MHB Problem (c) for Discrete Value Ring

[cbarker1]

Problem (c) for Discrete Value Ring for a unit

I am stuck in the middle of a proof. Here is the background information from Dummit and Foote Abstract Algebra 2nd ed.:

Let $K$ be a field. A discrete valuation on $K$ on a function $\nu$: $K^{\times} \to \Bbb{Z}$ satisfying

1. $\nu(a\cdot b)=\nu(a)+\nu(b)$ [i.e. $\nu$ is a homomorphism from the multiplication group of nonzero elements of $K$ to $\Bbb{Z}$]
2. $\nu$ is surjective, and
3. $\nu(x+y)\ge \min{[\nu(x),\nu(y)]}$, for all $x,y\in K^{\times}$ with $x+y\ne 0$

The set $R=\left\{x\in K^{\times}| \nu(x)\ge 0\right\} \cup \left\{0\right\}$.
From part b.) Prove that nonzero element $x\in K$ either $x$ or $x^{-1}$ is in $R$.
Proof: Suppose $x\in K^{\times}$. Assume $x\notin R$. Then $\nu(x)<0$, we have
\begin{align*}
&0=\nu(1)\\
&=\nu(xx^{-1})\\
&=\nu(x)+\nu(x^{-1})
\end{align*}
the last line implies $\nu(x^{-1})>0$. Therefore, $x^{-1} \in R$. QED

Here is the question: Prove that an element $x$ is a unit of $R$ if and only if $\nu(x)=0$.

Proof: Suppose $x$ is a unit of $R$. By part (b), $\nu(x)>0$...

Thanks
CBarker1

 

[Euge]

If $x$ is a unit of $R$, then $\nu(x),\, \nu(x^{-1}) \ge 0$; since $\nu(x) + \nu(x^{-1}) = 0$ one deduces $\nu(x) = 0$. Conversely, if $\nu(x) = 0$, then by the equation $\nu(x) + \nu(x^{-1}) = 0$ it follows that $\nu(x^{-1}) = 0$. Thus $x^{-1}\in R$.