# How to show that an object is a discrete valuation ring? (Fulton, Exercise 2.14)

I need some help to solve the following problem that appears on page 31 of the book of William Fulton entitled Algebraic Curves.

Exercise :

Let $V = \mathbb{A}^1$, $\Gamma (V) = k[X]$, $K = k(V) = k(X)$.

1) For each $a \in k = V$, show that $\mathcal{O}_a ( V )$ is a discrete valuation ring with uniformizing parameter $t = X-a$.

2) Show that $\mathcal{O}_{\infty} = \{ \ \dfrac{F}{G} \in k(X) \mid \mathrm{deg} ( G ) \geq \mathrm{deg} (F) \ \}$ is also a discrete valuation ring, with uniformizing parameter $t = \dfrac{1}{X}$.

• Ok. So perhaps it is given $\;k\;$ is an algebraically closed field? – Timbuc Apr 5 '15 at 15:17

1) The discrete valuation at $a\in k$ of $\frac {c_r(X-a)^r+\dots+c_n (X-a)^n}{b_s(X-a)^s+\dots+b_m(X-a)^m} \; (r\leq n, s\leq m; c_r,b_s\neq 0)$ is $r-s$.
In particular $v_a(X-a)=1$.

2) The discrete valuation at $\infty$ of $\frac {c_0 +\dots+c_n X^n}{b_0+\dots+b_mX^m} \; (c_n,b_m\neq 0)$ is $m-n$.
In particular $v_\infty(\frac 1X)=1$.

Remark
Algebraic closedness of $k$ is irrelevant in the above.
However if $k$ is not algebraically closed there are more discrete valuations than those I have described, namely the $v_{p(X)}$ corresponding to monic irreducible polynomials $p(X)\in k[X]$ of degree $\gt1$.
Those new valuations are described by a formula similar to that in 1), which corresponds to the case $p(X)=X-a$.

• In my book. We need to show that there is an irreducible element $t \in R$ such that every non zero : $z \in R$ may bee written uniquely in the form $z = ut^n$ and $u$ a unit in $R$, and $n$ a nonnegative integer. How to show that ? – Bryan261 Apr 5 '15 at 17:00
• In 1) $t$ is $X-a$ and in 2) $t$ is $\frac 1X$. Since $n$ is the valuation (respectively $r-s$ and $m-n$) you should be able to extract $u$ from the very explicit formula in the answer. Or just write brutally $u=\frac {z}{t^n}$ since you know $z,t$ and $n$ ...Also: beware that there is a clash in notation between your $n$ and my $n$ :-) – Georges Elencwajg Apr 5 '15 at 17:13
• You are welcome, Bryan. – Georges Elencwajg Apr 5 '15 at 17:57