While studying cohomology theories on the stable homotopy setting, I have come up with the following basic question:


Consider the additive formal group law, $F\_a$, and the multiplicative formal group law, $F\_m$, both defined over a ring $R$. 


If $R$ is a $\mathbb{Q}$-algebra, the exponential series defines a morphism of formal group laws 
$$F\_a \to F\_m \quad , \quad x \mapsto e^x := \sum\_{k=0}^\infty x^k / k! $$ and it is not difficult to check that any other morphism $\,f\colon F\_a \to F\_m\,$ is of the form $f (x) = e^{ax}$, for some $a\in R$. 


I can extend the previous reasoning in case $R$ is an integral domain–of any characteristic–, and I am wondering how many morphisms of formal group laws $F\_a \to F\_m$ are there over a general ring. That is to say:


The question is to describe all the series $\,f \in R[[x]]\,$ such that
$$ f (s+t) = f(s)\cdot f(t) \qquad \mbox{ and } \qquad f(0) = 1 \ . 
$$ 




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On the other hand, I am also interested in endomorphisms of the multiplicative group law $F\_m$. The obvious ones are rising to the $k^{th}$ power, for $k \in \mathbb{Z}$: 


$$ F\_m \to F\_m \quad , \quad x \mapsto (1+x)^k \ , $$ and the analogous question would be:


Describe all the series $\,f \in R[[x]]\,$ such that
$$ f( x + y + xy) = f(x) \cdot f(y) \quad \mbox{ and } \quad f(0) = 1 \ . $$


I think that there only exist these "rising to the $k^{th}$-power" maps, no matter the ring of coefficients.



Morphisms of formal group laws $\,F_a \rightarrow F_m\,$ and $\,F_m\to F_m$
Let $D$ be the divided power ring 
$$ D = \mathbb{Z}[a\_0,a\_1,a\_2,\dotsc]/(a\_0-1,a\_na\_m-(n,m)a\_{n+m}) $$
(where $(n,m)$ denotes the binomial coefficient $(n+m)!/(n!\,m!)$). Then ring maps from $D$ to $R$ biject with power series $f(t)=\sum\_ka\_kx^k\in R[[x]]$ satisfying $f(s+t)=f(s)f(t)$ and $f(0)=1$, and thus with homomorphisms $F\_a\to F\_m$ over $R$. Now $\mathbb{Q}\otimes D$ is just the polynomial ring $\mathbb{Q}[a\_1]$ and you recover the description that you mentioned for homomorphisms $F\_a\to F\_m$ over $\mathbb{Q}$-algebras. On the other hand, for any prime $p$ one can check that
$$ D/p = \mathbb{Z}/p[a\_1,a\_p,a\_{p^2},\dots]/(a\_{p^i}^p=0) $$
This gives rise to a description of the exponential series over any $\mathbb{Z}/p$-algebra $R$. More explicitly, given $b\in R$ with $b^p=0$ we get an exponential series 
$$ E\_b(x)=\sum\_{k=0}^{p-1}(bx)^k/k! $$
and we find that every exponential series hs the form
$$ f(x) = \prod\_i E\_{b\_i}(t^{p^i}) $$
for some sequence of coefficients $b\_i$ such that $b\_i^p=0$.


