Given an algebraic alpha be of degree n over F, show at most. .

[barbiemathgurl]

let alpha be algebraic over F of degree n, show that there exists at most n isomorphisms mapping F(alpha) onto a subfield of bar F (this means the algebraic closure).

thanx

 

[StatusX]

Let phi be one of the isomorphisms, and apply phi to the minimal polynomial for alpha. This should give you a condition on phi(alpha).

 

[Kummer]

let alpha be algebraic over F of degree n, show that there exists at most n isomorphisms mapping F(alpha) onto a subfield of bar F (this means the algebraic closure).

thanx

Please, reread your questions. This is not the first time you made an error. I was lucky to understand what you asked.

I presume you want to show there are at most n isomorphisms mapping F(a) onto a subfield of bar(F) leaving F fixed.

If \phi:F(a) \mapsto E is an isomorphism for E\leq \bar F then a\mapsto b where a\mbox{ and }b are conjugates. Conversely, if a\mapsto b and their are conjugates then by the Conjugation Isomorphism Theorem there is exactly one such isomorphism leaving F fixed. Let the irreducible monic polynomial for a be c_0+c_1x+...+c_nx^n then there are at most n zeros, and hence at most n conjugate elements.