= 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 $x\circ\phi=R\circ x$ gives $2\deg\phi=3\cdot2$, so \b[$\deg\phi=3$]. This is the <degree of an isogeny from its x-coordinate map>.
In characteristic three, direct differentiation gives
$$
R'(x)=\frac{H(x)}{(x-1)^3}.
$$
The supplied $y$-coordinate is $H(x)y/(x-1)^3$, so the <invariant differential on an elliptic curve> satisfies
$$
\phi^*\omega=\frac{R'(x)dx}{2H(x)y/(x-1)^3}=\omega.
$$
In particular $\phi$ is separable. Its <dual isogeny> satisfies $\widehat\phi\circ\phi=[3]$. Since $[3]^*\omega=0$ and $\phi^*\omega=\omega$, the scalar by which $\widehat\phi$ 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). Consequently
$$
(m\phi+n\widehat\phi)^*\omega=m\omega.
$$
The <differential criterion for separability of an isogeny> therefore gives \b[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 $[m]\phi$ and $[n]\widehat\phi$ in a nontrivial vanishing relation would force $m=\pm n$, and then cancellation would force $\phi=\pm\widehat\phi$, contradicting their different differential scalars.
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