Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 / 2

An isogeny of elliptic curves over $k$ is a nonconstant $k$-morphism $\phi :E_1\to E_2$ which sends $O_1$ to $O_2$. A nonconstant morphism of smooth projective curves is finite and surjective. Its compatibility with the elliptic curve group law from Riemann-Roch can be seen without assuming it in the definition.

Identify a point $P$ with the degree-zero divisor class $[P-O_1]$. A finite morphism pushes forward divisors, and pushes a principal divisor to the principal divisor of the norm of its rational function. It therefore induces a group homomorphism on degree-zero Picard groups of a curve. On geometric points, the pushforward of $[P-O_1]$ is $[\phi (P)-O_2]$. Since point addition is addition of these divisor classes, this provesThe equality is defined over $k$, so in particular it holds on the groups of $k$-rational points and after every field extension. The argument also covers inseparable isogenies of elliptic curves; separability is not part of the definition.

The squares in ${\mathbb{F}}_5$ are $0,1,4$. The following table counts affine solutions at each $x$, with one point at infinity to be added:Hence $\#E_1({\mathbb{F}}_5)=\#E_2({\mathbb{F}}_5)=8$. In characteristic five, a finite point has order two precisely when $y=0$. The first cubic has three roots $0,1,4$, giving four elements killed by two, including $O$. The second cubic has only the root $3$, giving two elements killed by two. An isomorphism of groups would preserve these counts, so the groups are not isomorphic.

More precisely $(2,1)\in E_1$ doubles to $(0,0)$ and has order four; the independent two-torsion point $(1,0)$ gives $C_4\times C_2$. On $E_2$, the point $(0,1)$ doubles to $(4,1)$, which doubles to $(3,0)$; it therefore has order eight. Thus the eight-point isogenous elliptic curves over F5 have point groups

Put $u=x_1-1$, $v=y_1$. Then $v^2=u(u^2+3u+2)$. After using the curve equation, the displayed rational functions becomeSet $X_0=X+2=u+3+2/u$. Direct substitution into $v^2=u(u^2+3u+2)$ givesover ${\mathbb{F}}_5$. Replacing $X_0$ by $X+2$ makes this $Y^2=X^3-X+1$. This verifies the target equation explicitly, and identifies the functions with the two-isogeny formula after translating the source's two-torsion point to zero and then translating the target coordinate.

We must also check the omitted affine points. At $T=(1,0)$, $v$ is a local parameter, and $u=v^2/(x_1(x_1+1))$ has order two, since $x_1(x_1+1)=2$ at $T$. Thus $X$ has a pole of order two and $Y$ one of order three. At $O_1$, the usual pole orders on a smooth cubic in Weierstrass form give poles two and three as well: the correction terms have higher order and do not cancel their leading terms. There are no other poles. A rational map from a smooth curve into a projective curve extends across these points; the indicated pole orders show that both $T$ and $O_1$ map to $O_2$. Hence this is a nonconstant identity-preserving morphism, and part (i) makes it an isogeny of elliptic curves.

The pullback of the pole divisor $2O_2$ of $X$ is $2O_1+2T$. Degrees of divisors multiply by the morphism degree, so $2\mathrm{deg}\phi =4$. We obtainThe degree is prime to five, so the elliptic isogeny is separable. Equivalently $dX/Y=d{x}_1/y_1$ is a nonzero pullback of an invariant differential.

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 ${\Delta }_2=-368=2$. ThusThe j-invariant of an elliptic curve is preserved by every isomorphism over the algebraic closure. Since these values differ, the two curves are not geometrically isomorphic, even though the preceding part gives a degree-two elliptic isogeny.