I'm having trouble proving this. This is what I have so far:

Let $F$ be a field.
Let $v(x) \rightarrow 1$ for all $x$ not equal to $0$.

So if we let $x$ be in $F$ where $x$ not zero then we can write $x$ as: $x=qy+r$ for some $y$ in $F$. If $r$ not zero then $V(r)=1$.

Not sure what to do from here. I know eventually I'm supposed to get no remainder $(r=0)$ but I'm stuck at applying the definition and valuation map.


1 Answer 1


You're probably overthinking it. Let $q = x \cdot y^{-1}$, which always exists, because $F$ is a field, and $y \ne 0$. So let $r$ be zero, and you don't need to worry at all about the valuation.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .