Understanding the Evaluation Homomorphism and Isomorphism in Extension Fields

[ehrenfest]

[SOLVED] simple extensions

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[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.

 

[ehrenfest]

anyone?

 

[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.

 

[ehrenfest]

That makes sense.