{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe follow the source, but we build the isomorphism from a generator of $F^\\times$. This avoids identifying the two prime subfields by hand.\n\n1. Let $F$ and $F'$ be [[finite fields|08RTDHX1SK]] with $\\abs{F} = \\abs{F'} = q$. By the [[prime power theorem|0H5WWC8RZX]], $q = p^n$ and $q = p'^{n'}$, where $p$ and $p'$ are the characteristics.\n2. By unique factorisation of $q$ into primes, $p = p'$ and $n = n'$.\n3. Let $P$ and $P'$ be the [[prime subfields|0VPFBAY7MW]] of $F$ and $F'$. As in the prime power theorem, the maps $\\Z \\to P$, $k \\mapsto k \\cdot 1$, and $\\Z \\to P'$ are surjective ring homomorphisms with kernel $p\\Z$.\n4. By the [[first isomorphism theorem|0ZMYX7GMH0]], both $P$ and $P'$ are isomorphic to $\\Z/p\\Z$. So $\\iota \\colon P \\to P'$, $k \\cdot 1 \\mapsto k \\cdot 1$, is a ring isomorphism.\n5. Applying $\\iota$ to the coefficients gives a ring isomorphism $P[x] \\to P'[x]$, written $g \\mapsto g^\\iota$. It preserves degrees and fixes $x^q - x$.\n6. By the [[cyclicity of $F^\\times$|0EBZ3G455G]], there is $\\alpha \\in F$ that generates $F^\\times$. Every nonzero element of $F$ is a power of $\\alpha$, so $F = P(\\alpha)$.\n7. Let $g$ be the [[minimal polynomial|0GK1WB9R98]] of $\\alpha$ over $P$. Then $\\deg g = [P(\\alpha) : P] = [F : P] = n$, by the [[degree of a simple extension|0N5QW9K7MH]] and the prime power theorem.\n8. Since $\\alpha^q = \\alpha$ by [[the equation $x^q = x$|0QWR8WTZEW]], the polynomial $g$ [[divides|06K7SARZKR]] $x^q - x$. Write $x^q - x = g h$ with $h \\in P[x]$ of degree $q - n$.\n9. Applying $\\iota$ gives $x^q - x = g^\\iota h^\\iota$ in $P'[x]$. Every $b \\in F'$ is a root of $x^q - x$ by the same result. So $g^\\iota(b) h^\\iota(b) = 0$ for each of the $q$ elements $b \\in F'$.\n10. The polynomial $h^\\iota$ is nonzero of degree $q - n \u003c q$, so it has [[fewer than $q$ roots|0KJVQV3E0F]] in $F'$. So $g^\\iota(\\beta) = 0$ for some $\\beta \\in F'$.\n11. The polynomial $g$ is monic and [[irreducible|0NFEG48RDR]]. The isomorphism of step 5 preserves factorisations and degrees, so $g^\\iota$ is monic and [[irreducible|00X2A9G9W3]] in $P'[x]$.\n12. The minimal polynomial of $\\beta$ over $P'$ divides $g^\\iota$. By step 11 it equals $g^\\iota$.\n13. By the [[structure of simple extensions|0ZN39QDTP9]], $F = P(\\alpha) \\cong P[x]/(g)$ and $P'(\\beta) \\cong P'[x]/(g^\\iota)$. The isomorphism of step 5 maps the ideal $(g)$ onto $(g^\\iota)$, so it induces $P[x]/(g) \\cong P'[x]/(g^\\iota)$.\n14. Composing, we get a ring isomorphism $\\sigma \\colon F \\to P'(\\beta)$. So $P'(\\beta)$ has $q$ elements.\n15. Since $P'(\\beta) \\subseteq F'$ and $\\abs{F'} = q$, we get $P'(\\beta) = F'$. So $\\sigma$ is an isomorphism from $F$ onto $F'$.\n\n## Idea\n\nBoth fields are generated over their copy of $\\F_p$ by a single element. The minimal polynomial of a generator of $F$ divides $x^q - x$, and $x^q - x$ vanishes on all of $F'$. So that polynomial also has a root in $F'$, and the two simple extensions it defines are isomorphic.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "12E20"
    ],
    "name": "Proof of the uniqueness of finite fields of a given order",
    "proves": "010PKCQPXP",
    "sources": [
      {
        "locator": "Existence and uniqueness",
        "url": "https://en.wikipedia.org/wiki/Finite_field"
      }
    ],
    "tag": "ACWQ5B20KG",
    "uses": [
      "08RTDHX1SK",
      "0H5WWC8RZX",
      "0VPFBAY7MW",
      "0ZMYX7GMH0",
      "0EBZ3G455G",
      "0GK1WB9R98",
      "0N5QW9K7MH",
      "06K7SARZKR",
      "0NFEG48RDR",
      "00X2A9G9W3",
      "0QWR8WTZEW",
      "0KJVQV3E0F",
      "0ZN39QDTP9"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/ACWQ5B20KG/",
  "links": [
    {
      "field": "proves",
      "target": "010PKCQPXP"
    },
    {
      "field": "uses",
      "target": "08RTDHX1SK"
    },
    {
      "field": "uses",
      "target": "0H5WWC8RZX"
    },
    {
      "field": "uses",
      "target": "0VPFBAY7MW"
    },
    {
      "field": "uses",
      "target": "0ZMYX7GMH0"
    },
    {
      "field": "uses",
      "target": "0EBZ3G455G"
    },
    {
      "field": "uses",
      "target": "0GK1WB9R98"
    },
    {
      "field": "uses",
      "target": "0N5QW9K7MH"
    },
    {
      "field": "uses",
      "target": "06K7SARZKR"
    },
    {
      "field": "uses",
      "target": "0NFEG48RDR"
    },
    {
      "field": "uses",
      "target": "00X2A9G9W3"
    },
    {
      "field": "uses",
      "target": "0QWR8WTZEW"
    },
    {
      "field": "uses",
      "target": "0KJVQV3E0F"
    },
    {
      "field": "uses",
      "target": "0ZN39QDTP9"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/ACWQ5B20KG.md",
  "msc": [
    "12E20"
  ],
  "name": "Proof of the uniqueness of finite fields of a given order",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/ACW/ACWQ5B20KG.md",
  "status": null,
  "tag": "ACWQ5B20KG",
  "uri": "m2://ACWQ5B20KG",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-uniqueness-of-finite-fields-of-a-given-order/",
  "writing": {
    "tag": "ACWQ5B20KG",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}