The minimal polynomial is irreducible

Statement

Let $L/K$ be a field extension and let $a \in L$ be algebraic over $K$. Then the minimal polynomial $m_a$ of $a$ over $K$ is irreducible over $K$.

Remarks

The proof: if $m_a = g h$ with $g, h \in K[x]$ of smaller degree, then $g(a) h(a) = 0$ in the field $L$. So $g(a) = 0$ or $h(a) = 0$, and this contradicts the minimality of $\deg m_a$. Conversely, a monic irreducible polynomial $f \in K[x]$ with $f(a) = 0$ is the minimal polynomial of $a$. For example, $x^3 - 2$ is irreducible over $\mathbb{Q}$, so it is the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$.

Explanations

L1 human

The minimal polynomial cannot be factored over the base field. If it could, one of the factors would already vanish at the element, and that factor would be a smaller polynomial.