{c}

Let $E$ be an elliptic curve over $\mathbb F_q$ and let $\pi$ be its <Frobenius isogeny>. The fixed points of $\pi$ are exactly $E(\mathbb F_q)$, and separability of $1-\pi$ gives
$$
\#E(\mathbb F_q)=\deg(1-\pi)=q+1-a,
$$
Here <Trace of Frobenius> is the integer $a$. Since $\deg\pi=q$, the quadraticity of degree gives, for every pair of integers $m,n$,
$$
0\leq\deg(m-n\pi)=m^2-amn+qn^2.
$$
If the discriminant $a^2-4q$ were positive, this homogeneous quadratic would be negative for some real ratio $m/n$, hence for a nearby rational ratio and then for some pair of integers. Therefore
$$
a^2\leq4q.
$$
This is the <Hasse theorem for elliptic curves>,
$$
\left|\#E(\mathbb F_q)-(q+1)\right|\leq2\sqrt q.
$$

The <Frobenius isogeny> satisfies
$$
\pi^2-a\pi+q=0.
$$
If $\alpha,\beta$ are the roots of $X^2-aX+q$, then $|\alpha|=|\beta|=\sqrt q$ and
$$
\#E(\mathbb F_{q^n})=q^n+1-\alpha^n-\beta^n.
$$
The <zeta function of an elliptic curve over a finite field> is
$$
Z(E/\mathbb F_q,T)
=\exp\left(\sum_{n\geq1}\#E(\mathbb F_{q^n})\frac{T^n}{n}\right).
$$
Substitution of the point-count formula and $-\log(1-z)=\sum_{n\geq1}z^n/n$ gives the rational function
$$
Z(E/\mathbb F_q,T)
=\frac{(1-\alpha T)(1-\beta T)}{(1-T)(1-qT)}
=\frac{1-aT+qT^2}{(1-T)(1-qT)}.
$$
The bounds $|\alpha|=|\beta|=\sqrt q$ are the <Riemann hypothesis for an elliptic curve over a finite field>. The relations $\alpha\beta=q$ and the displayed formula also give the functional equation
$$
Z(E/\mathbb F_q,1/(qT))=Z(E/\mathbb F_q,T).
$$

Hasse theorem and zeta functions

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 4 / Solution / Hasse theorem and zeta functions

{c}
{title2=$a_q$}

For an elliptic curve over $\mathbb F_q$, the trace of Frobenius is
$$
a_q=q+1-\#E(\mathbb F_q).
$$
It is the trace of the <Frobenius isogeny of an elliptic curve> on every prime-to-$q$ Tate module.

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 Incoming links: Trace of Frobenius