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!


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.