Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 3

For a short Weierstrass equation $E:y^2=x^3+ax+b$ in characteristic different from two, the invariant differential on an elliptic curveis nonzero and regular, including at $O$.

Write $X=x\circ \varphi$. Since $X(-P)=X(P)$, it is an even rational function and therefore belongs to $k(x)$; write $X=p(x)/q(x)$ with coprime polynomials $p,q$. The pullback of a nonzero invariant differential is invariant, so for some $c\in k^{\times }$,Sincewe obtainThus

Take $(x,y)\in E_r(K)\setminus \{O\}$. Because $v(x)<0$ and $p,q$ have degrees $d,d-1$, their leading terms dominate, givingThe polynomial $p^{\prime }q-p{q}^{\prime }$ has degree $2d-2$ with nonzero leading coefficient, the same degree as $q^2$. Their quotient consequently has valuation zero at $x$, soThe isogeny sends $O$ to $O$, hence $\varphi (E_r(K))\subseteq E_r^{\prime }(K)$.

Use the alternative affine coordinatesso that $x=t/w$ and $y=-1/w$. The equation becomesRecursive coefficient comparison gives a unique series $w(t)\in t^{3}k[[t]]$. Since $v(w(t))=3v(t)$ and hence $v(x)=-2v(t)$, $v(y)=-3v(t)$, this gives the formal coordinates on a short Weierstrass curve identification

A one-dimensional commutative formal group law over a ring $R$ is a power series $F(X,Y)\in R[[X,Y]]$ satisfying $F(X,0)=X$, commutativity, and associativity. A morphism $f:F\to G$ is a series $f(T)\in TR[[T]]$ satisfying

Express the isogeny of part b in the formal coordinates of part d and setPart c shows that points approaching $O$ map to points approaching $O^{\prime }$, so $f(t)\in tk[[t]]$. Since $\varphi$ is a group homomorphism, applying the parameter $t^{\prime }$ to $\varphi (P+Q)=\varphi (P)+\varphi (Q)$ givesThus $f$ is the formal-group morphism induced by an isogeny $\hat{E}\to \widehat{E^{\prime }}$.