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The $q$-power Frobenius isogeny sends affine coordinates to $(x^q,y^q)$, has degree $q$, and fixes exactly the points defined over ${\mathbb{F}}_q$, including the origin. ConsequentlyThe degree is $q$ because the $q$-power Frobenius on a smooth curve has function-field inseparable degree $q$. By the permitted separability of $1-\varphi$, this separable isogeny has exactly $\mathrm{deg}(1-\varphi )$ geometric kernel points; there is no inseparable multiplicity to remove. Part(i) therefore givesThis relates the elliptic-curve point count over a finite field to the integer trace, rather than treating the point count as an independently assumed definition of the trace.

An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

Write $N=\#E({\mathbb{F}}_q)$. The Hasse bound isWe prove it using isogeny degrees of endomorphisms, with no assumption about a pre-existing Frobenius eigenvalue estimate. The standard elliptic isogeny facts we use are these: isogeny degree is multiplicative under composition; $\mathrm{deg}[m]=m^2$; assigning isogeny degree zero to the zero homomorphism makes isogeny degree a quadratic form, satisfying the degree parallelogram law; and the isogeny degree of a separable isogeny equals the number of its geometric kernel of an isogeny points.

Let $\varphi$ be the $q$-power Frobenius isogeny. It has isogeny degree $q$ and zero differential. Therefore $1-\varphi$ has differential the identity and is a separable isogeny. Its kernel of an isogeny consists exactly of the points fixed by Frobenius, including the identity, soPut $t=q+1-N$. Polarization of the isogeny degree quadratic form gives, for all integers $m,n$,The cross coefficient is fixed by the case $m=n=1$, where the isogeny degree is $N=1-t+q$; the coefficients of $m^2,n^2$ are $\mathrm{deg}1=1$ and $\mathrm{deg}\varphi =q$.

Every isogeny degree is nonnegative. For $n\ne 0$, divide by $n^2$ to obtain $u^2-tu+q\ge 0$ for every rational $u=m/n$. Continuity and density of the rationals imply the same inequality for real $u$. Its minimum, attained at $u=t/2$, is $q-t^2/4$, so $t^2\le 4q$. This is precisely the stated bound. Equality is allowed; a zero value of the quadratic form can occur when the two endomorphisms involved are rationally dependent.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.

The Hasse theorem for elliptic curves statesWe prove it by the positive degree of an isogeny. Let $\pi$ be the $q$-power Frobenius isogeny, of degree $q$. Its fixed points are precisely $E({\mathbb{F}}_q)$, so they form the kernel of $1-\pi$. Since the differential of $\pi$ is zero, the differential of $1-\pi$ is the identity; thus $1-\pi$ is a separable isogeny. Its degree is the number of its geometric kernel points. ThereforeWe use the standard degree parallelogram law, together with $\mathrm{deg}[m]=m^2$: degree, extended by zero at the zero endomorphism, is a positive quadratic form on the endomorphism ring of an elliptic curve. Put $t=1+q-N$, the Trace of Frobenius. Polarizing the degree form gives, for every pair of integers $m,n$,Indeed, the values on $1$ and $\pi$ are $1$ and $q$, and the cross coefficient is fixed by $\mathrm{deg}(1-\pi )=1+q-t$. These are standard degree facts about isogenies of elliptic curves, which suffice without an assumption about characteristic or ordinariness.

For $n\ne 0$, division by $n^2$ shows $x^2-tx+q\ge 0$ at every rational $x=m/n$. Density of the rational numbers and continuity of this polynomial make it nonnegative on the real line. Evaluating at its minimum $x=t/2$ gives $q-t^2/4\ge 0$, whence $|t|\le 2\sqrt{q}$. Since $N=q+1-t$, this proves the estimate. This is the degree-form proof of the Hasse bound; it proves both sides simultaneously by positivity of one degree form.

Write the cubic in the Weierstrass equation of an elliptic curve as $f(x)=x^3+a_2{x}^2+a_{4}x+a_6$. A suitable invariant differential on an elliptic curve isIt is regular and nonzero wherever $y\ne 0$. At a point with $y=0$, nonsingularity gives $f^{\prime }(x)\ne 0$, and differentiating the equation shows $\omega =dy/f^{\prime }(x)$, again regular and nonzero. At the identity $O$, use the parameter $t=-x/y$: the expansions start $x=t^{-2}+\cdots$, $y=-t^{-3}+\cdots$, so $\omega =(1+\cdots )dt$. Thus it is a nowhere-vanishing regular differential on the whole smooth projective curve.

Here is a direct proof of translation invariance of a Weierstrass differential. Fix $Q=(u,v)$, let $P=(x,y)$ vary, and write $P+Q=(X,Y)$. On the open set where the chord-and-tangent group law uses an ordinary chord, put $h=(y-v)/(x-u)$. The addition formulas are $X=h^2-a_2-x-u$ and $Y=h(x-X)-y$. The line-intersection identity isDifferentiate this polynomial in $T$ at $T=x$ to obtain $f^{\prime }(x)-2yh=(x-u)(x-X)$. Differentiating $h$ along the curve now gives $dh/dx=(x-X)/(2y)$, and henceThis proves translation invariance on a dense open set. Since the translation on an elliptic curve is an automorphism and both sides are regular differentials, it proves invariance everywhere, including the exceptional addition cases. Translation by $O$ is the identity.

Translation invariance also gives ${\mu }^{\ast }\omega ={\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$ for the addition map: its differential on a tangent pair is the sum of the two translated tangent vectors. Therefore $[j{]}^{\ast }\omega =j\omega$ for every integer $j$.

Let $d_j=\mathrm{deg}[j]$, with the zero endomorphism assigned degree zero. The assumed degree identity, applied to $[j]$ and $[1]$, givesInduction gives $\mathrm{deg}[j]=j^2$ for positive $j$; negation is an automorphism, so the result holds for negative $j$ as well. This is quadratic degree recursion for elliptic multiplication.

For a prime $\ell \ne p$, the invariant differential on an elliptic curve has nonzero pullback under $[\ell ]$, so that map is a separable isogeny. Translation identifies all its fibres and their local multiplicities, so its geometric kernel has exactly its degree ${\ell }^2$ distinct points. It is killed by $\ell$, and thereforeThe algebraic closure is essential here; the assertion is not generally true for the rational-point subgroup over ${\mathbb{F}}_p$ itself. This is prime-to-characteristic geometric torsion.

For $[p]$, the pullback differential vanishes. Its total degree is $p^2$, and its inseparable degree is at least $p$, so its kernel has at most $p$ geometric points. Thus $E({\mathbb{F}}_p)[p]$ has dimension at most one over ${\mathbb{F}}_p$, while every other prime-torsion subgroup has dimension at most two. The rational-point group is finite and abelian. The structure theorem for finite abelian groups says that the number of cyclic factors of an $\ell$-primary component equals the dimension of its subgroup killed by $\ell$. Each component therefore has at most two factors. Combining the smaller primary factors into one cyclic group and the larger ones into another yieldsOne may also choose $p\nmid m$, since the $p$-primary component is cyclic. This explains two generators for elliptic curves over finite fields.

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.

Let $\pi$ be the Frobenius isogeny and let $a=p+1-N$ be its Trace of Frobenius. Its characteristic equation is ${\pi }^2-[a]\pi +[p]=0$, so the trace of the square of an elliptic-curve endomorphism is $a^2-2p$. The elliptic-curve point count over a finite field therefore givesThe prime-square point-count formula is consequentlyOne can also obtain it directly from the degree of an isogeny: the maps $1-\pi$ and $1+\pi$ are separable isogenies, with degrees $p+1-a$ and $p+1+a$. Their composition is $1-{\pi }^2$, whose kernel of an isogeny is exactly $E({\mathbb{F}}_{p^2})$. Multiplicativity of degree gives the same product.

Because $\varphi$ is a separable isogeny,The degree-zero divisorcorresponds under $E_1\simeq {\mathrm{Pic}}^0(E_1)$ to the sum of all elements of the finite abelian group $E_1[\varphi ]$. Pairing every $T$ with $-T$ shows that this sum is the sum of the elements in $E_1[\varphi ]\cap E_1[2]$. It vanishes when that intersection has one element and also when it has four elements, since the sum of the four elements of $(\mathbb{Z}/2\mathbb{Z}{)}^2$ is zero. The principal divisor criterion on an elliptic curve therefore gives a rational function $g$ with $\mathrm{div}(g)=D$.

The divisor $D$ is invariant under $[-1]$, so $[-1{]}^{\ast }g/g$ has zero divisor and is constant. Applying $[-1]$ twice shows that this constant has square one; hence $[-1{]}^{\ast }g=\pm g$.

For a short Weierstrass equation of an elliptic curvethe multiplication-by-two isogeny hasThus one may take $g=y$, and $[-1{]}^{\ast }y=-y$. For multiplication by three, the third division polynomial of an elliptic curvevanishes simply at the eight nonzero points of $E[3]$ and has a pole of order eight at $O_E$. Hence $g={\psi }_3(x)$ has divisor $[3{]}^{\ast }(O_E)-9(O_E)$ and satisfies $[-1{]}^{\ast }g=g$. Both signs occur.

If $\ell$ is a prime different from the characteristic, multiplication by $\ell$ on an elliptic curve has degree ${\ell }^2$ and nonzero differential. It is a separable isogeny, so its geometric kernel has ${\ell }^2$ points, all killed by $\ell$. This gives the displayed group over an algebraic closure; the rational subgroup over the base field may be smaller.