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$\sum\limits_{n=0}^{\infty}\dfrac{1}{n!}$ doesn't converge in $\mathbb{Q}_p$, however, $e^p:=\sum\limits_{n=0}^{\infty}\dfrac{p^n}{n!}$ does converge for $p\neq 2$. So my question is,

Are $e^p\in\mathbb{Q}_p$ for $p\neq2$ (and $e^4\in\mathbb{Q}_2$) known to be non-algebraic numbers?

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    $\begingroup$ Yes, there are known p-adic analogues of Lindemann-Weierstrass, Gelfond-Schneider, Baker,... $\endgroup$
    – Felipe Voloch
    Commented Nov 3, 2012 at 16:18
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    $\begingroup$ So, $\exp(3)$ is a $3$-adic number. Does it have (one or more) cube roots in $\mathbb Q_3$ ? And if so, is one of them naturally singled out as something to call $e$ ? If not, perhaps write $\exp(3)$ and not $e^3$ then... $\endgroup$
    – Gerald Edgar
    Commented Nov 3, 2012 at 20:19
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    $\begingroup$ @Gerald. No cube root. If $a \in \mathbb{Q}_3, a \equiv 1 \mod 3$, then $a^3 \equiv 1 \mod 9$. On the other hand, $\exp(3) \equiv 1+3 \mod 9$. $\endgroup$
    – Felipe Voloch
    Commented Nov 3, 2012 at 21:29

1 Answer 1

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According to the last paragraph in Section 3 of the paper "Transcendental numbers in the p-adic domain" by William W. Adams (Amer. J. of Math., Vol. 88, 1966):

http://www.jstor.org/discover/10.2307/2373193?uid=3738736&uid=2&uid=4&sid=21101389828747

the answer is yes. More specifically, they prove that if $a \in \mathbf{Q}_p$ is a non-zero element, algebraic over $\mathbf{Q}$, such that $\left| a \right| < p^{-1/(p-1)}$, then $\exp(a) \in \mathbf{Q}_p$ is transcendental over $\mathbf{Q}$ (ibid.).

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    $\begingroup$ Of course, the bound on $\left| a \right|$ is there just to make sure that the infinite sum for $\exp$ converges. $\endgroup$
    – R.P.
    Commented Nov 3, 2012 at 16:33
  • $\begingroup$ Why not include additional information (besides a link) ... "Transcendental Numbers in the P-Adic Domain" William W. Adams, American Journal of Mathematics Vol. 88, No. 2 (Apr., 1966), pp. 279-308. $\endgroup$
    – Gerald Edgar
    Commented Nov 3, 2012 at 20:15

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