Number Theory: An Introduction to Mathematics

6 Locally Compact Valued Fields 285

is complete, it now follows thatFcontains (a copy of)Rand that the absolute value
onFreduces to the usual absolute value onR.IfFcontains an elementisuch that
i^2 =−1, thenFcontains (a copy of)Cand, by Proposition 21, the absolute value on
Freduces to the usual absolute value onC.
We now show that ifa∈Fand|a|<1, then 1−ais a square inF.LetBbe the
set of allx∈Fsuch that|x|≤|a|and, for anyx∈B, put

Tx=(x^2 +a)/ 2.

Then alsoTx∈B,since

|Tx|≤(|x|^2 +|a|)/ 2 ≤(|a|^2 +|a|)/ 2 ≤|a|.

Moreover, the mapTis a contraction since, for allx,y∈B,

|Tx−Ty|=|x^2 −y^2 |/ 2 =|x−y||x+y|/ 2 ≤|a||x−y|.

SinceFis complete andBis a closed subset ofF, it follows from the contraction

principle (Proposition I.26) that the mapT has a fixed pointx ̄ ∈ B. Evidently

x ̄=(x ̄^2 +a)/2and

1 −a= 1 − 2 x ̄+ ̄x^2 =( 1 − ̄x)^2.

We show next that, if the polynomialt^2 +1 does not have a root inF, then the

valuation onFcan be extended to the fieldE=F(i),wherei^2 =−1. Eachγ∈Ehas

a unique representationγ=a+ib,wherea,b∈F. We claim that|γ|=

√

|a^2 +b^2 |

is an extension toEof the given valuation onF.

The only part of this claim which is not easily established is the triangle inequality.

To prove it, we need only show that

| 1 +γ|≤ 1 +|γ| for everyγ∈E.

That is, we need only show that

|( 1 +a)^2 +b^2 |≤ 1 + 2

√

|a^2 +b^2 |+|a^2 +b^2 | for alla,b∈F.

Since, by the triangle inequality inF,

|( 1 +a)^2 +b^2 |≤ 1 + 2 |a|+|a^2 +b^2 |,

it is enough to show that

|a|≤

√

|a^2 +b^2 | for alla,b∈F

or, since we may supposea=0,

1 ≤| 1 +c^2 | for everyc∈F.

Assume, on the contrary, that| 1 +c^2 |<1forsomec∈F. Then, by the previous

part of the proof,

−c^2 = 1 −( 1 +c^2 )=x^2 for somex∈F.

Sincec=0, this implies that− 1 =i^2 for somei∈F, which is a contradiction.