I am struggling to understand the proof of the following proposition Let $A=\{x\in K|v(x)\ge 0\}$ for a field $K$ be a discrete valuation ring. Let $t\in A$ s.t. $v(t)=1$. Then any element $x\in A$ has a unique representation as $x=t^nu$ where u is a unit and $n\in \mathbb{N}$
Proof: Claim 1) an element $u\in A$ is a unit iff $v(u)=0$
Proof: for any $x\in K$, $v(x)=v(1\cdot x)=v(1)+v(x)$ which forces $v(1)=0$ this I get but the next one is sort of mystery...
then $0=v(1)=v(x\cdot x^{-1})=v(x)+v(x^{-1})$ assuming $x\neq 0$ but this implies $v(x^{-1})=-v(x)$ thus $v(x)=0$
how does this imply that $u\in A^{\star}$?
for the second part
take $x\in A$ let $v(x)=n$ say $u=xt^{n}$ then $v(u)=v(xt^{-n})=v(x)-nv(t)=0\Rightarrow u\in A^{\star}$ and $x=t^{n}u$
Any help appreciated
Let $x\in K$. If $x\in A$ then $\nu(x)\ge 0$, so that $\nu(x^{-1})=-\nu(x)\le 0$. But $x$ is a unit in $A$ if and only if with $x\in A$ also $x^{-1}\in A$, that is, if and only if $\nu(x)\ge 0$ and $\nu(x)\le 0$, i.e., if and only if $\nu(x)=0$.