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A nonzero isogeny of elliptic curves is separable exactly when its pullback of an invariant differential on an elliptic curve is nonzero. For a map of smooth curves in characteristic $p$, a zero differential means the function-field extension has a nontrivial inseparable part. The criterion turns sums of elliptic endomorphisms into computations of their tangent scalars.

For a Weierstrass equation of an elliptic curve, use the local parameter $t=-x/y$ and put $w=-1/y$. The formal kernel of a minimal Weierstrass equation construction gives $w=t^3+a_1{t}^4+(a_1^2+a_2)t^5+(a_1^3+2a_1{a}_2+a_3)t^6+O(t^7)$. Substitution into the elliptic curve group law yields the displayed initial coefficients and the formal inverse $-t/(1-a_{1}t-a_{3}w(t))$. In characteristic zero, the invariant differential on an elliptic curve becomes $(1+a_{1}t+(a_1^2+a_2)t^2+O(t^3))dt$, giving the formal logarithm $t+a_1{t}^2/2+(a_1^2+a_2)t^3/3+O(t^4)$. The addition coefficients are integral polynomial expressions and hence also apply in positive characteristic of a field.

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

The mapis an automorphism of the elliptic curve over $K$, fixes its identity point, and squares to $[-1]$. For the invariant differential on an elliptic curve $\omega =dx/(2y)$, it satisfies $[i{]}^{\ast }\omega =i\omega$. Thus $\mathbb{Z}[i]\subseteq {\mathrm{End}}_K(E)$, with the indicated embedding of $K$.

Under analytic complex uniformization of an elliptic curve $E(\mathbb{C})\simeq \mathbb{C}/L$, this automorphism acts as multiplication by $i$, so $L$ is stable under $\mathbb{Z}[i]$. It is a rank-one torsion-free module over the Gaussian integers. Since that ring is a principal ideal domain, $L=\Omega \mathbb{Z}[i]$. Every analytic endomorphism is multiplication by an $a\in \mathbb{C}$ with $aL\subseteq L$, which here says exactly $a\in \mathbb{Z}[i]$. Therefore the geometric endomorphism ring of an elliptic curve is $\mathbb{Z}[i]$, and all its elements are generated over $\mathbb{Z}$ by the displayed $K$-defined automorphism. Thus ${\mathrm{End}}_K(E)=\mathbb{Z}[i]$.

If an endomorphism is defined over $\mathbb{Q}$, its pullback on the rational differential $\omega$ has scalar in $\mathbb{Q}$. The differential action is injective in characteristic zero, because a nonzero isogeny of elliptic curves is separable and has nonzero differential. Its scalar must therefore lie in $\mathbb{Q}\cap \mathbb{Z}[i]=\mathbb{Z}$. Conversely integer multiplication is defined over $\mathbb{Q}$. Hence

Put $\lambda =1+i$. The valuations of the differences of the three nonidentity fourth roots of unity from $1$ areFor instance $2=-i{\lambda }^2$, and the other two differences are associates of $\lambda$. None is divisible by ${\lambda }^3$. The group $(\mathbb{Z}[i]/{\lambda }^3{)}^{\times }$ has $8-4=4$ elements, so the four units map bijectively to it. Every ideal prime to $\lambda$ consequently has a unique primary Gaussian integer generator.

The Grössencharacter of the square-lattice elliptic curve is explicitlyFor integral ideals prime to $2$, the generator is an odd Gaussian integer; extend multiplicatively to fractional ideals prime to $2$. Products of primary generators are primary, so this is a Hecke character with ideal infinity type $(1,0)$: on a principal ideal with generator $a\equiv 1(\mathrm{mod}{\lambda }^3)$, its value is $a$. Its conductor divides ${\lambda }^3$, and cannot divide ${\lambda }^2$, since $-1\equiv 1(\mathrm{mod}{\lambda }^2)$ would force ${\psi }_E((-1))=-1$, whereas the unit ideal has value $1$. Thus the displayed conductor is exact.

We next verify directly that this character gives the Frobenius values of this particular curve, fixing the unit ambiguity. The pointsatisfies, by the chord-and-tangent addition law,For the first addition, $[i]P=(-i,1+i)$ and the chord has slope $-1$. Since $P$ is killed by ${\lambda }^3$ but not by ${\lambda }^2$, it generates the eight-element $\mathbb{Z}[i]$-module $E[{\lambda }^3]\simeq \mathbb{Z}[i]/{\lambda }^3$. Thus all this torsion is rational over $K$.

The model has discriminant $64$, so it has good reduction at every $\gamma \ne (\lambda )$. First suppose $\gamma$ lies over a split rational prime $p$. Then its residue field is ${\mathbb{F}}_p$, $p\ge 5$, and $\pi ={\psi }_E(\gamma )$ has norm $p$. Reduction of $[\pi ]$ still has degree $p$ and has zero differential, since its differential scalar $\pi$ belongs to $\gamma$. It is therefore purely inseparable of degree $p$. Factoring out the Frobenius isogeny givesfor an automorphism $u$ of the reduction. In characteristic at least five, the automorphisms of this model fixing $O$ are the four Gaussian units: comparing coefficients in $(x,y)\mapsto (v^{2}x,v^{3}y)$ gives $v^4=1$. Prime-to-$p$ torsion reduces injectively. Both $[\pi ]$ and ${\mathrm{Frob}}_p$ fix the reduction of $E[{\lambda }^3]$, the former by the primary congruence and the latter by its rationality over the residue field. Thus $u$ fixes this entire module. The unit congruences proved above force $u=1$.

For an inert rational prime $p\equiv 3(\mathrm{mod}4)$, we have $\gamma =(p)$, $N\gamma =p^2$, and its primary generator is $\pi =-p$, because $4\mid p+1$ implies ${\lambda }^3\mid (-p-1)$. To determine Frobenius without excluding $p=3$, writeReplacing $x$ by $-x$ changes each nonzero quadratic-character value to its negative, because $-1$ is a nonsquare. Hence $a_p=0$. The Frobenius characteristic equation ${\mathrm{Frob}}_p^2-[a_p]{\mathrm{Frob}}_p+[p]=0$ gives ${\mathrm{Frob}}_{p^2}=[-p]$. This proves the asserted Frobenius value at inert primes as well, and identifies the displayed Hecke character with that of $E$.

Finally, the formal parameter $t=-x/y$ is defined over the local ring at $\gamma$. Since the reduced endomorphism is the $q$-power Frobenius isogeny, with $q=N\gamma$, its pullback sends $t$ to $(-x/y{)}^q=t^q$. Passing to the completion at the identity gives the equality of formal series

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.

Put $N(x)=x^3+x+2$, $H(x)=x^3+2x+1$ and $R(x)=N(x)/(x-1{)}^2$, all in characteristic three. Since $N(1)=1$, the numerator and denominator have no common factor, and $R$ has degree three. The map $x:E\to {\mathbb{P}}^1$ has degree two. Comparing the degrees in $x\circ \varphi =R\circ x$ gives $2\mathrm{deg}\varphi =3\cdot 2$, so $\mathrm{deg}\varphi =3$. This is the degree of an isogeny from its x-coordinate map.

In characteristic three, direct differentiation givesThe supplied $y$-coordinate is $H(x)y/(x-1{)}^3$, so the invariant differential on an elliptic curve satisfiesIn particular $\varphi$ is separable. Its dual isogeny satisfies $\hat{\varphi }\circ \varphi =[3]$. Since $[3{]}^{\ast }\omega =0$ and ${\varphi }^{\ast }\omega =\omega$, the scalar by which $\hat{\varphi }$ pulls back the differential is zero.

Pullback on the elliptic invariant differential is additive for sums of homomorphisms, by the addition identity proved in (a). ConsequentlyThe differential criterion for separability of an isogeny therefore gives separability exactly when $3\nmid m$, with $n$ arbitrary. Such a map is automatically nonzero. When $3\mid m$, every nonzero resulting map is inseparable; the zero map is not a separable isogeny. In fact the zero map occurs only at $(m,n)=(0,0)$: equality of the degrees of $[m]\varphi$ and $[n]\hat{\varphi }$ in a nontrivial vanishing relation would force $m=\pm n$, and then cancellation would force $\varphi =\pm \hat{\varphi }$, contradicting their different differential scalars.

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 short Weierstrass equation $E:y^2=x^3+ax+b$ in characteristic different from two, the invariant differential on an elliptic curveis nonzero and regular, including at $O$.

For a fixed point $Q$ on an elliptic curve, translation is the automorphism $P\mapsto P+Q$. Its inverse is translation by $-Q$. Translations let local computations of an invariant differential on an elliptic curve control every point of the curve.