I want to show that the $p$-adic exponential function is not periodic (or, equivalently, that there is no $x \in \mathbb{Z}_p, x \neq 0$ with $\exp(x)=1$). I read that this can be showed with Strassmans theorem, but I am not sure why (and how) this theorem may be applied to the series defining $\exp$. Can anyone help me?
$p$-adic exponential and Strassmans theorem
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power-series
p-adic-number-theory
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1You can show the image of the exponential is contained within the radius of convergence of the logarithm and vice versa. Therefore the exponential and logarithm are inverse of each other, and both are homeomorphisms (from a disk to another), hence injective. – 2017-01-10
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0thanks for your answer. Can this also be showed with Strassmans theorem? – 2017-01-10
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0I think you can apply Strassman's theorem to show that the series $\exp(x)-1 = x + \frac{x^2}{2} + \ldots$ has at most $1$ zero. – 2017-01-11
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0But why can Strassmans's theorem be applied for this series? Don't the coefficients have to converge to $0$ in $p$-adic metric? – 2017-01-16
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0@JoelCohen, what you say is correct, so long as you stay within $\Bbb Q_p$. Once the ramification gets great enough, logarithm is much more convergent than exponential. – 2017-01-17
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1@Lubin : you are right! Actually I shouldn't have mentioned the "vice versa" part at all, because it is not needed. We just need the logarithm to be convergent on the image of the exponential. – 2017-01-17
1 Answers
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It’s pretty well known that the exponential series $\exp(x)$ is convergent only for $v_p(x)>1/(p-1)$. If you do take $x$ in that range, $v_p\bigl(\exp(x)-1\bigr)=v_p(x)$, because the first term in the series for $\exp(x)-1$ dominates. Thus no possibility that the value of the exponential can be $1$ for nonzero $x$.
I’m not aware of any extension of the exponential beyond the domain of convergence of the series.
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0thank you. I read in many places (for example [here](https://en.wikipedia.org/wiki/P-adic_exponential_function) that this can be proved with Strassmans theorem an I'm really curious how this works – 2017-01-17
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0I never heard of Strassmann’s Theorem, looked it up, saw that it’s an easy corollary of Weierstrass Preparation. Anyhow, from the exponential series $\exp(x)-1$, you can get one with coeffs going to zero by looking instead at $\exp(px)-1$. Just use well-known facts about the $p$-adic valuation of the factorials. (special argument necessary for $p=2$) – 2017-01-18
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0thank you, now I was able to show it – 2017-01-19