= Solution
{c}
Put $U(T)=F_X(0,T)$. The <formal group> identity $F(X,0)=X$ gives $U(0)=1$, so $U$ is a <unit in a ring> in $R[[T]]$, even if $R$ has <zero divisors>.
A differential $h(T)dT$ is invariant under formal translation if
$$
h(F(X,T))F_X(X,T)=h(X)
$$
as an identity of <formal power series>. Setting $X=0$ gives $h(T)U(T)=h(0)$. Consequently every <invariant differential of a formal group law>, if it exists, must have the form $h(0)U(T)^{-1}dT$. In particular there can be only one with constant coefficient one.
Existence follows directly from associativity. Differentiate
$$
F(Z,F(X,T))=F(F(Z,X),T)
$$
with respect to $Z$ at zero, obtaining
$$
U(F(X,T))=F_X(X,T)U(X).
$$
Every displayed $U$ is a <unit in a ring>, so
$$
U(F(X,T))^{-1}F_X(X,T)=U(X)^{-1}.
$$
This is precisely invariance of $U(T)^{-1}dT$. Thus
$$
\boxed{\omega=F_X(0,T)^{-1}dT,\qquad
\{\text{invariant differentials}\}=R\omega.}
$$
Normalization means its constant coefficient is one. This proves both existence and uniqueness over an arbitrary <commutative ring>, without division by any <integer>.
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