284 VI Hensel’sp-adic Numbers
Proposition 21Let F be a complete field with respect to the absolute value||and
let the field E be a finite extension of F. Then there is at most one extension of the
absolute value on F to an absolute value on E , and E is necessarily complete with
respect to the extended absolute value.
Proof Lete 1 ,...,enbeabasisforE, regarded as a vector space overF.Thenany
a∈Ecan be uniquely expressed in the form
a=α 1 e 1 +···+αnen,
whereα 1 ,...,αn ∈ F. By Lemma 7, for any extended absolute value there exist
positive real numbersσ,μsuch that
σ|a|≤max
i
|αi|≤μ|a| for everya∈E.
It follows at once thatEis complete. For ifa(k)is a fundamental sequence, thenαi(k)is
a fundamental sequence inFfori= 1 ,...,n.SinceFis complete, there existαi∈F
such thatαi(k)→αi(i= 1 ,...,n)and thena(k)→a,wherea=α 1 e 1 +···+αnen.
It will now be shown that there is at most one extension toEof the absolute value
onF.Sincewesawin§4 that the trivial absolute value onEis the only extension of
the trivial absolute value onF, we may assume that the given absolute value onFis
nontrivial. For a fixeda∈E, consider the powersa,a^2 ,.... For eachkwe can write
ak=α 1 (k)e 1 +···+αn(k)en.
Since|a|<1 if and only if|ak|→0, it follows from the remarks at the beginning
of the proof that|a|<1 if and only if|αi(k)|→ 0 (i= 1 ,...,n). This condition is
independent of the absolute value onE. Thus if there exist two absolute values,|| 1
and|| 2 , which extend the absolute value onF,then|a| 1 <1 if and only if|a| 2 <1.
Hence, by Proposition 3, there exists a positive real numberρsuch that
|a| 2 =|a|ρ 1 for everya∈E.
In factρ=1, since for somea∈Fwe have|a| 2 =|a| 1 >1.
6 LocallyCompactValuedFields................................
We prove first a theorem of Ostrowski (1918):
Theorem 22A complete archimedean valued field F is (isomorphic to) either the real
fieldRor the complex fieldC, and its absolute value is equivalent to the usual absolute
value.
Proof Since the valuation on it is archimedean, the fieldFhas characteristic 0 and
thus containsQ. Since an archimedeanabsolute value onQis equivalent to the usual
absolute value, by replacing the given absolute value onFby an equivalent one we
may assume that it reduces to the usual absolute value onQ. Since the valuation onF