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I would like to illustrate my lecture on p-adic numbers with some elementary results. I proved that the series $e^p=\sum_{n\ge0}\frac{p^n}{n!}$ converges in $\mathbb Q_p$ for every prime $p$.

Now I would like to teach that $e^p$ is irrational in $\mathbb Q_p$ by elementary methods (this is true by Mahler result on transcendence of $e^x$ in $p$-adic fields).

Do you know such an elementary proof. I can't find one in the literature...

Good catch! I meant $\frac{p^n}{n!}$ instead of $\frac1n$. I edited it.

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    $\begingroup$ The sum of $1/n$ doesn't converge $p$-adically, since the $n$'th term doesn't go to $0$. Also, you certainly don't want to start with $n=0$. The series $\sum_{n\ge1} x^n/n$ converges if and only if $|x|_p<1$. And for $e^x$, the series $\sum_{n\ge0}x^n/n!$ converges for $|x|_p<1$ if $p$ is odd, otherwise you need $|x|_2<\frac12$. $\endgroup$
    – Joe Silverman
    Commented Feb 12, 2014 at 3:25
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    $\begingroup$ You should call the constant $e^p$. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 12, 2014 at 3:37
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    $\begingroup$ Please make sure you tell the students that this $p$-adic $e^p$ has nothing at all to do with the real number $e$. But then it also means that this application is less exciting than others, since the $p$-adic $e^p$ (or $e^4$ when $p = 2$) is some new, unfamiliar number in the student's experience. Why not use $p$-adics to show any prime factor in the denominator of $\binom{r}{n}$, for $r \in {\mathbf Q}$, must be a prime in the denominator of $r$? That's a nice use of $p$-adic limits and continuity to say something about rational numbers. $\endgroup$
    – KConrad
    Commented Feb 12, 2014 at 9:16
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    $\begingroup$ @Gerald Edgar: just out of curiosity: how would you go about proving that, for $p$ a prime such that $p\equiv 1\pmod{4}$, the $p$-adic expansion of $\sqrt{-1}$ is aperiodic? Personally, I would prove this as a corollary to the fact that $\sqrt{-1} \notin \mathbb{Q}$. However, you seem to be saying that there is a more direct proof. $\endgroup$
    – R.P.
    Commented Feb 13, 2014 at 0:04
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    $\begingroup$ Does it converge for $p=2$? It seems that no. $\endgroup$
    – Fedor Petrov
    Commented Apr 26, 2016 at 17:25

1 Answer 1

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Maybe I harp on this too much, but I think the $p$-adic exponential is not nearly as interesting as the logarithm. It’s defined on a much larger subgroup than the exponential, it’s a homomorphism (unlike the complex logarithm), and its roots are the $p$-power roots of unity. As a function on $1+D$, where $D$ is the open unit disk in $\Bbb C_p$, it’s onto $\Bbb C_p$, so that even though for $z\in\Bbb C_p$ you won’t usually be able to define $e^z$, you can find $\alpha\in\Bbb C_p^*$ such that $\log\alpha=z$.

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    $\begingroup$ But do you have a simple/elementary proof of the fact that $log(1+p)$ is irrational? Or of the fact that $log(1+x)$ is irrational for every rational $p$-adic $x$ such that $|x|<1$? (the latter would of course answer OP's question) I'd love to see a short and simple proof of the irrationality of just about any $p$-adic "of this sort" (and that is, in fact, transcendental). $\endgroup$
    – Gro-Tsen
    Commented Apr 28, 2016 at 18:29
  • $\begingroup$ Indeed I don’t. The burden of my argument was that it makes better sense to ask about $\log(1+p)$ than $\exp p$. $\endgroup$
    – Lubin
    Commented Apr 29, 2016 at 5:23

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