---
tag: R15G1M6B31
kind: proof
name: Proof that the minimal polynomial is irreducible
msc: [12F05]
proves: 0NFEG48RDR
uses: [0WQ1A8SHTZ, 0NPW1DWECM, 0GK1WB9R98, 00X2A9G9W3, 0BDX4MBKT4, 0T233AZ1GE, 019NFGFPR5, 0125QVNCFQ, 0QA8CHH8MG]
methods: [0J0XBKG64S]
sources:
  - url: https://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)
    locator: Irreducibility
---

## Proof

1. Let $L/K$ be a [[field extension|0WQ1A8SHTZ]], let $a \in L$ be [[algebraic|0NPW1DWECM]] over $K$, and let $m_a$ be its [[minimal polynomial|0GK1WB9R98]].
2. The units of $K[x]$ are the nonzero constants. Indeed, $u v = 1$ forces $\deg u = 0$ by the [[degree of a product|0T233AZ1GE]], and a nonzero constant $c$ has the inverse $c^{-1}$.
3. The only [[monic|0BDX4MBKT4]] 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|019NFGFPR5]].
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 $a$|0125QVNCFQ]] 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|0QA8CHH8MG]], $g(a) = 0$ 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|00X2A9G9W3]] over $K$.

## Idea

A factorisation of $m_a$ into two factors of smaller degree would give, after evaluation at $a$, a product of two elements of a field equal to $0$. One factor must then vanish at $a$. After scaling it to be monic, it would be a polynomial of smaller degree with root $a$.