= Solution
Work in characteristic different from two, as in the number-field application below. Nonsingularity is equivalent to $b(a^2-4b)\ne0$. The chord through $P=(x,y)$ and $T=(0,0)$ has slope $y/x$. Using $y^2/x^2=x+a+b/x$ in the <elliptic-curve addition formula> gives
$$
\boxed{x^{\prime}=\frac b x,\qquad y^{\prime}=-\frac{by}{x^2}.}
$$
These formulas hold for $P\ne O,T$; addition interchanges $O$ and $T$.
It follows that $\xi=x+a+b/x=y^2/x^2$ and $\eta=y(1-b/x^2)$. The relation is
$$
\boxed{\eta^2=\xi\bigl(\xi^2-2a\xi+a^2-4b\bigr).}
$$
For a direct verification, observe that
$$
(\xi-a)^2-4b=(x-b/x)^2,
\qquad
\eta^2=\frac{y^2}{x^2}(x-b/x)^2.
$$
Therefore the <two-isogeny formula> is
$$
\boxed{\begin{aligned}
E^{\prime}&:Y^2=X(X^2-2aX+a^2-4b),\\
\phi(x,y)&=\left(x+a+\frac b x,\ y\left(1-\frac b{x^2}\right)\right),\\
\phi(O)&=\phi(T)=O.
\end{aligned}}
$$
The target <elliptic curve> is nonsingular because its corresponding coefficient product is $16b(a^2-4b)\ne0$. The <rational map of projective varieties> extends over the exceptional points to a morphism of smooth projective <algebraic curves>; at both $O$ and $T$ its affine coordinates tend to infinity, giving the displayed values. A nonconstant morphism between <elliptic curves> sending $O$ to $O$ is a <group homomorphism>, so this is an <isogeny of elliptic curves>.
One can also see the quotient directly: translation by $T$ leaves $\xi,\eta$ invariant. The equation $x^2+(a-\xi)x+b=0$ makes the source <function field> a degree-two extension of the target <function field>; its nontrivial automorphism is translation by $T$. Equivalently, the <degree of an isogeny from its x-coordinate map> is two. The <kernel of an isogeny> is precisely $\{O,T\}$. \b[Thus $\phi$ is a separable isogeny of degree two.]
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