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
$$
f(F(X,Y))=G(f(X),f(Y)).
$$

Express the isogeny of part b in the formal coordinates of part d and set
$$
f(t)=-\frac{X(t,w(t))}{Y(t,w(t))}.
$$
Part c shows that points approaching $O$ map to points approaching $O'$, so $f(t)\in tk[[t]]$. Since $\phi$ is a group homomorphism, applying the parameter $t'$ to $\phi(P+Q)=\phi(P)+\phi(Q)$ gives
$$
f(F_E(t_1,t_2))=F_{E'}(f(t_1),f(t_2)).
$$
Thus $f$ is the <formal-group morphism induced by an isogeny> $\widehat E\to\widehat{E'}$.

Solution

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

An isogeny $\phi:E\to E'$ sends the identity to the identity, so substituting the formal coordinates into $t'=-x'/y'$ gives a series $f(t)\in tK[[t]]$. Compatibility with the elliptic-curve group laws gives
$$
f(F_E(X,Y))=F_{E'}(f(X),f(Y)),
$$
making $f$ a morphism of formal group laws.


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