# Commutative Algebra, Notes 6: Valuations

Definition 1.  Let $F$ be a field.  An absolute value on $F$ is a map $|\cdot|:F\to\mathbb{R}$ such that for all $x,y\in F$

1. $|x|\geq 0$ and $|x|=0$ iff $|x|=0.$
2. $|xy|=|x||y|.$
3. $|x+y|\leq|x|+|y|.$

We will call the pair $(F,|\cdot|)$ an absolute field.  The absolute value that sends all nonzero elements to $1$ is called the trivial absolute value.  One can also verify that the function $d(x,y)=|x-y|$ defines a metric on $F .$  Two absolute values are dependent if their induced topologies are the same, and independent otherwise.  Note that

$|1|=|1^2|=|1|^2$ and $|1|=|(-1)^2|=|-1|^2.$

Hence $|1|=1,$ and thus $|-1|=\sqrt{1}=1$ since $|x|\geq 0.$  Similarly

$|-x|=|-1||x|=|x|$

and $1=|1|=|xx^{-1}|=|x||x^{-1}|\Rightarrow |x^{-1}|=|x|^{-1}.$

Proposition 2.  Let $|\cdot|_1$ and $|\cdot|_2$ be two nontrivial absolute values on $F .$  Then $|\cdot|_1$ and $|\cdot|_2$ are dependent iff $|x|_1<1\Rightarrow |x|_2<1.$  If they are dependent, then there exist some $\lambda>0$ such that $|\cdot|_1=|\cdot|_2^\lambda.$

Theorem 3.  (Approximation Theorem) (Artin-Whaples).  Let $F$ be a field and $|\cdot|_1,...,|\cdot|_n$ be nontrivial pairwise independent absolute values on $F .$  Then there exist elements $x_1,...,x_n\in F$ such that for any $\varepsilon>0,$ there exists an $x\in F$ such that

$|x-x_i|_i<\varepsilon$

for all $1\leq i\leq n.$

An absolute field is complete if all of its Cauchy sequences converge to a point in the field.  One can show that each absolute field has a unique completion field up to isometry.

Definition 4.  An abelian group $G$ is an ordered group if it has a partial ordering such that $x\leq y\Rightarrow xg\leq yg$ for all $x,y,g\in G.$

Proposition 5.  $(G,\leq)$ is an ordered group iff $G=\{1\}\sqcup S\sqcup S^{-1}$ for a multiplicative subset $S\subset G$ where $x<1 for all $x\in S$ and $y\in S^{-1}.$

For an ordered group $G,$ we will hereafter use $G$ to denote $G[0]$ where $0g=0$ and $0 for all $g\in G.$

Definition 6.  Let $F$ be a field and $G$ be an ordered group.  A valuation on $F$ is a map $|\cdot|:F\to G$ such that

1. $|x|=0$ iff $x=0.$
2. $|xy|=|x||y|.$
3. $|x+y|\leq\max\{|x|,|y|\}.$

We will call the triple $(F,G,|\cdot|)$ a valuation field.  The image of $|\cdot|$ on $F$ is an ordered subgroup of $G.$  Two valuations are equivalent if there is an isomorphism $\lambda:F\to G$ that respects order and value.  One can also verify that $|\pm x|=|x|$ and that if $|x|<|y|,$ then $|x+y|=|y|.$

Definition 7.  A subring $R$ of a field $F$ is a valuation ring if for all $x\in F,$ $x\in R$ or $x^{-1}\in R.$

Proposition 8.  If $R\leq F$ is a valuation ring, then it induces a valuation field.

Proof.  Note that $R$ is a local ring, for, if $x,y$ are not units in $R,$ then if $x/y\in R,$  we have

$1+x/y=(x+y)/y\in R.$

So if $x+y$ is a unit in $R$, then $1/y\in R,$ which is a contradiction.  Hence $x+y$ is not a unit.  Furthermore if $x$ is not a unit, then for all $r\in R$ we have that $rx$ is not a unit.  So the nonunits form an ideal in $R,$ which is necessarily maximal (and uniquely so).  So $R$ is a local ring.  Let us denote this ideal by $\mathfrak{m}.$  Thus we can write

$F^*=\mathfrak{m}^*\sqcup U\sqcup\mathfrak{m}^{*^{-1}}$

where $U$ is the group of units of $R.$  We now give $F$ a valuation.  We will define our group $G$ as the quotient group $F^*/U.$  In turn, we define $|x|=xU$ with $|1|=U:=1.$  When then define $|0|=0,$ $|x|<1$ iff $x\in\mathfrak{m}^*,$ and $|x|>1$ iff $x\in\mathfrak{m}^{*^{-1}}.$  Observe that

$|xy|=xyU=xUyU=|x||y|$

and

$|x+y|=(x+y)U\leq\max\{xU,yU\}=\max\{|x|,|y|\}.$

Hence $(F,F^*/U,|\cdot|)$ is a valuation field.

We thus hereafter refer to a valuation ring as the restriction of the induced valuation field to the ring.

Proposition 9.  Let $F\leq K$ be fields where $F$ is a valuation field.  Then there exists an extension making $K$ a valuation field.

Proof Idea.  Let $R=\{x\in F:|x|\leq 1\}$ and $U=\{x\in F:|x|=1\}.$  We will call $R$ the valuation ring of $F .$  One first takes the morphism $\varphi:R\to R/(R-U)$ and extends it to a valuation ring $D$ in $K.$  One can construct an order preserving monomorphism

$\psi:F^*/U\to K^*/U'$

where $U'$ is the maximal ideal in $D.$  We then define $|x|_K$ as before, which is seen to agree with $|x|_F.$

Proposition 10.  Let $F\leq K$ be fields and $[K:F]=n.$  Let $F$ be a $G$-valued valuation field and $G'$ be the extension group to the induced valuation on $K.$  Then $[G':G]\leq n.$

We will call $[G':G]$ the ramification index of $F$ in $K.$

Definition 11.  A valuation is discrete if its codomain is a cyclic group.

It turns out that if $F$ is a valuation field and $\mathfrak{m}$ is the maximal ideal of its valuation ring, then there exists an element $\pi\in\mathfrak{m}$ such that $|\pi|$ is a generator of $G.$  Such an element is called a local parameter of $|\cdot|.$  One can also show that $\mathfrak{m}$ is a principal ideal generated by $\pi.$  Moreover every element $x\in F$ can be written as

$x=u\pi^r$

for some unit $u\in R$ integer $r.$  $r$ is called the order of $x$ at $|\cdot|.$  If $r>0,$ we say $x$ has a zero of order $r,$ and if $r<0,$ then we say $x$ has a pole of order $r.$

Proposition 12.  Let $F\leq K$ be fields where $K$ is a finite extension of $F .$  Further suppose that $F$ is a complete discrete valuation field (i.e. induced metric space is complete) and that $R,R'$ and $\mathfrak{m},\mathfrak{m}'$ are the corresponding valuation rings and maximal ideals after extending the valuation.  Then

$[K:F]=[G':G][R'/\mathfrak{m}',R/\mathfrak{m}].$

[1] Lang, Serge.  Algebra.  Revised Third Edition.  Springer-Verlag.  2000.