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Let $R$ be a ring and $v: R \to \mathbb Z \cup \{+\infty\}$ a map that meets following axioms:

  1. $v(a) = +\infty \iff a=0$
  2. $v(ab) = v(a)+v(b)$
  3. $v(a+b) \geq \min\{v(a),v(b)\}$

I have to show that $v(a) \neq v(b)$ implies $v(a+b) = \min\{v(a),v(b)\}$, but I have no idea where to begin, can anyone give me some hints?

flawr
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1 Answers1

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Assume $v(a)>v(b)$ and $v(a+b)>v(b)$. Then $$ v(b)=v((a+b)+(-a))\ge\min\{v(a+b),v(-a)\}=\min\{v(a+b),v(a)\}>v(b).$$

Hagen von Eitzen
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    You should also prove that $v(-a)=v(a)$, but it's easy. – egreg Oct 27 '15 at 09:58
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    @egreg This would just have been my follow up question, I tried to prove this but I just do not get anywhere, can you give me another hint how to prove $v(-a) = v(a)$? – flawr Oct 27 '15 at 10:04
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    @hagenvoneitzen Thank you very much for your elegant answer so far! – flawr Oct 27 '15 at 10:05
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    @flawr Prove $v(1)=0$, then $v(-1)=0$ follows; therefore $v(-a)=v((-1)a)=v(-1)+v(a)$. – egreg Oct 27 '15 at 10:21
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    Thank you very much! This seems so easy when you know the solution, but it seems you just have to see it or you'll never see it =) – flawr Oct 27 '15 at 10:31