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 \(\deg(f) = \deg(h) + \deg(g)\).

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.