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


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


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<y 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<g 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




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


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


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


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