I need help proving this theorem:
Given the field extension: $\mathbf{K} \subseteq \mathbf{L}$, for $\alpha \in \mathbf{L}$ and $g(x) \in \mathbf{K}[x]$, $\alpha$'s minimal polynomial over $K$,
and $f(x) \in \mathbf{L}[x]$, $\alpha$'s minimal polynomial over $L$,
then the degree of $g$ is bigger than the degree of $f$ and $f(x)$ divides $g(x)$.
Best Answer
Because $\mathbf{K}\subseteq\mathbf{L}$, you also have $\mathbf{K}[x]\subseteq\mathbf{L}[x]$, so that $g\in \mathbf{L}[x]$ and $g(\alpha)=0$, and therefore (because $f$ is the minimal polynomial of $\alpha$ over $\mathbf{L}$) we must have $f\mid g$, and hence also $\deg(f)\leq\deg(g)$.