Annihilating Polynomial of Minimal Degree is Irreducible
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then $f$ is irreducible in $K \sqbrk x$.
Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$.
- $0 = \map f \alpha = \map g \alpha \map h \alpha$
Therefore, as $\map g \alpha, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Hence the result, by Proof by Contradiction.