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  "body": "\n## Proof\n\n1. Let $K$ be a [[field|0QA8CHH8MG]] and $f \\in K[x]$.\n2. The ring $K[x]$ is a [[Euclidean domain|0DQG39K2NH]]. So it is a [[principal ideal domain|0QKVE1GD9A]].\n3. Case $f = 0$. Then $(f) = \\set{0}$, and the [[quotient ring|0VFW4Y69BT]] $K[x]/(f)$ is isomorphic to $K[x]$ through $g + (0) \\mapsto g$.\n4. In $K[x]$ the element $x$ is not a [[unit|019NFGFPR5]]: $x v = 1$ would give $1 + \\deg v = 0$ by the [[degree of a product|0T233AZ1GE]]. So $K[x]/(0)$ is not a field.\n5. Also $0$ is not [[irreducible|00X2A9G9W3]], since an irreducible polynomial is nonzero by definition. So both sides of the equivalence are false, and the equivalence holds.\n6. Case $f \\ne 0$. By definition, $f$ is irreducible over $K$ if and only if it is an [[irreducible element|052SDEPZH7]] of $K[x]$.\n7. By Step 2 and the [[theorem on principal ideal domains|095ZHQ7WCJ]], this holds if and only if $(f)$ is a [[maximal ideal|09J6KGT65G]] of $K[x]$.\n8. Since $K[x]$ is commutative, [[$K[x]/(f)$ is a field if and only if $(f)$ is maximal|0WD3SE7S1X]].\n9. Steps 6 to 8 give: $K[x]/(f)$ is a field if and only if $f$ is irreducible over $K$.\n\n## Idea\n\nIn a principal ideal domain, irreducible elements are exactly the nonzero generators of maximal ideals. A quotient by an ideal is a field exactly when the ideal is maximal. The zero polynomial is checked separately.\n",
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