{
  "aliases": null,
  "backlinks": [
    {
      "field": "uses",
      "target": "08YERZDM9X"
    },
    {
      "field": "uses",
      "target": "0CDAHW576M"
    }
  ],
  "body": "\n## Statement\n\nLet $F$ be a [[finite field|08RTDHX1SK]] of order $q$.\nThen $a^q = a$ for every $a \\in F$, and\n$$x^q - x = \\prod_{a \\in F} (x - a) \\quad \\text{in } F[x].$$\n\n## Remarks\n\nThe proof: the [[group of units|05THZKR9AN]] $F^\\times$ has order $q - 1$, so $a^{q-1} = 1$ for $a \\neq 0$, by [[Lagrange's theorem|3C53TD]]. Then $a^q = a$ for all $a$, also for $a = 0$.\nSo the $q$ elements of $F$ are $q$ distinct roots of $x^q - x$, which has degree $q$. This gives the factorisation.\nFor $q = p$ prime, $a^p = a$ in $\\mathbb{Z}/p\\mathbb{Z}$ is Fermat's little theorem.\n\n## Explanations\n\n### L1 human\n\nIn a field with $q$ elements, raising to the power $q$ changes nothing.\nSo the polynomial $x^q - x$ has every element of the field as a root.\n",
  "commit": "6ac761d40018146541484a9e8b9eafd6ecc303dc",
  "front": {
    "kind": "statement",
    "msc": [
      "12E20"
    ],
    "name": "Every element of a finite field of order q satisfies x^q = x",
    "role": "proposition",
    "sources": [
      {
        "locator": "Multiplicative structure",
        "url": "https://en.wikipedia.org/wiki/Finite_field"
      }
    ],
    "tag": "0QWR8WTZEW",
    "uses": [
      "08RTDHX1SK",
      "05THZKR9AN"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/0QWR8WTZEW/",
  "links": [
    {
      "field": "uses",
      "target": "08RTDHX1SK"
    },
    {
      "field": "uses",
      "target": "05THZKR9AN"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/0QWR8WTZEW.md",
  "msc": [
    "12E20"
  ],
  "name": "Every element of a finite field of order q satisfies x^q = x",
  "prov": null,
  "role": "proposition",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/0QW/0QWR8WTZEW.md",
  "status": null,
  "tag": "0QWR8WTZEW",
  "uri": "m2://0QWR8WTZEW",
  "url": "https://math2.tamnd.com/wiki/every-element-of-a-finite-field-of-order-q-satisfies-x-q-x/",
  "writing": {
    "tag": "0QWR8WTZEW",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}