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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

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Since there is demonstrated that the function ord: K → Z∪{∞} is an order function on K? – user67293 Mar 18 '13 at 20:51
Is this a homework problem? – user38268 Mar 18 '13 at 21:14

$\phi(1) = 2 \phi(1)$ 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.

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When showing that $R$ is a ring, how do we get that $R$ is closed under additive inverses? Given $b \in R$, how do we prove $-b \in R$ – aherring Apr 2 '15 at 22:21

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