= Two-isogeny formula
{title2=$\phi:E\longrightarrow E^{\prime},\quad\deg\phi=2$}
Assume $\operatorname{char}K\ne2$, $b\ne0$ and $b'=a^2-4b\ne0$. Then
$$
E:y^2=x(x^2+ax+b),\qquad E':Y^2=X(X^2-2aX+b')
$$
are <elliptic curves>, and
$$
\phi(x,y)=(x+a+b/x,\ y(1-b/x^2))
$$
is a degree-two <separable isogeny> with <kernel> $\{O,(0,0)\}$. Substitution verifies the target equation. The poles at $O$ and $(0,0)$ extend to the target <point> at infinity, giving a <morphism> of the full <projective curves>. This is a special case of <Vélu formulas>.
Applying the same construction to $E'$ gives a curve with coefficients $4a,16b$. Rescaling that curve to $E$ supplies the <dual isogeny>
$$
\widehat\phi(X,Y)=\left(\frac{X-2a+b'/X}{4},\ \frac{Y(1-b'/X^2)}8\right),\qquad\widehat\phi\phi=[2].
$$
The factors four and eight are essential to this dual normalization. The <characteristic> restriction permits these divisions, while $bb'\ne0$ excludes singular cubic models.
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