 Solution

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.