# Thread: algebraic elements and evaluation homomorphism

1. ## algebraic elements and evaluation homomorphism

Let F be a subfield of E, $\displaystyle \alpha\in E$and $\displaystyle \alpha$is an algebraic over $\displaystyle F$. Let $\displaystyle \phi_\alpha$be the evaluation homomorphism. $\displaystyle F(\alpha)$is the minimal field containing $\displaystyle F,\alpha$
True or false: $\displaystyle \phi_\alpha(F[x])\cong F(\alpha)$?
One main problem is:$\displaystyle \phi_\alpha(F[x])$is a field????

2. Originally Posted by ynj
Let F be a subfield of E, $\displaystyle \alpha\in E$and $\displaystyle \alpha$is an algebraic over $\displaystyle F$. Let $\displaystyle \phi_\alpha$be the evaluation homomorphism. $\displaystyle F(\alpha)$is the minimal field containing $\displaystyle F,\alpha$
True or false: $\displaystyle \phi_\alpha(F[x])\cong F(\alpha)$?
One main problem is:$\displaystyle \phi_\alpha(F[x])$is a field????
yes! let $\displaystyle f(x) \in F[x]$ be the minimal polynomial of $\displaystyle \alpha,$ i.e. $\displaystyle f$ is monic and it has the smallest degree among all $\displaystyle g(x) \in F[x]$ with this property that $\displaystyle g(\alpha)=0.$ then $\displaystyle \ker \phi_{\alpha} = <f(x)>$ and, since

$\displaystyle \phi_{\alpha}:F[x] \longrightarrow F[\alpha]$ is surjective, we have $\displaystyle \frac{F[x]}{<f(x)>} \cong \phi_{\alpha}(F[x]) = F[\alpha].$ finally, since $\displaystyle f(x)$ is irreducible and $\displaystyle F[x]$ is a PID, the ideal $\displaystyle <f(x)>$ is maximal and thus $\displaystyle F[\alpha]$ is a field. as a result

$\displaystyle F[\alpha]=F(\alpha).$ the converse is also true: if $\displaystyle F[\alpha]$ is a field, then $\displaystyle \alpha$ is algebraic over $\displaystyle F.$ the reason is that if $\displaystyle \alpha$ is transcendental over $\displaystyle F,$ then $\displaystyle F[\alpha] \cong F[x]$ and obviously $\displaystyle F[x]$ is never a field.

3. emm...I ignored the fact that $\displaystyle f(x)$is irreducible..