Homework Help: Simple extensions

1. Feb 12, 2008

ehrenfest

[SOLVED] simple extensions

1. The problem statement, all variables and given/known data

My book (Farleigh) says the following (I am paraphrasing):

Let E be an extension field of a field F, and let $\alpha \in E$. Suppose $\alpha$ is algebraic over a field F. Let $\phi_{\alpha}$ be the evaluation homomorphism of F[x] into E with $\phi_{\alpha}(a)=a$ for $a \in F$ and $\phi_{\alpha}(x)=\alpha$ .

Suppose \alpha is algebraic over F. Then $F[x]/<irr(\alpha,F)>$ is isomorphic to the image of $\phi_{\alpha}[F[x]]$ in E.

What I do not understand is how this can be true when $\alpha$ is in F. When $\alpha$ is in F, $\phi_{\alpha}[F[x]]$ = F, but I think it is pretty clear that $F[x]/<irr(\alpha,F)>=F[x]/<x-\alpha>$ will not equal F because x - \alpha will generate a nontrivial ideal. What is wrong here!!!!

Also, I do not understand why the stuff in bold is necessary. I thought that was obvious from the definition of an evaluation map.

2. Relevant equations

3. The attempt at a solution

Last edited: Feb 12, 2008
2. Feb 12, 2008

ehrenfest

I am starting to kind of see why F[x]/<x-alpha> is isomorphic to F, but can someone explain/prove that please? It is probably really easy.

3. Feb 13, 2008

ehrenfest

anyone?

4. Feb 13, 2008

morphism

If alpha is in F, then the evaluation at alpha homomorphism from F[x] to F is surjective and its kernel is <x-alpha>. So by the first isomorphism theorem, F[x]/<x-alpha> =~ F.

And about the bold stuff: it's probably there because that's the definition of phi_alpha.

5. Feb 13, 2008

ehrenfest

That makes sense.