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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.

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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 $\phi(\pi^{-1}z)$?

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  • $\begingroup$ 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. $\endgroup$ – Gauss57 Feb 27 at 15:38

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