Endomorphisms of the multiplicative formal group law Is there a simple description of the ring of endomorphisms $\mathrm{End}(\mathbb{G}_m)$ of the formal group law $$\mathbb{G}_m(X,Y) = X + Y + XY,$$ at least over a ring of characteristic zero? I'm working with coefficients in $\mathbb{Z}_p$, $p$ a prime.
It contains at least the polynomials $(1+X)^n - 1$, $n \in \mathbb{N}$, but I'm not sure what else.
 A: I’m not sure of the right answer to a question of this generality. As mercio correctly notes, if you’re looking for $\Bbb Z_p$-endomorphisms, the answer is $\Bbb Z_p$ (we’re referring to the ring of $p$-adic integers here, of course). And if you’re looking for the $\mathscr O$-endomorphisms, where that’s the ring of integers in an algebraic extension of $\Bbb Q_p$, the answer is still $\Bbb Z_p$, as you can read in the text of Fröhlich. But if $R$ is any subring of $\mathbb Q$, it looks to me as if $\text{End}_R(\Bbb G_m)$ will again be $R$. What about if you take a subring $R$ of an algebraic number field $K$? I think the $R$-endomorphisms of the multiplicative formal group might turn out to be $R\cap\mathbb Q$. It might be a fun exercise to work this out.
A: I assume first that $k$ is a field of characteristic $0$ and that we are looking at $\Bbb G_m(k)$
Suppose $f(X) = \sum_{i \ge 1} a_i X^i \in k[[X]]$ is an endomorphism of $\Bbb G_m(k)$.
Then $f(X)+f(Y)+f(X)f(Y) - f(X+Y+XY) = \sum_{i,j \ge 1} c_{i,j}X^iY^j$ where $c_{i,j}$ is a polynomial expression in the $a_k$. More precisely, $c_{i,j} = a_ia_j - \sum_{0 \le k \le i,j} a_{i+j-k}\frac{(i+j-k)!}{(i-k)!(j-k)!k!}$
First, we notice that we have a lot more equations than what we need.
By simply looking at the equations $c_{i,1}=0$ we obtain $a_{i+1} = \frac 1 {i+1}a_i(a_1-i)$ and this tells us that there is at most one solution for each $a_1$.
Secondly, for $m \in \Bbb Z$, $f(X)= (1+X)^m-1$ gives the solution to this system of equation with $a_1 = m$ (keep in mind that $(1+X)^{-1} = 1-X+X^2-X^3  + \ldots$). This means that the equations $c_{i,j} = 0$ for $i,j>1$, when we express $a_i$ for $i \ge 2$ in terms of $a_1$, return a polynomial equation on $a_1$ which has infinitely many solutions, hence the zero polynomial : so they are consequences of the equations $c_{i,1}=0$.
Hence the endomorphism ring of $\Bbb G_m(k)$ is isomorphic to $(k,+,\times)$ :
If $f(X) = \sum a_iX^i$, $f$ is entirely determined by $a_1$, and conversely, for any $a \in k$, there is the endomorphism $f_a(X) = aX + a(a-1)X^2/2 + a(a-1)(a-2)X^3/6 + \ldots $, which is everything we would want $(1+X)^a-1$ to be if it made sense.
Composition of endomorphisms correspond to multiplication in $k$ :  $f_a \circ f_b = f_{ab}$.
Addition in $k$ corresponds to the addition (via the formal group law) of $f_a$ with $f_b$ : $f_{a+b} = f_a + f_b + f_af_b$

If we work over an $R$ who is only an integral domain of characteristic $0$, then $End(\Bbb G_m(R))$ is the subring of $R$ consisting of the $a_1$ generating a sequence $(a_n)$ of elements of $R$.
If $R = \Bbb Z[X]$ for example, then $End(\Bbb G_m(R))= \Bbb Z$. If $a_1$ is a nonconstant polynomial with leading coefficient $c$ then $a_n$ is a polynomial with leading coefficient $c^n/n!$, which is not an integer when $n$ is large enough.
If $R = \Bbb Z_p$, then we still have $End(\Bbb G_m(R)) = R$ : given some $n \ge 1$, $a(a-1)\ldots(a-n+1) \in n!R$ for $a\in \Bbb Z$, and $\Bbb Z$ is dense in $R$ so this is true for $a \in R$.
