= Prime-multiplication coefficient splitting for formal groups
{title2=$[p](T)=pf(T)+g(T^p)$}
For a <formal group law> over any <commutative ring> $R$, the <invariant differential of a formal group law> satisfies $[p]^*\omega=p\omega$. Its coefficient is a <unit in a ring>, so $[p]'(T)\in pR[[T]]$. If $[p](T)=\sum a_mT^m$, this means $ma_m\in pR$. When $p\nmid m$, the <integer> <Bézout's identity> implies $a_m\in pR$, even if $p$ is a <zero divisor>. Put those coefficients into $pf(T)$ and the exponents divisible by $p$ into $g(T^p)$.
Back to article page
