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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 $\mathrm{deg}{\varphi }_i=q$.

With zero degree assigned to the zero element of the homomorphism group of elliptic curves, the degree parallelogram law makes this symmetric pairing additive in both arguments. Its diagonal is the degree of an isogeny; it is positive definite on every finite-dimensional rational span after extension to the real numbers. The integral polarization $2B(\varphi ,\psi )$ is integer-valued, while $B$ can be half-integral. For instance, if an elliptic curve has an automorphism $\omega$ with ${\omega }^2+\omega +1=0$, then $1+\omega =-{\omega }^2$ has degree one, so $B(1,\omega )=-1/2$.

On the product of an elliptic curve with itself, let $D_{+}$ and $D_{-}$ be the diagonal and the graph of negation. The principal divisor of $x(P)-x(Q)$ is $D_{+}+D_{-}-2(\{O\}\times E)-2(E\times \{O\})$: equality of the coordinates means $Q=\pm P$, while $x$ has a double pole at $O$. Pulling back along $(\varphi ,\psi )$ and taking degrees proves the parallelogram law for the degree of an isogeny when $\varphi ,\psi$ are nonzero and $\varphi \ne \pm \psi$. The exceptional cases follow from $\mathrm{deg}[2]=4$. The same argument uses the quotient by negation in characteristic two. Thus degree, with degree zero assigned to the zero map, is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$.

The group ${\mathrm{Hom}}_K(E_1,E_2)$ consists of morphisms of elliptic curves defined over $K$ that preserve the group operation, with pointwise addition. Its zero element is the constant map to the identity. Every nonzero element is an isogeny of elliptic curves: its image is a closed connected subgroup of the target, so nonconstancy makes it the whole curve, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny extends to this group by assigning degree zero to the zero homomorphism.

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.

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.

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 $\pi ,{\pi }^{\prime }$ be the $p$-power Frobenius isogenies of $E,E^{\prime }$. Since $\psi$ is defined over ${\mathbb{F}}_p$, it commutes with Frobenius:Taking degrees, using multiplicativity and cancelling the nonzero degree of an isogeny $\mathrm{deg}\psi$, givesBoth differences are separable, so the preceding proof identifies their degrees with rational point counts. Therefore

Isogenous elliptic curves can have different rational point groups. Here is an explicit example. Over ${\mathbb{F}}_7$, takeThe mapextends across $O$ and $(0,0)$ to a degree-$2$ isogeny of elliptic curves with those two points as its kernel. Substitution verifies the target equation, or this follows from the two-isogeny formula with $a=0,b=-1$. Both curves are smooth modulo $7$.

For $x=0,1,\dots ,6$, the numbers of affine points with that abscissa on $E$ are $(1,1,0,0,2,2,1)$, and on $E^{\prime }$ they are $(1,0,2,2,0,0,2)$. Adding the point at infinity gives $8$ on each. The first curve has four rational points of order dividing $2$, from $O$ and the three roots $0,1,-1$. The second has only two: its quadratic factor $X^2+4$ has no root modulo $7$, since $3$ is not a square. By the classification of finite abelian groups,They are thus isogenous with equal orders and nonisomorphic groups.

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.

For a nonzero isogeny of elliptic curves $\varphi :E_1\to E_2$, its degree of an isogeny is the degree of the induced extension of function fields. Since $\varphi$ commutes with negation, its first coordinate is a rational function of $x_1$ alone. Write it in lowest terms asThe coordinate maps $x_i:E_i\to {\mathbb{P}}^1$ are finite morphisms of degree two. Multiplying function field degrees in the commutative diagram $x_2\circ \varphi =R\circ x_1$ gives the degree of an isogeny from its x-coordinate map:This is the total degree, so the argument also covers an inseparable isogeny of elliptic curves. Counting distinct points in the kernel of an isogeny would give only its separable degree. The zero group homomorphism is assigned degree zero separately.

On a Short Weierstrass form $y^2=x^3+ax+b$ in characteristic of a field other than two, the elliptic-curve addition formula givesThere is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f^{\prime }(r{)}^2$, which is nonzero because the elliptic-curve discriminant is nonzero. The numerator has degree four and the denominator degree three. HenceFor completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general Weierstrass equation of an elliptic curve usewhere $b_2=a_1^2+4a_2$, $b_4=a_1{a}_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2{a}_6+4a_2{a}_6-a_1{a}_3{a}_4+a_2{a}_3^2-a_4^2$. The resultant of this numerator and denominator is ${\Delta }^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.

Work first where all three coordinates are finite and the joining line is not vertical. By the chord-and-tangent group law, the third intersection of $y=\lambda x+\mu$ with the elliptic curve is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic polynomialIts roots are $x_1,x_2,x_3$, counted with intersection multiplicity. Comparing its coefficients with the elementary symmetric polynomials provesThis symmetric-coordinate identity for elliptic-curve addition is an identity of rational functions. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.

Here is how the identity yields the quadratic form for the degree of an isogeny. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomesBoth $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following divisor on an algebraic curve calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.

On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the principal divisor identity iswhere $D_{+}$ is the diagonal and $D_{-}$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at 2-torsion points.

For nonzero isogenies of elliptic curves $\varphi ,\psi :E_1\to E_2$ with $\varphi \ne \pm \psi$, pull this identity back by $(\varphi ,\psi )$. Taking degrees of the resulting principal divisor on $E_1$ gives the divisor proof of the degree parallelogram law:If one map is zero, this is immediate. If $\varphi =\pm \psi$, it follows from $\mathrm{deg}([2]\circ \varphi )=4\mathrm{deg}\varphi$. This also proves the parallelogram law for a general Weierstrass equation of an elliptic curve in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.

Put $q(\varphi )=\mathrm{deg}\varphi$ and $q(0)=0$. Applying the parallelogram law to $n\varphi ,\varphi$ gives the recurrence $q((n+1)\varphi )+q((n-1)\varphi )=2q(n\varphi )+2q(\varphi )$, and hence $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$. Its polarization $B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$ is symmetric and integer-valued. For fixed $\psi$, the parallelogram law makes $B(\cdot ,\psi )$ odd and gives $B(x+y,\psi )+B(x-y,\psi )=2B(x,\psi )$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has bilinearity, and $q$ is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$, with $q(\varphi )>0$ whenever $\varphi \ne 0$.

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.

The Hasse theorem for elliptic curves states that every elliptic curve $E/{\mathbb{F}}_q$ satisfiesLet $\pi$ be the Frobenius isogeny and put $a=\mathrm{tr}(\pi )$. Its fixed points are exactly $E({\mathbb{F}}_q)$, so separability of $1-\pi$ givesThe degree of an isogeny is a nonnegative quadratic form on the endomorphism ring, and for all integers $m,n$,If $a^2>4q$, this quadratic polynomial has two real roots and takes a negative value at some rational $m/n$ between them, hence after clearing denominators at some integer pair $(m,n)$. Therefore $a^2\le 4q$, and substituting $a=q+1-\#E({\mathbb{F}}_q)$ proves the bound. This is the degree-form proof of the Hasse bound.

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.