Number Theory: An Introduction to Mathematics

276 VI Hensel’sp-adic Numbers

We now show that, for any nonzeroa∈M, there is a positive integerksuch that
|a|=|π|k. In fact we can choosekso that

|π|k+^1 <|a|≤|π|k.

Then|π|<|aπ−k|≤1, which implies|aπ−k|=1 and hence|a|=|π|k. Thus
the value group of the valued fieldFis the infinite cyclic group generated by|π|.The
final statement of the lemma follows immediately.

It is clear that if an absolute value||on a fieldFis discrete, then its extension to
the completionF ̄ofFis also discrete. Moreover, ifπis a generating element for the
valuation ideal ofF, then it is also a generating element for the valuation ideal ofF ̄.
Suppose now that not only isM =(π)a principal ideal, but the residue field
k=R/Mis finite. Then there exists a finite setS⊆R, with the same cardinality as
k, such that for eacha∈Rthere is a uniqueα∈Sfor which|α−a|<1. Since the
elements ofkare the cosetsα+M,whereα∈S, we callSaset of representativesinR
of the residue field. It is convenient to chooseα=0 as the representative forMitself.
Under these hypotheses a rather explicit representation for the elements of the
valued field can be derived:

Proposition 13Let F be a field with a non-archimedean absolute value||, and let R
and M be the corresponding valuation ring and valuation ideal. Suppose the absolute
value is discrete, i.e. M=(π)is a principal ideal. Suppose also that the residue field
k=R/M is finite, and let S⊆R be a set of representatives of k with 0 ∈S.
Then for each a ∈F there exists a unique bi-infinite sequence(αn)n∈Z,where
αn∈S for all n∈Zandαn= 0 for at most finitely many n< 0 , such that

a=

∑

n∈Z

αnπn.

If N is the least integer n such thatαn= 0 ,then|a|=|π|N. In particular, a∈Rif
and only ifαn= 0 for all n< 0.
If F is complete then, for any such bi-infinite sequence(αn), the series

∑

n∈Zαnπ

n

is convergent with sum a∈F.

Proof Supposea∈Fanda=0. Then|a|=|π|Nfor someN∈Zand hence
|aπ−N|=1. There is a uniqueαN∈Ssuch that|aπ−N−αN|<1. Then|αN|= 1 ,
|aπ−N−αN|≤|π|and

aπ−N=αN+a 1 π,

wherea 1 ∈R. Similarly there is a uniqueαN+ 1 ∈Ssuch that

a 1 =αN+ 1 +a 2 π,

wherea 2 ∈R. Continuing in this way we obtain, for any positive integern,

a=αNπN+αN+ 1 πN+^1 +···+αN+nπN+n+an+ 1 πN+n+^1 ,