# Minimal polynomial over an extension field divides the minimal polynomial over the base field

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 $g(x)$.

• As you assume $\alpha\in \mathbf L$, isn't automatically $f(x)=x-\alpha$? Jul 14, 2013 at 14:55
• Sorry, I meant $\mathbf{K} \subseteq \mathbf{L} \subseteq \mathbf{M}$ and $\alpha \in \mathbf{M}$. Jul 14, 2013 at 15:04
• Of course we cannot conclude from the stated facts that degree of $g$ is strictly "bigger" than degree of $f$. Indeed your assumptions allow $\mathbf{K} = \mathbf{L}$, so we cannot rule out the possibility $g=f$. Jul 14, 2013 at 15:36

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 $\deg(f)\leq\deg(g)$.
• You go straight from "what we assume" to "what we want to be true". "Because $f$ is the minimal polynomial of $\alpha$ over $L$" can have many implications. Not all of them will get us where we want to be, so be specific: which properties of the minimal polynomial are you using? Apr 5, 2020 at 22:10
• The property that "if $g\in \mathbf{L}[x]$ has $g(\alpha)=0$, then the minimal polynomial of $\alpha$ over $\mathbf{L}$ must divide $g$" - in other words, the property that the minimal polynomial is a generator of the ideal $\{g\in \mathbf{L}[x]:g(\alpha)=0\}$ in the ring $\mathbf{L}[x]$. Apr 6, 2020 at 16:49