= Solution
For a <Short Weierstrass form> $y^2=x^3+Ax+B$, use $c_4=-48A$ and $\Delta=-16(4A^3+27B^2)$. Both curves have $A=-1$, so $c_4=48=3$ in $\mathbb F_5$. For $E_1$, $B=0$ gives $\Delta_1=64=4$; for $E_2$, $B=1$ gives $\Delta_2=-368=2$. Thus
$$
\boxed{j(E_1)=3^3/4=3,\qquad j(E_2)=3^3/2=1\quad\text{in }\mathbb F_5.}
$$
The <j-invariant of an elliptic curve> is preserved by every isomorphism over the algebraic closure. Since these values differ, \b[the two curves are not geometrically isomorphic], even though the preceding part gives a degree-two <elliptic isogeny>.
Back to article page
