 Solution

Let ${\omega }_G=h_G(T)dT$. The pullback through the formal group homomorphism isThe identity $f(F(X,T))=G(f(X),f(T))$ and the chain rule show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of ${\omega }_F$. Its coefficient at zero is $h_G(0)f^{\prime }(0)=f^{\prime }(0)$, so

Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q^{\prime }(0)=p$. If ${\omega }_F=h(T)dT$ with $h(0)=1$, the pullback formula givesbecause $h(q(T))$ is a unit in a ring. Write $q(T)={\sum }_{m\ge 1}{a}_m{T}^m$. Then $m{a}_m\in pR$ for every $m$. If $p\nmid m$, choose integers $u,v$ with $um+vp=1$. It follows thatChoose $b_m\in R$ with $a_m=p{b}_m$ for those $m$ not divisible by $p$, and defineCoefficient comparison now provesNo cancellation of $p$ was used, so this prime-multiplication coefficient splitting for formal groups remains valid when $p$ is a zero divisor in $R$.