at 4:33

$\begingroup$ @dhy: there is only one point $\infty$ over $\mathbb{F}_3$, i believe it is degree 1, so there is one effective divisor of degree two $2\infty$. What are another three points of degree 2 in this case? $\endgroup$– ZolaEllipticDec 25, 2017 at 5:01

2$\begingroup$ The three $\mathbb{F}_9$points are the points corresponding to $(0,i),(1,i)$, and $(2,i),$ where $i$ is a square root of $1$ in $\mathbb{F}_9$. $\endgroup$– dhyDec 25, 2017 at 5:16
2 Answers
For a nonsingular cubic curve $X \subset \mathbb{P}^2$ over a finite field $\mathbb{F}_q$, if $a_1$ is the number of $\mathbb{F}_q$points of $X$, and $a_2$ is the number of effective degree two divisors on $X$, there is a relation $$a_2 = (q+1)\cdot a_1.$$ So in the example you consider, $a_1 = 1$ implies $a_2 = 4$.
This is a consequence of the following geometric observation: any degree two effective divisor determines a line in $\mathbb{P}^2$ passing through it, and the third intersection point of the line with the curve. Conversely, any rational point on $X$ and a line $L$ through it determines a degree two divisor obtained as a residual intersection of $L$ with $X$. The $(q+1)$ multiple in the formula comes from the fact that there are $q+1$ lines through a point in $\mathbb{P}^2$.
This geometric observation works in higher dimension and for singular cubics as well and allows to relate the symmetric square of the cubic to the Fano variety of lines on it in the Grothendieck ring of varieties thus relating their number of rational points over finite fields, Hodge structures and so on, see our paper with Sergey Galkin: https://arxiv.org/pdf/1405.5154.pdf. What I explained above is formula (5.2) from the paper with $d=1$ and the two rightmost terms being equal to zero.
UPDATE: When applying the procedure above to the curve $y^2 = x^3  x  1$, one finds that lines through the point at infinity are given by the equation $x = \alpha$, where $\alpha = 0,1,2$, and intersecting these lines with the curve one finds points $(0, \pm i), (1, \pm i), (2, \pm i)$ as in dhy's comments. The last divisor is $2 \cdot \infty$ obtained from the line infinity triple tangent to the curve.
RiemannRoch together with the fact, that there is a unique point implies that Picard group is cyclic and line bundles are uniquely determined by their degree. So you have a unique line bundle of degree two, that gives you a hyperelliptic doublecovering map to $P^1$, and effective divisors of degree two are exactly the fibers of this map. Concretely this map is given by the projection from the unique $F_3$point.
Also note that $13 = P^2(F_3)$, so all effective divisors of degree three are just given by thirteen lines in $P^2(F_3)$.