Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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

share|improve this question

1 Answer

up vote 2 down vote accepted

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

share|improve this answer
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 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 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. – YACP Jan 19 at 15:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.