15/05/06 09:47:47.72 SXg2oFIb.net
>>328 つづき
おっちゃんのために
PAUL B. YALE，Pomona College >>258 より
P137
THEOREM 2. Any isomorphism between sub fields of C extends IQ, the identity map on Q.
THEOREM 3. The only isomorphisms between sub fields of C whose domains include R and which map R into R are IR, Ic, and complex conjugation.
Proof. Let Φ be such an isomorphism, i.e., assume R⊆domain Φ and x∈R
implies Φ(x) ∈R. We first show that Φ preserves order in R. If x <y, then there
is a real number ω such that ω≠O and y -x =ω^2. But then Φ(y) -Φ(x) = [Φ(W)]^2
with Φ(ω) ∈R and Φ(ω)≠O. Hence Φ(y) -Φ(x) is positive, i.e., Φ(x) <Φ(y).
Now assume a∈R, but that a≠Φ(a). Choose a rational number, q, between a and Φ(a).
Since Φ(q) =q by Theorem 2, the order between a and q is reversed by Φ
and we have a contradiction. Hence a∈R implies Φ(a) =a, i.e., IR⊆Φ.
If Φ≠IR, then the domain of Φ is a sub field of C containing R as a proper subset.
In any such subfield we can find a complex number, a+bi, with b≠O as
well as all real numbers. But, since x+yi=x+y([(a+bi)-a]/b), this implies
that the subfield is C itself. Thus the domain of Φ is C. Consider Φ(i). Since
i^2=-1, [Φ(i)]^2=Φ(-1)=-1. The only roots of x^2=-1 are ±i; hence Φ(i)
= ±i. If Φ(i) =i, then Φ = Ic, and if Φ(i) = -i, then Φ is complex conjugation.
Theorem 2 implies that Q has no nontrivial automorphism, and Theorem 3
implies the same for R. Theorem 3 also implies that a nontrivial automorphism
of a subfield of R cannot be extended to an automorphism of R. For example,
the automorphism σ defined just before Theorem 2 cannot be extended to an automorphism of R.