Minimal Polynomial is Irreducible

Proof

Let $f\in \mathbb{F}[X]$ be the minimal polynomial of $\alpha$. If $f$ were reducible, then there exists a non-trivial factorisation $f(X)=g(X)h(X)$ where neither $g$ nor $h$ are constant polynomials. Then $f(\alpha )=g(\alpha )h(\alpha )=0$ and therefore $g(\alpha )=0$ or $h(\alpha )=0$ by the zero product property in the field $\mathbb{F}$. Then one of $g$ or $h$ contradicts the minimality of the degree of $f$ given that $\mathrm{deg}(f)=\mathrm{deg}(h)+\mathrm{deg}(g)$.

Proof

Let $f\in \mathbb{F}[X]$ be a monic, irreducible polynomial for which $\alpha$ is a root. However the minimal polynomial then divides this polyonmial, but since $f$ is irreducible, this implies that $f$ is indeed that minimal polynomial. This is only up to a constant multiple, but we impose both polynomials must be monic.