# Proving that “Giving a DVR with quotient field K is equivalent to defining an order function on K”

I'm trying to solve problem 2.28 in Fulton's Algebraic Curves.

The problem is the following:

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'm having difficulties showing that R is a DVR. So far, I've demonstrated that R is indeed an integral domain, and that if $$u$$ is a unit in R, then $$\phi(u) = 0$$, hence it remains to show that if $$z$$ is a non-unit in R, then $$\phi(z) > 0$$, and that the ideal of non-units in R is maximal and principal, aswell as that R is noetherian (or any other elementary characterization of DVR'S).

I'd apreciate some help with solving this, as the question is referred to later in the text. Thanks.

## 1 Answer

If $$\phi(z)=0$$ then by property ii) of the order function we must have $$\phi(z^{-1})=0$$ as well and thus $$z^{-1}\in R$$ by definition of $$R$$.

Thus, non-units in $$R$$ must be positive valued by $$\phi$$.

Next, choose $$\pi\in\mathfrak m$$ such that $$\phi(\pi)$$ is minimal (without loss of generality $$\phi(\pi)=1$$ after renormalization). Pick any $$z\in\mathfrak m$$. What can we say about $$\phi(\pi^{-1}z)$$?

• I see. I was able to show that any element of R is expressible as $u\pi^{n}$ for some unit $u$ in R and a natural number (including $0$) $n$. Thank you for the hint. – Gauss57 Feb 27 at 15:38