Frobenius isogeny of an elliptic curve ($\pi$)

For an elliptic curve over ${\mathbb{F}}_q$, the Frobenius isogeny sends $(x,y)$ to $(x^q,y^q)$. If $a=q+1-\#E({\mathbb{F}}_q)$, then

If an isogeny of elliptic curves $\pi :E_1\to E_2$ satisfies $\pi {\varphi }_1=-{\varphi }_2\pi$ for their Frobenius isogenies, then $\pi (1-{\varphi }_1)=(1+{\varphi }_2)\pi$. Multiplicativity of the degree of an isogeny gives $\mathrm{deg}(1-{\varphi }_1)=\mathrm{deg}(1+{\varphi }_2)$. The degree parallelogram law then gives the displayed point-count identity, since $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$.

For an elliptic curve over ${\mathbb{F}}_q$, the trace of Frobenius isIt is the trace of the Frobenius isogeny of an elliptic curve on every prime-to-$q$ Tate module.

For odd $p$, let $y^2=f(x)$ be a nonsingular cubic Weierstrass equation of an elliptic curve over ${\mathbb{F}}_p$. Its Trace of Frobenius is $a_p=-{\sum }_x\chi (f(x))$, where $\chi$ is the Legendre symbol extended by $\chi (0)=0$. By the Euler criterion, reduce this sum to $f(x{)}^{(p-1)/2}$. The sum of a positive power $x^j$ over ${\mathbb{F}}_p$ is zero unless $p-1$ divides $j$, when it is $-1$. Since the polynomial has degree $3(p-1)/2<2(p-1)$, only its $x^{p-1}$ coefficient contributes.

If $\alpha ,\beta$ are the roots of $T^2-aT+q$, where $a=q+1-\#E({\mathbb{F}}_q)$, then

For an elliptic curve over ${\mathbb{F}}_q$, the Frobenius isogeny has characteristic polynomial $T^2-tT+q$. Taking traces of its powers gives this recurrence. If $\alpha ,\beta$ are the roots, then $t_n={\alpha }^n+{\beta }^n$ and $\#E({\mathbb{F}}_{q^n})=q^n+1-t_n$. The Hasse bound implies $|\alpha |=|\beta |=\sqrt{q}$, including the repeated-root case.

At an odd good prime, use a model $y^2=x^3-d^{2}x$ with $d$ a unit. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels their Legendre symbols, so the trace is zero. If $p\equiv 1(\mathrm{mod}4)$, the four distinct rational 2-torsion points force the group order to be divisible by four, whereas $p+1$ is two modulo four. The trace cannot vanish. This proves the converse without an explicit complex-multiplication character-sum evaluation.

For an elliptic curve over a finite field of characteristic $p$, its rational-point group is finite and abelian. Prime-to-characteristic geometric torsion bounds every other prime-torsion dimension by two. Multiplication by $p$ has degree $p^2$ and inseparable degree at least $p$, so the $p$-torsion dimension is at most one. The structure theorem for finite abelian groups therefore gives two cyclic factors with $m\mid n$ and $p\nmid m$.

For odd $p$ and $k\not\equiv 0(\mathrm{mod}p)$, take $E:y^2=x^3+kx$. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels the Legendre symbols. If $p\equiv 1(\mathrm{mod}4)$, the coefficient formula for trace of Frobenius modulo p givesThe binomial coefficient is nonzero because every factorial involved is shorter than $p$. Thus the Trace of Frobenius is not zero in the second case.

For an elliptic curve over ${\mathbb{F}}_p$, put $N=\#E({\mathbb{F}}_p)$ and $a=p+1-N$. The trace of the square of an elliptic-curve endomorphism gives $\mathrm{tr}({\pi }^2)=a^2-2p$ for the Frobenius isogeny. Therefore $\#E({\mathbb{F}}_{p^2})=p^2+1-a^2+2p=N(2p+2-N)$.

For $E:y^2=x^3-x-1$ over ${\mathbb{F}}_3$, the Trace of Frobenius is three andIn particular, $\#E({\mathbb{F}}_{3^r})=3^r+1$ exactly when $r\equiv 3(\mathrm{mod}6)$.

For an elliptic curve over ${\mathbb{F}}_q$,

For Frobenius $\pi$, put $a=\mathrm{tr}(\pi )$ and note that $\mathrm{deg}\pi =q$. Nonnegativity of the quadratic formfor all integers $m,n$ forces its discriminant to be nonpositive, so $a^2\le 4q$. Since $\#E({\mathbb{F}}_q)=\mathrm{deg}(1-\pi )=q+1-a$, this is the Hasse bound.

This finite-field function supplies a local factor of a Hasse–Weil zeta function when the curve is a reduction of a curve over a number field.

If $a=q+1-\#E({\mathbb{F}}_q)$, then

If $1-aT+q{T}^2=(1-\alpha T)(1-\beta T)$ is the numerator of the zeta function, the Riemann hypothesis in this case is $|\alpha |=|\beta |=\sqrt{q}$. It is equivalent to the Hasse theorem for elliptic curves.