Number Theory: An Introduction to Mathematics

6 Locally Compact Valued Fields 289

thus has prime characteristicp. It follows from Proposition 4that the restriction toQ
of the absolute value onEis (equivalent to) thep-adic absolute value. Hence, sinceE
is necessarily complete,Qp⊆E.IfEwere infinite-dimensional as a vector space over
Qpthen, by Lemma 26, it would not be locally compact. HenceEis a finite extension
ofQp.

We consider next locally compact valued fields of prime characteristic.

Proposition 28A valued field F with prime characteristic p is locally compact if and
only if F is isomorphic to the field K((t))of formal Laurent series over a finite field
K of characteristic p, with the absolute value defined in example(iv)of§1. The finite
field K is the residue field of F.

Proof We need only prove the necessity of the condition, since (Corollary 25) we have
already established its sufficiency. SinceFhas prime characteristic, the absolute value
onFis non-archimedean. Hence, by Proposition 23 and Lemma 12, the absolute value
onFis discrete and the valuation idealMis a principal ideal. Letπbe a generating
element forM. By Proposition 23 also, the residue fieldk=R/Mis finite. Evidently
the characteristic ofkmust also bep.Letq=pfbe the number of elements ink.
SinceFhas characteristicp,foranya,b∈F,

(b−a)p=bp−ap

and hence, by induction,

(b−a)p

n

=bp

n

−ap

n

for alln≥ 1.

The multiplicative group ofkis a cyclic group of orderq−1. Choosea∈Rso that
a+Mgenerates this cyclic group. Then|aq−a|<1. By what we have just proved,

aq

n+ 1

−aq

n

=(aq−a)q

n

,

and hence(aq

n
)is a fundamental sequence, by Lemma 10. SinceFis complete, by
Proposition 23, it follows thataq

n

→α∈R. Moreoverαq=α,since

lim

n→∞

(aq

n

)q= lim

n→∞

aq

n+ 1

,

andα−a∈M,sinceaq
n+ 1
−aq
n
∈Mfor everyn≥0. Henceα=0andαq−^1 =1.
Moreoverαj=1for1≤j<q−1, sinceαj ≡ajmodM. It follows that the set
Sconsisting of 0 and the powers 1,α,...,αq−^1 is a set of representatives inRof the
residue fieldk.
SinceFhas characteristicp,αgenerates a finite subringKofR. In factKis a
field, sinceβq=βfor everyβ∈Kand soββq−^2 =1ifβ=0. SinceS⊆Kand the
polynomialxq−xhas at mostqroots inK, we conclude thatS=K. ThusKhasq
elements and is isomorphic to the residue fieldk.
Every elementaofFhas a unique representation

a=

∑

n∈Z

αnπn,