# algebraic elements and evaluation homomorphism

• Aug 22nd 2009, 03:39 AM
ynj
algebraic elements and evaluation homomorphism
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????
• Aug 22nd 2009, 04:17 AM
NonCommAlg
Quote:

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} = $ and, since

$\phi_{\alpha}:F[x] \longrightarrow F[\alpha]$ is surjective, we have $\frac{F[x]}{} \cong \phi_{\alpha}(F[x]) = F[\alpha].$ finally, since $f(x)$ is irreducible and $F[x]$ is a PID, the ideal $$ 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.
• Aug 22nd 2009, 04:36 AM
ynj
emm...I ignored the fact that $f(x)$is irreducible..