# 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!

## 1 Answer

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 $$\log_F^{-1}\circ\log_{\mathcal M}\>$$