Any ideas?
f(x) belongs to F[x] and alpha is root of f(x) is some field extension of F. Show irreducible poly. g(x) belonging to F[x] has property that h(x) belonging to F[x] also has alpha as a root then g(x)/h(x) in F[x].
Let $\displaystyle F$ be a field and $\displaystyle E$ be its extension field. Let $\displaystyle \alpha \in E$ be algebraic over $\displaystyle F$ (how do we know it is algebraic?). Next define the evaluation homomorphism:
$\displaystyle \phi_{\alpha} : F[x]\mapsto E$. We know that the $\displaystyle \ker \phi_{\alpha} = \{p(x) \in F[x]| p(\alpha) = 0\}$ is a principal ideal. So $\displaystyle \ker \phi_{\alpha} = \left< q(x) \right>$ for some $\displaystyle q(x) \in F[x]$. This polynomial $\displaystyle q(x)$ must be of minimial degree because otherwise if $\displaystyle q_1(x)$ is smaller degree then we require that $\displaystyle q_1(x) = q(x)\cdot q_2(x)$ for some $\displaystyle q_2(x) \in F[x]$ because the ideal is principal. But that is a contradiction because $\displaystyle \mbox{deg} q_1 (x) < \mbox{ deg} q(x)q_2(x)$. Thus, $\displaystyle q(x)$ is of minimal degree. Next we claim that $\displaystyle q(x)$ is irreducible because if not $\displaystyle q(x) = r(x)s(x)$ and so one of the polynomial has the property that (WLOG)$\displaystyle r(\alpha)=0$ but that cannot be because $\displaystyle r(x)$ has a smaller degree.
So we see that all the polynomial having this zero, i.e. $\displaystyle \ker \phi_{\alpha}$ are generated by an irreducible polynomial $\displaystyle q(x)$. And so if $\displaystyle g(x)\in \ker \phi_{\alpha}$ then $\displaystyle g(x) = q(x)\cdot q_3(x)$ for some $\displaystyle q_3(x)$. This shows that $\displaystyle q(x)$ divides $\displaystyle g(x)$.
1)Okay define $\displaystyle N= \{ f(x) \in F[x] | f(\alpha) = 0\}$.
2)Convince yourself that $\displaystyle N\triangleleft F[x]$.
3)Since $\displaystyle F[x]$ is a PID it follows $\displaystyle N=\left< q(x) \right>$ for some $\displaystyle q(x)\in F[x]$.
4)Use a similar argument as above.
Note, this is basically the same approach except kernels eliminate the need to prove #1 which is a little lengthy but straightforward checking of the field axioms.