An order function on a field $K$ is a function $\phi:K\to \mathbb{Z} \cup {\{\infty}\}$ satisfying:
i) $\phi(a) = \infty$ if and only if $a=0$.
ii) $\phi(ab) = \phi(a) + \phi(b)$.
iii) $\phi(a+b) \geq \min( \phi(a), \phi(b))$.
Show that $R=\{z \in K \mid \phi(z) \geq 0\}$ is a DVR with maximal ideal $\mathfrak m = \{z\mid \phi(z)>0\}$, and quotient field $K$. Conversely, show that if $R$ is a DVR with quotient field $K$, then the function $\operatorname{ord}: K \to\mathbb{Z} \cup {\{\infty}\}$ is an order function on $K$.
Giving a DVR with quotient field $K$ is equivalent to defining an order function on $K$.
I need help please. Thank you
$\phi(1) = 2 \phi(1)$ so $\phi(1) = 0$, hence $\phi(x x^{-1}) = \phi(x) + \phi(x^{-1}) = 0$.
So if $\phi(x) = 0$, then $\phi(x^{-1}) = 0$.
The axioms (i), (ii), (iii) tell you that $\rm R$ is a ring, and what I said before tells you that elements $x \in \rm R$ that are not in the ideal $\{\phi(x) > 0\}$ are invertible in $\rm R$. So $\rm R$ is a DVR.