# 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.