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Apply the dual isogeny degree identity to $1-\varphi$. Its dual is $1-\hat{\varphi }$, soRearrangement givesThe right side is multiplication by an integer, proving the assertion. In particular the trace of an elliptic-curve endomorphism isThe argument includes $\varphi =0$ and $\varphi =1$ with the zero-map convention. Multiplying $\hat{\varphi }=[t]-\varphi$ by $\varphi$ also yields the useful characteristic relation

Let $t=q+1-\#E({\mathbb{F}}_q)$. The stated characteristic equation for the Frobenius isogeny isTake $\alpha ,\beta$ to be the complex roots of $T^2-tT+q$. They satisfy $\alpha +\beta =t$ and $\alpha \beta =q$. By the Hasse bound, $t^2\le 4q$. If the inequality is strict, the roots are complex conjugates and have squared modulus $q$. In the equality case both roots are the same real number $\pm \sqrt{q}$. Thus in every caseTo obtain every extension-field count without assuming distinct roots, set $t_n=\mathrm{tr}({\pi }^n)$ and $t_0=2$. The dual isogeny satisfies $\hat{\pi }\pi =[q]$ and $\pi +\hat{\pi }=[t]$. Consequently the sums ${\pi }^n+{\hat{\pi }}^n=[t_n]$ satisfyAlternatively, apply additivity of the trace of an elliptic-curve endomorphism directly to the characteristic equation multiplied by ${\pi }^{n-2}$. The sequence ${\alpha }^n+{\beta }^n$ has the same recurrence and initial values, so $t_n={\alpha }^n+{\beta }^n$ by induction. This is the Frobenius trace recurrence, valid also at a repeated root.

The fixed points of ${\pi }^n$ are $E({\mathbb{F}}_{q^n})$, and $1-{\pi }^n$ is again separable. Since $\mathrm{deg}{\pi }^n=q^n$, the trace-degree identity givesFactoring and using $\beta =q/\alpha$ yieldsThis is the elliptic-curve point count over a finite field formula and the Riemann hypothesis for an elliptic curve over a finite field. In particular the powers on both factors are necessary; they are damaged in the converted TeX but present in the original PDF.

The degree of an isogeny $\varphi :E\to E^{\prime }$ is the degree of the induced finite extension of function fields. Saying that degree is a quadratic form on $\mathrm{End}(E)$ meansand thatis bilinear, equivalentlyThe trace of an elliptic-curve endomorphism isThe relation ${\varphi }^2-[\mathrm{tr}\varphi ]\varphi +[\mathrm{deg}\varphi ]=0$ impliesthe trace of the square of an elliptic-curve endomorphism.

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve, put $a=\mathrm{tr}(\pi )$, and note that $\mathrm{deg}\pi =q$ andFor integers $m,n$, quadraticity givesThis binary quadratic form cannot have positive discriminant, since rational numbers $m/n$ are dense, so $a^2-4q\le 0$. Substitution proves the bound. This is the degree-form proof of the Hasse bound.

Both endpoints occur. The curve $E:y^2=x^3-x$ over ${\mathbb{F}}_3$ is supersingular with trace zero. Over ${\mathbb{F}}_9$, its Frobenius is $[-3]$, soIts nontrivial quadratic twist over ${\mathbb{F}}_9$ has the opposite trace and therefore haspoints.

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve. Its fixed points are exactly $E({\mathbb{F}}_q)$, and $1-\pi$ is separable because its differential is the identity. ThereforeFor every endomorphism $\alpha$, the dual isogeny gives $\alpha \hat{\alpha }=[\mathrm{deg}\alpha ]$, and the degree satisfies the parallelogram lawConsequently, if $a=1+q-\mathrm{deg}(1-\pi )$ is the trace of $\pi$, thenThis is nonnegative for every pair of integers $m,n$. By approximating the minimizing real ratio $m/n=a/2$, the quadratic polynomial can be nonnegative for all rational ratios only if its discriminant is nonpositive. Thus $a^2\le 4q$, and the displayed point-count identity proves Hasse's theorem.

An explicit example with nonisomorphic groups occurs over ${\mathbb{F}}_{43}$. PutOn $E_1$, the point $(30,19)$ has order seven and its nonzero cyclic subgroup has $x$-coordinates $30,31,40$. Vélu formulas give the quotient coefficientsso the degree-seven quotient iswhich is $E_2$. Direct quadratic-residue counts give $\#E_1({\mathbb{F}}_{43})=\#E_2({\mathbb{F}}_{43})=49$. The points killed by seven on $E_1$ are only the displayed kernel and $O$, while $(1,8)$ and $(3,2)$ are independent seven-torsion points on $E_2$. Hence

Finally, let $p<43$ and suppose two curves are linked by an isogeny of degree seven. The isogeny and its dual induce inverse isomorphisms on every prime-to-seven primary subgroup, so nonisomorphic rational point groups would require $49$ to divide their common order. Since this order is less than $57$ by Hasse, one group would then be cyclic of order $49$ and the other would contain all of $E[7]$. The Weil pairing on rational seven-torsion forces ${\mu }_7\subset {\mathbb{F}}_p$, hence $p\equiv 1(\mathrm{mod}7)$ unless $p=7$. The only prime below $43$ congruent to one modulo seven is $29$, but Hasse gives $\#E({\mathbb{F}}_{29})<49$; for $p=7$ the same inequality is immediate. No such pair exists below $43$.

The Hasse theorem for elliptic curves states that, for an elliptic curve over ${\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=q+1-\#E({\mathbb{F}}_q)$. The fixed points of $\pi$ are $E({\mathbb{F}}_q)$, and $1-\pi$ is separable, soHence the trace of an elliptic-curve endomorphism iswhile $\mathrm{deg}\pi =q$.

The degree on $\mathrm{End}(E)$ is a nonnegative quadratic form. Polarization and the identities for the dual isogeny give, for integers $m,n$,If $a^2>4q$, this real binary quadratic form is indefinite. An open cone on which it is negative contains a nonzero rational point and therefore a nonzero integer point, contradicting nonnegativity of isogeny degree. Thus $a^2\le 4q$, which is exactly the claimed inequality.

For an endomorphism $\varphi$ of an elliptic curve, define its trace of an elliptic-curve endomorphism byPolarizing the quadratic form $\mathrm{deg}$ shows that this is the integer for whichwhere $\hat{\varphi }$ is the dual isogeny. ConsequentlyAlsoThe left side is $[\mathrm{tr}({\varphi }^2)]$, so