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Consider the $p$-adic exponential defined over $\mathbb C_p$. One knows $\exp$ is analytic in the domain $\mathcal D=\{z\in\mathbb C_P\mid v_p(z)>\frac1{p-1}\}$. Does it exist an element $z_0\in\mathcal D$ such that $\exp(z_0)=0$? Thanks in advance.

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The exponential function satisfies $\exp \left({x + y}\right) = \exp\left(x \right) \exp\left(y \right)$ for $x, y$ in the convergence domain. It also satisfies $\exp \left( 0 \right) = 1$. So if $\exp \left( z_0 \right) = 0$, then $0 = \exp \left( z_0 \right) \cdot \exp\left(-z_0 \right) = \exp \left( z_0 + (-z_0) \right) = \exp\left( 0 \right) = 1$, which is a contradiction.

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  • $\begingroup$ But what if $\exp\left(-z_0 \right) = \infty$? $\endgroup$
    – Pablo H
    Commented Aug 12, 2020 at 15:21
  • $\begingroup$ $\exp(-z_0)=\infty$ is impossible in the domain on which $\exp$ is analytic, by definition of analytic (and $\mathcal D$ is invariant under $z_0\mapsto-z_0$). $\endgroup$
    – Greg Martin
    Commented Aug 12, 2020 at 18:59
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It’s hard to see in what sense it could be true that the exponential function is “defined over $\Bbb C_p$”, since the logarithm is defined on the whole open unit disk there, and has so very many zeros.

If you look closely, you can see that for all $z\in\mathcal D$, we have $v_p(e^z - 1)=v_p(z)$. This obtains quite independently of any multiplicativity of the exponential.

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    $\begingroup$ The OP surely just means by "defined over $\mathbf C_p$" that its domain is being considered on $\mathbf C_p$, in the same way one might refer to the exponential function on $1 + 5\mathbf Z_5$ as the exponential function "defined over/on $\mathbf Q_5$". I agree this use of "defined over" is awkward, and the idea that an exponential function in any setting might take the value $0$ is peculiar. $\endgroup$
    – KConrad
    Commented Aug 12, 2020 at 1:45
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    $\begingroup$ You don't mean $v_p(e^z) = v_p(z)$ (try $z = 0$!), but rather $v_p(e^z - 1) = v_p(z)$. $\endgroup$
    – KConrad
    Commented Aug 12, 2020 at 1:46
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    $\begingroup$ Of course, @KConrad. You may imagine that I was thinking of the formal-group formulation. I’ll correct. $\endgroup$
    – Lubin
    Commented Aug 12, 2020 at 1:58

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