# A question regarding non archimedean absolute values

I don't understand an equation I am reading in my notes:

Suppose, $|\cdot |$ is a nonarchimedean absolute value on a field $K$ complete wrt this absolute value. Suppose, $|a_0|>|a_i|$ for all $i>0$. Then, $|a_0+a_1+a_2+...|=|a_0|$. I understand that $|a_0+a_1+a_2+...|\leq max\{|a_i|:i\geq 0\}=|a_0|$ by ultrametric inequality. I am not sure how to get the reverse inequality.

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Let $a = \sum a_i$. You can apply the argument to $a_0 = a- (a_1 + \dots + a_n)$. We have then that $$|a_0| \leq \max(|a|, |a_1 + \dots a_n|) = |a|$$ because $|a_0|$ is bigger than $|a_1 + \dots + a_n|$, and consequently the second term cannot contribute to the maximum. This implies that $|a| \geq |a_0|$, which is the reverse inequality you wanted.

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That's what I was attempting. However I am not sure why, $|a_0|>|a_1+a_2+...|$. – Derek Scavo Dec 5 '10 at 7:16
@Derek: This follows from $|a_1 + \dots + a_n| \leq \max_{i>0} |a_i| < |a_0|$, which is in the hypotheses. – Akhil Mathew Dec 5 '10 at 7:17
Damn, I should have seen that. Thanks a lot. – Derek Scavo Dec 5 '10 at 7:18
I think this argument has circular reasoning in it (the part where you say "consequently the second term cannot contribute") so I did the proof thoroughly in an own answer. – born Sep 2 '12 at 20:36

$\newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\jleq}[1]{\justifyed{#1}{\leq}} \newcommand{\jeqtxt}[1]{\jeq{\text{#1}}} \newcommand{\jl}[1]{\justifyed{#1}{<}} \newcommand{\justifyed}[2]{\stackrel{#1}{#2}} \newcommand{\jeq}[1]{\justifyed{#1}{=}} \newcommand{\jleqref}[1]{\jleqtxt{\ref{#1}}} \newcommand{\jeqref}[1]{\jeqtxt{\ref{#1}}} \newcommand{\jleqtxt}[1]{\jleq{\text{#1}}} \newcommand{\df}{\mathrel{\mathop:}=} \newcommand{\norm}[1]{\left|#1\right|}$ Isosceles triangle principle: Let $\norm{\cdot}$ be a non-Archimedean norm on a field $\mathbb{F}$. Then $\forall\ x,y\in\mathbb{F}$ with $\norm{x}<\norm{y}$ it holds that $\norm{x+y}=\norm{y}$.

Proof: Let $x,y\in\mathbb{F}$ with $\norm{x}<\norm{y}$. We now calculate \begin{align} \norm{x-y}\jleqtxt{non-Arch.}\max\set{\norm{x},\norm{y}}=\norm{y}=\norm{x-(x-y)}\jleqtxt{non-Arch.}\max\set{\norm{x},\norm{x-y}}. \end{align} From this equation we get that $\norm{y}\leq\max\set{\norm{x},\norm{x-y}}$ but since we assumed $\norm{x}<\norm{y}$, we have that $\norm{y}\nleq\norm{x}$ so $\max\set{\norm{x},\norm{x-y}}=\norm{x-y}$ actually. Together with the equation above, we get that $\norm{x-y}=\norm{y}$.

Remark: The name "Isosceles Triangle Principle" comes from the geometric interpretation: If $x$ and $y$ are two points of a triangle which has the origin $0$ as it's third point, the sides have the length $x$, $y$ and $x-y$. The principle then says that the third side is as long as the longer one of the others, so every "triangle" is isosceles with respect to a non-Archimedean norm.

Lemma: Let $\norm{\cdot}: \mathbb{F}\rightarrow\mathbb{R}_{\geq 0}$ be a non-Archimedean norm and let $\forall\ i\in \set{0, \ldots, n}: a_i\in\mathbb{F}$. If $\forall\ 0<i\leq n: \norm{a_0}>\norm{a_i}$ it holds that $\norm{\sum_{i=0}^n a_i}=\norm{a_0}$.

Proof: We prove this statement by induction. For $n=0$, the statement is trivial. Now assume that the statement is proven for $n$ and let $a_{n+1}\in\mathbb{F}$ with $\norm{a_0}>\norm{a_{n+1}}$. Now set $x\df a_{n+1}$ and $y=\sum_{i=0}^n a_i$ and note that: $$\norm{y}\jeqtxt{I.H.}\norm{a_0}>\norm{a_{n+1}}=\norm{x}$$ and we can apply the Isosceles Triangle Principle (ITP): $$\norm{\sum_{i=0}^{n+1} a_i}=\norm{\sum_{i=0}^{n} a_i + a_{n+1}} = \norm{x+y}\jeqtxt{ITP}\norm{y}=\norm{\sum_{i=0}^n a_i}\jeqtxt{I.H.}\norm{a_0}.$$

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