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I'm reading a Dwork's book about G-functions and I'm stuck in one part who assert the following:

If $\pi$ is an element of $K$ such that $|\pi|=\max\{z\in G_K: z<1\}$ then $G_K = \langle|\pi|\rangle$

with $K$ a field with a non-Archimedean absolute value $|\cdot|$. Recall that an absolute value (non-Archimedean) is a function from $K$ to the positive real numbers with $0,$ such that

  1. $x=0$ iff $|x|$

for all $x,y\in K$ we have

  1. $|xy|=|x|y||$, and

  2. $|x+y|\leq\max\{|x|,|y|\}$.

$G_K=\{z\in\mathbb{R}:z=|y|, \text{ for some } y\in K, y\neq 0\}$ and we say that an absolute value is discrete if $G_K$ is a discrete subgroup of the topological group $R\setminus\{0\}$.

In general, I know the proof that every additive discrete subgroup must be cyclic, and in that case the generator are the least element. And the proof uses the division algorithm. I tried to "copy" some arguments to have a contradiction about the maximallity about |\pi| but I couldn't get anything.

I would appreciate any hint, Thanks

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Let $y\in K\setminus\{0\}$. Since $|\pi|<1$, the set $$\{n\in\Bbb Z:|y\pi^n|<1\}$$ is not empty and bounded below. Let $n\in\Bbb Z$ its smallest element. Then $1\leq |y\pi^{n-1}|$, from which $|\pi|\leq|y\pi^n|<1$ which implies $|\pi|=|y\pi^n|$, hence $|y|=|\pi^{1-n}|\in\langle|\pi|\rangle$.

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  • $\begingroup$ thank you!!! This is so creative $\endgroup$ – Camacho Camachito Aug 19 '18 at 20:29

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