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  "body": "\n## Statement\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $f \\in K[x]$.\nThen the [[quotient ring|0VFW4Y69BT]] $K[x]/(f)$ is a field if and only if $f$ is [[irreducible|00X2A9G9W3]] over $K$.\n\n## Remarks\n\nThe proof: $K[x]$ is a [[principal ideal domain|0RZYDVZ475]]. So $(f)$ is maximal if and only if $f$ is irreducible, by the [[theorem on PIDs|095ZHQ7WCJ]]. Then [[$R/M$ is a field iff $M$ is maximal|0WD3SE7S1X]].\nFor example, $\\mathbb{R}[x]/(x^2 + 1) \\cong \\mathbb{C}$ and $\\mathbb{Q}[x]/(x^2 - 2) \\cong \\mathbb{Q}(\\sqrt{2})$.\nThe field $\\mathbb{F}_2[x]/(x^2 + x + 1)$ has $4$ elements.\nIf $f$ is irreducible of degree $n$, then $K[x]/(f)$ is an extension of $K$ of degree $n$.\nThe class of $x$ in $K[x]/(f)$ is a root of $f$.\n\n## Explanations\n\n### L1 human\n\nYou can build new fields from polynomials.\nTake an irreducible polynomial $f$ and declare $f(x) = 0$.\nThe result is a field in which $f$ has a root, as the complex numbers come from $x^2 + 1 = 0$.\n",
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  "url": "https://math2.tamnd.com/wiki/k-x-f-is-a-field-iff-f-is-irreducible/",
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