Proof that the minimal polynomial is irreducible

Proof

1. Let $L/K$ be a field extension, let $a \in L$ be algebraic over $K$, and let $m_a$ be its minimal polynomial.
2. The units of $K[x]$ are the nonzero constants. Indeed, $u v = 1$ forces $\deg u = 0$ by the degree of a product, and a nonzero constant $c$ has the inverse $c^{-1}$.
3. The only monic polynomial of degree $0$ is $1$, and $1$ does not vanish at $a$. So $\deg m_a \ge 1$. Hence $m_a$ is nonzero and, by Step 2, not a unit.
4. Suppose that $m_a = g h$ with $g, h \in K[x]$, and that neither $g$ nor $h$ is a unit. We derive a contradiction.
5. By Step 2, $\deg g \ge 1$ and $\deg h \ge 1$. Since $\deg g + \deg h = \deg m_a$, both degrees are less than $\deg m_a$.
6. Evaluation at $a$ is a ring homomorphism $L[x] \to L$, and $K[x] \subseteq L[x]$. So $g(a) h(a) = m_a(a) = 0$.
7. Since $L$ is a field, $g(a) = 0$ or $h(a) = 0$. Say $g(a) = 0$.
8. Let $c \ne 0$ be the leading coefficient of $g$. Then $c^{-1} g$ is monic, vanishes at $a$, and has degree $\deg g < \deg m_a$.
9. This contradicts the minimality of $\deg m_a$. So $g$ or $h$ is a unit.
10. By Steps 3 and 9, $m_a$ is irreducible over $K$.