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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 [itex] \alpha \in E[/itex]. Suppose [itex]\alpha[/itex] is algebraic over a field F. Let [itex]\phi_{\alpha}[/itex] be the evaluation homomorphism of F[x] into Ewith [itex]\phi_{\alpha}(a)=a[/itex] for [itex]a \in F[/itex] and [itex]\phi_{\alpha}(x)=\alpha[/itex].

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

What I do not understand is how this can be true when [itex]\alpha[/itex] is in F. When [itex]\alpha[/itex] is in F, [itex]\phi_{\alpha}[F[x]][/itex] = F, but I think it is pretty clear that [itex]F[x]/<irr(\alpha,F)>=F[x]/<x-\alpha>[/itex] 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

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# Homework Help: Simple extensions

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