4 Non-Archimedean Valued Fields 275m/n,wheremandnare relatively prime integers,n>0andpdoes not dividen.The
valuation ideal isM=pRpand the residue fieldFp=Rp/pRpis the finite field with
pelements.
As another example, let F = K(t)be the field of rational functions with
coefficients from an arbitrary fieldK and let|| = ||t be the absolute value
considered in example (iii) of§1 for the irreducible polynomialp(t) =t.Inthis
case the valuation ringRis the set of all rational functionsf =g/h,wheregandh
are relatively prime polynomials andhhas nonzero constant term. The valuation ideal
isM=tRand the residue fieldR/Mis isomorphic toK,sincef(t)≡f( 0 )modM
(i.e.,f(t)−f( 0 )∈M).
LetF ̄be the completion ofF.IfR ̄andM ̄ are the valuation ring and valuation
ideal ofF ̄, then evidently
R=R ̄∩F, M=M ̄∩F.MoreoverRisdenseinR ̄since, if 0<ε≤1, for anyα∈R ̄there existsa∈ F
such that|α−a|<εand thena∈R(andα−a∈ M ̄). Furthermore the residue
fieldsR/MandR ̄/M ̄ are isomorphic. For the mapa+M→a+M ̄(a∈R)is an
isomorphism ofR/Monto a subfield ofR ̄/M ̄ and this subfield is not proper (by the
preceding bracketed remark).
The valuation ring of the fieldQpofp-adic numbers will be denoted byZpand
its elements will be calledp-adic integers. The ringZof ordinary integers is dense in
Zp, and the residue field ofQpis the finite fieldFpwithpelements, since this is the
residue field ofQ.
Similarly, the valuation ring of the fieldK((t))of all formal Laurent series is the
ringK[[t]] of all formal power series
∑
n≥ 0 αntn. The polynomial ringK[t]isdenseinK[[t]], and the residue field ofK((t))isK, since this is the residue field ofK(t)
with the absolute value||t.
A non-archimedean absolute value||on a fieldFwill be said to bediscreteif
there exists someδ∈( 0 , 1 )such thata∈Fand|a| =1 implies either|a|< 1 −δor
|a|> 1 +δ. (This situation cannot arise for archimedean absolute values.)
A non-archimedean absolute value need not be discrete, but the examples of non-
archimedean absolute values which we have given are all discrete.
Lemma 12Let F be a field with a nontrivial non-archimedean absolute value||,
and let R and M be the corresponding valuation ring and valuation ideal. Then the
absolute value is discrete if and only if M is a principal ideal. In this case the only
nontrivial proper ideals of R are the powers Mk(k= 1 , 2 ,...).
Proof Suppose first that the absolute value||is discrete and putμ=supa∈M|a|.
Then 0<μ<1 and the supremum is attained, since|an|→μimplies|an+ 1 an−^1 |→1.
Thusμ=|π|for someπ∈ M.Foranya∈Mwe have|aπ−^1 |≤1 and hence
a=πa′,wherea′∈R. ThusMis a principal ideal with generating elementπ.
Suppose next thatMis a principal ideal with generating elementπ.If|a|<1,
thena∈M. Thusa=πa′,wherea′∈R, and hence|a|≤|π|. Similarly if|a|>1,
thena−^1 ∈M. Thus|a−^1 |≤|π|and hence|a|≥|π|−^1. This proves that the absolute
value is discrete.