Definition of an invariant differential form on a formal group These lecture notes define an invariant differential form on a formal group as follows:
My question concerns the statement 

Equivalently, this can be restated as
  $$
P(F(T,S))F_X(T,S) = P(T).
$$

However from the definition
$$
\omega \circ F(T,S) = \omega(T),
$$
one only gets that
$$
P(F(T,S))F_X(T,S)dT = P(T)dT.
$$
It isn't clear to be how they are able to drop the $dT$ from the above statement. 
To be clear on definitions, the space of formal differential forms on a ring of power series $R\left[\left[T\right]\right]$ is the quotient of the $R\left[\left[T\right]\right]$-module spanned by the symbols $df$ for $f \in R\left[\left[T\right]\right]$, by the submodule generated by elements of the form $f'dT - df$ for $f \in R\left[\left[T\right]\right]$.
I would appreciate if anyone was able to help clear this up for me.
 A: I think I have managed to see why this is true formally now. I will provide my solution but leave the question open for the meantime.
Let $M$ be the $R\left[ \left[ T \right]\right]$-module of formal differential forms on a formally group $F$. Then $M$ is the quotient of the free $R\left[ \left[ T \right]\right]$-module $F$ on the set
$$
\left\{df \mid f \in R\left[ \left[ T \right]\right] \right\},
$$
by the submodule $S$ generated by
$$
\left\{f'dT - df \mid f \in R\left[ \left[ T \right]\right]  \right\}. 
$$
Thus if $P,Q \in R\left[ \left[ T \right]\right]$ satisfy $PdT = QdT$ in $M$, then $(P-Q)dT = 0$ in $M$, or equivelently when we view $(P-Q)dT$ as an element of $F$ we get $(P-Q)dT \in S$. There thus exists
$$
f_1, \dots, f_\ell, g_1, \dots, g_{\ell} \in R\left[ \left[ T \right]\right],
$$
such that
$$
(P-Q)dT = \sum_i g_i(f'_idT - df_i).
$$
We may assume, without loss of generality, that the $f_i$ are distinct, and that the $g_i$ are non-zero. Then since $F$ is free we must have $g_i = 0$ unless $f_i = T$, and so we must have $\ell = 1$, and $f_1 = T$. But then
$$
(P-Q)dT = g_1(dT - dT) = 0,
$$
and so $P = Q$ since $F$ is free. 
A: I think the answer is that if $\, f'dT - df = g'dT - dg \,$ then
$\, (f'-g')dT = (df - dg)\,$ and thus $\, df = dg\,$ iff $\, f'=g'\,$ and this is equivalent to $\, f=g+c\,$ where $\,c\,$ is a constant.
Since we assume that $\,P(0)=1\,$ then the constant must be $0$.
