1. ## Local fields

A valuation is a function $\displaystyle v: K(field)\to \mathbb R$ so that
$\displaystyle 1) v(0) = \infty\\ 2) v(xy) = v(x)+v(y)\\ 3) v(x+y)\geq min\{v(x),v(y)\}$

Consider the field $\displaystyle F(x)$ of rational functions in one variable over a field $\displaystyle F$.
Let $\displaystyle p\in F[x]$ be a monic irreducible polynomial.
Define the map $\displaystyle \nu_{p}: F(x)\to \mathbb Z\cup \{\infty\}$ as follows:
For a nonzero polynomial $\displaystyle f\in F[x]$ with $\displaystyle p^d||f$, that is,
$\displaystyle p^d|f$ and $\displaystyle p^{d+1}\nmid f,$ set $\displaystyle \nu_p(f) = d;$
For a nonzero rational function $\displaystyle f/g\in F(x)$, set $\displaystyle \nu_p(f/g)=\nu_p(f)-\nu_p(g);$

Set $\displaystyle \nu_p(0) =\infty.$

How can I see that this map $\displaystyle \nu_p$ is a valuation. My notes claims that its valuation ring is:

$\displaystyle \mathfrak o = \left\{ \frac{f}{g} : f,g\in F[x], \; p\nmid g\right\},$
and its maximal ideal is $\displaystyle \mathfrak m = \left\{ \frac{f}{g} : f,g\in F[x], \; p|f, \; p\nmid g \right\}.$
How can I see that $\displaystyle \mathfrak o$ is its valuation ring and $\displaystyle \mathfrak m$ is its maximal ideal?

2. Originally Posted by GaloisGroup
A valuation is a function $\displaystyle v: K(field)\to \mathbb R$ so that
$\displaystyle 1) v(0) = \infty\\ 2) v(xy) = v(x)+v(y)\\ 3) v(x+y)\geq min\{v(x),v(y)\}$

Consider the field $\displaystyle F(x)$ of rational functions in one variable over a field $\displaystyle F$.
Let $\displaystyle p\in F[x]$ be a monic irreducible polynomial.
Define the map $\displaystyle \nu_{p}: F(x)\to \mathbb Z\cup \{\infty\}$ as follows:
For a nonzero polynomial $\displaystyle f\in F[x]$ with $\displaystyle p^d||f$, that is,
$\displaystyle p^d|f$ and $\displaystyle p^{d+1}\nmid f,$ set $\displaystyle \nu_p(f) = d;$
For a nonzero rational function $\displaystyle f/g\in F(x)$, set $\displaystyle \nu_p(f/g)=\nu_p(f)-\nu_p(g);$

Set $\displaystyle \nu_p(0) =\infty.$

How can I see that this map $\displaystyle \nu_p$ is a valuation. My notes claims that its valuation ring is:

$\displaystyle \mathfrak o = \left\{ \frac{f}{g} : f,g\in F[x], \; p\nmid g\right\},$
and its maximal ideal is $\displaystyle \mathfrak m = \left\{ \frac{f}{g} : f,g\in F[x], \; p|f, \; p\nmid g \right\}.$
How can I see that $\displaystyle \mathfrak o$ is its valuation ring and $\displaystyle \mathfrak m$ is its maximal ideal?

It's hard to know from where to start: almost everything follows pretty straightforward from the definitions! Where are you stuck? Perhaps (3) of the definition of valuation is the hardest one here...?

About the local ring and its maximal ideal...I don't, ain't it obious or almost? Do you know how to characterize local rings and their (unique, of course) maximal ideal? Tell us what've you done.

Tonio

3. In definition of valuation how to prove the third condition, and why do we have that p divide f in maximal ideal condition? and why dont we have this condition in the valuation ring. I am defining maximal ideal as the set of elements x of a field for which v(x)>0, and valuation ring as set of elements x for which v(x)>=0. I can see that p should not divide g in the condition but what about f?

thanks

4. Originally Posted by GaloisGroup
In definition of valuation how to prove the third condition, and why do we have that p divide f in maximal ideal condition? and why dont we have this condition in the valuation ring. I am defining maximal ideal as the set of elements x of a field for which v(x)>0, and valuation ring as set of elements x for which v(x)>=0. I can see that p should not divide g in the condition but what about f?

thanks

Try to prove the following rather easy lemma. I think it'll help you out here

Lemma: If in a commutative unitary ring the set of all non-unit elements is an ideal then the ring is a local one and the above ideal is THE UNIQUE maximal one.

Tonio

5. But how can I show that this is a valuation?

6. Originally Posted by GaloisGroup
But how can I show that this is a valuation?
Suppose $\displaystyle \nu_p(f)=n\,,\,\nu_p(g)=m\,,\,\,n\geq m\,,\,n,m,\in\mathbb{N}\cup\{0\}$ , so that $\displaystyle p^n\mid\mid f\,,\,\,p^m\mid\mid g$ , and then

$\displaystyle f(x)=p(x)^nh(x)\,,\,\,g(x)=p(x)^mk(x)\Longrightarr ow f(x)+g(x)=p(x)^m[p(x)^{n-m}h(x)+k(x)]$

$\displaystyle \Longrightarrow \nu_p(f+g)\geq m=min\{\nu_p(f),\,\nu_p(g)\}$ ...

Tonio