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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.

An isogeny of elliptic curves is a nonconstant morphism preserving identity points; it is automatically a finite surjective group homomorphism. On the affine chart $v\ne 0$, putThe equation of $E$ is $t^3+d=s^3$, and the proposed map isIt lands on $E^{\prime }$ becauseThe rational formulas extend across $v=0$ to a morphism sending $O_E$ to $O_{E^{\prime }}$. It is nonconstant, hence an isogeny. On function fields, $t$ satisfies $t^3=y$, so the degree is at most three; generically the three cube roots give three distinct preimages. Equivalently, the points with $v=0$ form its three-element geometric kernel. Therefore

In characteristic zero, the isogeny of elliptic curves $\varphi :E\to E^{\prime }$ is finite and separable, soFor every $T\in E[\varphi ]$, translation ${\tau }_T(P)=P+T$ satisfies $\varphi {\tau }_T=\varphi$. It therefore induces a $K(E^{\prime })$-automorphism of $K(E)$. These translations are distinct, giving $\#E[\varphi ]$ automorphisms of an extension of the same degree. The extension is consequently Galois, andis an isomorphism.

Here descent is an arithmetic calculation with square classes and an isogeny of elliptic curves, rather than the gluing of objects along a covering.

For $E:y^2=x^3+a{x}^2+bx$ and its two-isogenous curve $E^{\prime }:y^2=x^3-2a{x}^2+(a^2-4b)x$, square-class maps on $E(\mathbb{Q})$ and $E^{\prime }(\mathbb{Q})$ determine the Mordell-Weil rank.

For $E:y^2=x(x^2+ax+b)$, set $b^{\prime }=a^2-4b$ and $E^{\prime }:Y^2=X(X^2-2aX+b^{\prime })$. The isogeny of elliptic curves with kernel $\{O,(0,0)\}$ is $\varphi (x,y)=(x+a+b/x,y(1-b/x^2))$. Its dual is $\hat{\varphi }(X,Y)=((X-2a+b^{\prime }/X)/4,Y(1-b^{\prime }/X^2)/8)$. The maps extend over their exceptional affine points and satisfy $\hat{\varphi }\varphi =[2]$.