Lindemann theorem for Artin-Hasse exponential

Question

Though the Lindemann--Weierstrass theorem is not known in the $p$-adic settings, its "Lindemann" part -- the transcendence of $\exp(a)$ for algebraic $a$ with $0<|a|_p<p^{-1/(p-1)}$ -- was shown by K. Mahler in 1932. Is there anything known (unconditionally!) about the transcendence of the values of the Artin--Hasse exponential $$ E_p(x)=\exp\biggl(x+\frac{x^p}p+\frac{x^{p^2}}{p^2}+\dots+\frac{x^{p^n}}{p^n}+\dots\biggr)\in\mathbb Z_p[[x]] $$ at algebraic $a$ with $0<|a|_p<1$?

Answer 1

The following will probably shed neither light nor heat on your question, but perhaps it does offer another way of looking at it:

I look at the Artin-Hasse series not as any kind of exponential but as the strict isomorphism between two height-one formal groups, namely the formal group of multiplication $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$ and its $p$-typicalization $\mathcal F_p$, whose logarithm is exactly the series displayed in your question, namely $\mathcal L_p(x)=\sum_0^\infty p^{-n}x^{p^n}$.

The Artin-Hasse series $E_p:\mathcal F_p\to\hat{\mathbf G}_{\mathrm m}$ is a strict isomorphism, and you’re asking whether it takes suitably small algebraic (over $\Bbb Q$) elements of $\Bbb Z_p$ to algebraic elements (of $\Bbb Z_p$).

Whether it makes sense to ask the same question of the (substitional) inverse of $E_p$ I’ll leave to others to judge; but I’ll go ahead and ask it. In particular, one might ask, just in characteristic $2$, whether the $2$-torsion element of $\hat{\mathbf G}_{\mathrm m}$, namely $-2$, has an algebraic or transcendental image under $E_2^{-1}$. This image, of course, is the unique $2$-torsion element of $\mathcal F_2$, and this is the unique $\Bbb Z_2$-rational root of $\mathcal L_2(x)$. I see no reason why this number should be algebraic; a proof either way would be wonderful.

Answer 2

Consider the Dwork exponential $E(x)=\exp(\pi(x-x^p))$ with $\pi=\sqrt[p-1]{-p}$. It is well known that $E(1)=\gamma_p$ a $p$-th root of $1$ and hence not transcendant.