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Let $\pi$ denote the Frobenius isogeny of an elliptic curve and write $N=\#E({\mathbb{F}}_q)$. Its degree of an isogeny is $q$. The isogeny of elliptic curves $1-\pi$ has differential equal to the identity because $d\pi =0$, so it is a separable isogeny. Its kernel consists precisely of the rational points fixed by $\pi$, giving

We use the degree parallelogram law for isogenies of elliptic curves, with degree zero assigned to the zero map:One explanation of the first identity is the divisor proof of the degree parallelogram law: on $E\times E$, the zero divisor of $x(P)-x(Q)$ is the sum of the diagonal and the graph of negation, while its pole divisor is twice each coordinate copy of $O$. Pulling the associated line bundle identity back by $(f,g)$ and taking degrees gives the identity, including exceptional cases by the line-bundle formulation. This works for a general Weierstrass equation of an elliptic curve, including characteristic two; $x$ is the quotient coordinate for negation. Polarization therefore makes degree a quadratic form on the endomorphism ring of an elliptic curve.

Set $a=q+1-N$, the Trace of Frobenius. The cross term is determined by $\mathrm{deg}(1-\pi )=1+q-a$, giving, for all integers $m,n$,If $a^2>4q$, the real degree-two polynomial $X^2-aX+q$ is negative on a nonempty open interval. That interval contains a rational $m/n$, contradicting the displayed nonnegativity after multiplication by $n^2$. Thus $a^2\le 4q$. Hasse's bounds areThe argument proves the Hasse theorem for elliptic curves without assuming its bound in advance.

For an elliptic curve over ${\mathbb{F}}_q$, where $q$ is a prime power, the Hasse theorem for elliptic curves isWe prove it from the degree form on elliptic-curve endomorphisms, including the required degree facts.

A nonzero homomorphism of elliptic curves is a finite surjective morphism; its degree is the degree of the induced extension of function fields. Set the degree of the zero homomorphism to $0$. The degree of an isogeny is also the degree of the pullback of any point divisor: locally, a finite map of smooth curves gives a module of rank equal to the function-field degree, and its fibre length is the sum of the local multiplicities. For a composition, the tower law for function fields givesIf the isogeny of elliptic curves is separable, its pullback of a nonzero invariant differential on an elliptic curve is nonzero. Translation invariance implies that its differential is nonzero at every point, so each fibre point has multiplicity $1$. Every fibre is a translate of the kernel. ThereforeFor an inseparable isogeny, multiplicities must be retained; counting geometric kernel points alone would be incorrect.

Here is a divisor proof of the degree parallelogram law. On $E\times E$, let $\Delta$ be the diagonal and ${\Delta }^{-}$ the locus $P=-Q$. A Weierstrass $x$-function has a double pole at $O$, and the equation $x(P)=x(Q)$ describes $P=Q$ or $P=-Q$. ConsequentlyThis remains valid in characteristic $2$ using a general Weierstrass equation of an elliptic curve: the $x$-map still has degree $2$ and its two points are exchanged by negation. The divisor identity gives an identity of line bundles. Pull it back by $P\mapsto (\alpha (P),\beta (P))$ and take degrees. The diagonal is the inverse image of $O$ under subtraction and the antidiagonal under addition, soUsing line bundles makes this pullback argument valid even when $\alpha =\beta$, $\alpha =-\beta$ or one map is zero, when directly substituting into the rational function would fail.

Let $d_n=\mathrm{deg}[n]$. Applying the degree parallelogram law to $[n]$ and $[1]$ gives $d_{n+1}+d_{n-1}=2d_n+2$, with $d_0=0,d_1=1$. Induction givesMore generally, the same recurrence and polarization show that the degree restricted to the subgroup generated by two endomorphisms is a quadratic form: its mixed coefficient is determined by their sum or difference. This can be checked directly by the second-difference recurrence in each integer variable and the parallelogram identity in the two diagonal directions.

Let $\pi$ be the $q$-power Frobenius isogeny of an elliptic curve. It has degree $q$: it is purely inseparable, has one geometric point in each fibre, and sends a local parameter at the rational point $O$ to its $q$th power, giving fibre multiplicity $q$. Its pullback of any differential is zero. On the other hand, the standard addition rule for an invariant differential on an elliptic curve givesHence $1-\pi$ is separable, and its kernel consists exactly of the points fixed by Frobenius, namely $E({\mathbb{F}}_q)$. Put $N=\#E({\mathbb{F}}_q)$ and $t=q+1-N$. Thenand the quadratic-form calculation gives, for all $m,n\in \mathbb{Z}$,If $t^2>4q$, the real polynomial $X^2-tX+q$ is negative on an open interval. That interval contains a rational $n/m$ with $m\ne 0$, contradicting the displayed nonnegativity after multiplication by $m^2$. Thus $t^2\le 4q$, which is precisely the Hasse theorem for elliptic curves.

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.

The Hasse theorem for elliptic curves states thatLet $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=p+1-\#E({\mathbb{F}}_p)$. The degree on $\mathrm{End}(E)$ is a positive-definite quadratic form, its associated bilinear form gives $\mathrm{tr}(\pi )=a$, and $\mathrm{deg}\pi =p$. Consequentlyfor all integers $m,n$. If $a^2-4p>0$, this real quadratic form is indefinite, so by density of rational slopes it is negative at some nonzero integer pair $(m,n)$, contradicting nonnegativity of the degree. Hence $a^2\le 4p$, which is the claimed bound.

Let $\pi$ be the Frobenius isogeny of an elliptic curve and putPart (a) givesLet $\alpha ,\beta$ be the two roots of $T^2-aT+p$. The points over ${\mathbb{F}}_{p^r}$ are exactly $\ker (1-{\pi }^r)$. Since the differential of $1-{\pi }^r$ is the identity, this isogeny is separable, and thereforeUsing $\mathrm{deg}\psi =(1-\psi )(1-\hat{\psi })$ in the endomorphism algebra gives the elliptic-curve point count over a finite fieldEquivalently, if $a_r={\alpha }^r+{\beta }^r$, thenand $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.

For $E:y^2=x^3-x$ over ${\mathbb{F}}_3$, each of $x=0,1,2$ gives the single affine point with $y=0$. Including $O$ gives $\#E({\mathbb{F}}_3)=4$, so the Frobenius trace is $a=3+1-4=0$. Therefore the Frobenius isogeny of an elliptic curve $\pi$ satisfiesOn the $11$-torsion, this reads ${\pi }^2=[-3]=[8]$. The element $8\in {\mathbb{F}}_{11}^{\times }$ has order ten and $8^5=-1$. Henceand no smaller positive power of $\pi$ is the identity on $E[11]$. A division field of an elliptic curve over a finite field has degree equal to the order of Frobenius on the torsion module, so

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.