an inverse of the Artin-Hasse exponential? In the p-adic world the Artin-Hasse exponential is the sollowing power series:
$$
  E_p(x)= \exp \left( \sum_{n=0}^{\infty}\frac{x^{p^n}}{p^n} \right)
$$
where $E_p(x)\in 1+x\mathbb{Z}_{(p)}[[x]]$ with radius of convergence $r=1$.
My question is : as the classical exponential it is possibile define a sort of 'logarithm' which invert this series?
Thanks for the suggestions!
 A: Let’s consider two formal groups over $\Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $\mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $\mathcal M$, the formal group of multiplication, ${\mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($\Bbb Q$-formal-group isomorphism with the additive formal group ${\mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+\cdots$, which you know all about; in particular, you know that its inverse is $\exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
\log_F(x)=x+\frac{x^p}p+\frac{x^{p^2}}{p^2}+\cdots=\sum_{n=0}^\infty\frac{x^{p^n}}{p^n}\,.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)\equiv x^p\pmod p$. When this happens, the formal groups are $\Bbb Z_p$-isomorphic, and indeed $\Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $\Bbb Z_{(p)}$-isomorphism $\psi:F\to\mathcal M$ such that $\psi'(0)=1$. Note that since it’s a $\Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $\log_F^{-1}\in\Bbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $\log_F^{-1}\circ\log_{\mathcal M}\>$  …
