The Isomorphism Extension Theorem

(Section 49) The Isomorphism Extension Theorem

(subsection) Extension theorem

(Isomorphism Extension Theorem) (Thm 49.3) Let E be an algrbraic extension of a field F. Let \sigma be an isomorphism of F onto a field F’. Let \bar{F’} be an algebraic closure of F’. Then \sigma can be extended to an isomorphism \tau of E onto a subfield of \bar{F’} s.t. \tau(a) = \sigma(a) for all a \in F.

(Cor 49.4) If E \le \bar{F} is an algebraic extension of F and \alpha , \beta \in E are conjugate over F, then the conjugation isomorphism \psi_{\alpha , \beta } : F(\alpha ) \to F(\beta ) , given by (Thm 48.3) , can be extended to an isomorphism of E onto a subfield of \bar{F}.

(Cor 49.5) Let \bar{F} and \bar{F’} be two algebraic closures of F. Then \bar{F} is isomorphic to \bar{F’} under an isomorphism leaving each element of F fixed.

(subsection) Index of a Field Extension

(Thm 49.7) Let E be a finite extension of a field F. let \sigma be an isomorphism of F onto a field F’ , and let \bar{F’} be an algebraic closure of F’. Then the number of extensions of \sigma to an isomorphism \tau of E onto a subfield of \bar{F’} is finite, and independent of F’, \bar{F’}, and \sigma. That is, the number of extensions is completely determined by the two fields E and F; it is intrinsic to them.

(Def 49.9) Let E be a finite extension of a field F. The number of isomorphisms of E onto a subfield of \bar{F} leaving F fixed is the index [E : F] of E over F.

(Cor 49.10) If F \le E \le K, where K is a finite extension field of the field F, then [K : F] = [K : E] [E : F] .