algebraic elements and evaluation homomorphism

**ynj** · August 22nd 2009, 02:39 AM

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

**NonCommAlg** · August 22nd 2009, 03:17 AM

Originally Posted by ynj

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

yes! let $f(x) \in F[x]$ be the minimal polynomial of $\alpha,$ i.e. $f$ is monic and it has the smallest degree among all $g(x) \in F[x]$ with this property that $g(\alpha)=0.$ then $\ker \phi_{\alpha} = <f(x)>$ and, since

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

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

**ynj** · August 22nd 2009, 03:36 AM

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

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