# Properties of valuation map

It seems not so hard to prove but how can we prove by induction. Let $K$ be a field and $\nu$ be a valuation map. If $a_{1} + a_{2} + ... + a_{n} = 0$ then prove that $\nu(a_{i}) = \nu(a_{j})$ for some $i \neq j$, where $a_{1}, a_{2},...,a_{n} \in K$. Thanks

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## 1 Answer

Prove by induction on $n$ the following: if $v(a_1)>\cdots >v(a_n)$, then $v(a_1+\cdots+a_n)=v(a_n)$.

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Thanks YACP. We have $a_{1} + a_{2} + ... + a_{n} = 0$. If $\nu(a_{1})> \nu(a_{2})>...> \nu(a_{n})$ then $\nu(a_{n}) = \nu(0) = \infty$, which is a contradiction because $a_{n}\neq 0$. Thus we must have $\nu(a_{i}) = \nu(a_{j})$ for some $i\neq j$. –  Rajesh Jan 19 '13 at 3:27
@ YACP thanks. Still I could not figure out where did you use $a_{1} + a_{2} + ... + a_{n} = 0$. –  Rajesh Jan 19 '13 at 15:50
@Rajesh I didn't use this (what I said is true for any elements $a_i\in K$). But I've let you the pleasure to finish the proof and you did this in your first comment. Hope all is clear now. –  user26857 Jan 19 '13 at 15:56