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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.

For an elliptic curve over a finite field of characteristic $p$, its rational-point group is finite and abelian. Prime-to-characteristic geometric torsion bounds every other prime-torsion dimension by two. Multiplication by $p$ has degree $p^2$ and inseparable degree at least $p$, so the $p$-torsion dimension is at most one. The structure theorem for finite abelian groups therefore gives two cyclic factors with $m\mid n$ and $p\nmid m$.