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Usually absolute value $v$ on a field $K$ is defined as a map $x\to |x|_v$ such that $\forall x,y\in K$ one has

$$\begin{align} &|x|_v=0\iff x=0\\ &|xy|_v=|x|_v|y|_v\\ &|x+y|_v\leq |x|_v+|y|_v \end{align}$$

If one has $|x+y|_v\leq \max(|x|_v,|y|_v)$ which is stronger than the triangle inequality, the absolute value is ultra metric or non Archimedean.

Now Marc Hindry in his “arithmétique” suggests a broader definition replacing the third inequality by

$$\exists C_v\gt 0,\,\,|x+y|_v\leq C_ v\max(|x|_v,|y|_v)$$

This definition covers the ultra metric case with $C_v=1$ and if the absolute value verifies the triangle inequality, it verifies the above with $C_v=2$.

How to prove the converse i.e a map such that $|x+y|_v\leq 2\max(|x|_v,|y|_v)$ verifies the triangle inequality.

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    $\begingroup$ Can you point out where (page) Marc Hindry makes that suggestion? $\endgroup$ – Chilote Dec 19 '17 at 1:09
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    $\begingroup$ « Arithmétique » published by C&M. chapter V.2 remarque 2.1.4 page 177 $\endgroup$ – marwalix Dec 19 '17 at 5:09
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I tried to solve it my self for hours but I realized that the problem is far from trivial. So after searching in some books I found the proof and it is not straightforward.

See chapter 3, 18.6-18.9 of the book: Topological Fields, S. Warner, Volume 157 of North-Holland Mathematics Studies, Elsevier, 1989

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  • $\begingroup$ Thanks I will try to find the reference $\endgroup$ – marwalix Dec 19 '17 at 15:10
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    $\begingroup$ Thanks ! I will try to write it down with the minimum required so the post is complete $\endgroup$ – marwalix Dec 20 '17 at 19:10
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To simplify the notations we will write $v(x)=|x|_v$. the definition of absolute value rewrites as follows

$$\begin{align} &v(x)0\iff x=0\tag{*}\label{*}\\ &v(xy)=v(x)v(y)\tag{**}\label{**}\\ &v(x+y)\leq2\max(v(x),v(y))\tag{***}\label{***} \end{align}$$

We will first prove by induction on $r$ that

$$\forall\left(x_i\right)_{1\leq i \leq 2^r},\,v\left(\sum_i x_i\right)\leq 2^r\max_{1\leq i\leq 2^r}\left(v(x_i)\right)\tag{1}\label{1}$$

With $\ref{1}$ in hand we will prove

$$\forall n\in\Bbb{N},\,v(n\cdot 1)\leq 2.n\tag{2}\label{2}$$

Indeed consider $n$ and $r$ such that $2^{r-1}\leq n\lt 2^r$.

Now apply $\ref{1}$ to the $(x_i)_{1\leq i\leq 2^r}$ such that $x_i=1$ for $1\leq i\leq n$ and $x_i=0$ for $n\lt i\leq 2^r$ and you get the result.

$\ref{2}$ combined with the inequality $\ref{1}$ will lead to the triangle inequality and complete the proof.

Indeed assume $\ref{1}$ and $\ref{2}$ and take $n=2^r-1$. One has

$$\begin{align} v(1+x)^n&\stackrel{\ref{**}}=v\left((1+x)^n\right)=v\left(\sum_{k=0}^n{n\choose k}x^k\right)\\ &\stackrel{\ref{1}}\leq 2^r\max_k\left(v\left({n\choose k}x^k\right)\right)\\ &\stackrel{\ref{**}}=2^r\max_k\left(v\left({n\choose k}\cdot 1\right)v\left(x^k\right)\right)\\ &\stackrel{\ref2}\leq 2\cdot2^r\max_k\left({n\choose k}v(x)^k\right)\\ &\leq 2^{r+1}\sum_{k=0}^n{n\choose k}v(x)^k=2^{r+1}\left(1+v(x)\right)^n \end{align}$$

Now take the $n^{th}$ root and let $r$ go to infinity, one gets

$$v(1+x)\leq\lim_{r\to\infty}2^{r+1\over 2^r-1}\left(1+v(x)\right)=1+v(x)$$

We can now conclude.

$$v(x+y)=v(x(1+yx^{-1})=v(x)v(1+yx^{-1})\leq v(x)(1+v(yx^{-1})=v(x)+v(y)$$

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