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  "body": "\n## Proof\n\n1. 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]].\n2. 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 $c^{-1}$.\n3. 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]].\n4. 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.\n5. 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 $\\deg m_a$.\n6. [[Evaluation at $a$|0125QVNCFQ]] is a ring homomorphism $L[x] \\to L$, and $K[x] \\subseteq L[x]$. So $g(a) h(a) = m_a(a) = 0$.\n7. Since $L$ is a [[field|0QA8CHH8MG]], $g(a) = 0$ or $h(a) = 0$. Say $g(a) = 0$.\n8. Let $c \\ne 0$ be the leading coefficient of $g$. Then $c^{-1} g$ is monic, vanishes at $a$, and has degree $\\deg g \u003c \\deg m_a$.\n9. This contradicts the minimality of $\\deg m_a$. So $g$ or $h$ is a unit.\n10. By Steps 3 and 9, $m_a$ is [[irreducible|00X2A9G9W3]] over $K$.\n\n## Idea\n\nA 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 $a$.\n",
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