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

1. Jul 8, 2007

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

2. Jul 8, 2007

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).

3. Jul 8, 2007

Kummer

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.

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