{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Let $p$ be an odd prime, $a$ a positive integer with $p \\nmid a$, and $h = (p-1)/2$. Let $E = \\set{2, 4, \\dots, p - 1}$, a set of $h$ even numbers.\n2. For $u \\in E$, let $r(u)$ be the least positive residue of $au$ modulo $p$. By [[Euclid's lemma|051VYKA1EX]], $p \\nmid au$, so $1 \\le r(u) \\le p - 1$.\n3. Let $e(u) = r(u)$ if $r(u)$ is even, and $e(u) = p - r(u)$ if $r(u)$ is odd. In both cases $e(u)$ is even with $1 \\le e(u) \\le p - 1$, because $p$ is odd. So $e(u) \\in E$.\n4. In both cases $r(u) \\equiv (-1)^{r(u)} e(u) \\pmod p$.\n5. The map $u \\mapsto e(u)$ is injective on $E$. Suppose $e(u) = e(v)$. Then $r(u) \\equiv \\pm r(v)$, so $au \\equiv \\pm av$, and $p \\mid u - v$ or $p \\mid u + v$ by Euclid's lemma.\n6. Here $\\abs{u - v} \u003c p$. Also $u + v$ is even with $4 \\le u + v \\le 2p - 2$, and the only multiple of $p$ in this range is the odd number $p$. So $u - v = 0$.\n7. Since $E$ is finite, $u \\mapsto e(u)$ is a bijection of $E$. So $\\prod_{u \\in E} e(u) = \\prod_{u \\in E} u$.\n8. By step 4 and [[products of congruences|072D5G4SK6]], $a^h \\prod_{u \\in E} u = \\prod_{u \\in E} au \\equiv \\prod_{u \\in E} r(u) \\equiv (-1)^{\\sum_u r(u)} \\prod_{u \\in E} u \\pmod p$.\n9. Each $u \\in E$ is less than $p$, so $p \\nmid \\prod_{u \\in E} u$ by Euclid's lemma. Hence $a^h \\equiv (-1)^{\\sum_u r(u)} \\pmod p$.\n10. By the [[division algorithm|0CHQKZFK5S]], $au = p q(u) + r(u)$ with $q(u) = \\lfloor au/p \\rfloor$, since $0 \u003c r(u) \u003c p$ and the [[floor function|0YVRNEDSVG]] picks the integer part.\n11. Here $au$ is even and $p$ is odd. So $0 \\equiv q(u) + r(u) \\pmod 2$, and $r(u) \\equiv q(u) \\pmod 2$. Thus $(-1)^{\\sum_u r(u)} = (-1)^{\\sum_u \\lfloor au/p \\rfloor}$.\n12. By [[Euler's criterion|0M7DEDKSYB]], $\\left(\\frac{a}{p}\\right) \\equiv a^h \\equiv (-1)^{\\sum_u \\lfloor au/p \\rfloor} \\pmod p$.\n13. Both sides are $\\pm 1$, since $p \\nmid a$ and by the definition of the [[Legendre symbol|01J8AMV9RS]]. Two signs that agree modulo $p \\ge 3$ are equal, which proves the lemma.\n\n## Idea\n\nFold each residue $r(u)$ to an even number by replacing it with $p - r(u)$ when it is odd. The folded numbers are again the even numbers $2, \\dots, p - 1$, so the sign picked up is $(-1)^{\\sum r(u)}$. The parity of $r(u)$ equals the parity of $\\lfloor au/p \\rfloor$, because $au$ is even and $p$ is odd.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "11A15"
    ],
    "name": "Proof of Eisenstein's lemma",
    "proves": "0XSDBCP4RZ",
    "sources": [
      {
        "locator": "Proof of Eisenstein's lemma",
        "url": "https://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity"
      }
    ],
    "tag": "3SMMPJ6DVN",
    "uses": [
      "01J8AMV9RS",
      "0M7DEDKSYB",
      "051VYKA1EX",
      "0CHQKZFK5S",
      "0YVRNEDSVG",
      "072D5G4SK6",
      "0TPW25CBN6"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/3SMMPJ6DVN/",
  "links": [
    {
      "field": "proves",
      "target": "0XSDBCP4RZ"
    },
    {
      "field": "uses",
      "target": "01J8AMV9RS"
    },
    {
      "field": "uses",
      "target": "0M7DEDKSYB"
    },
    {
      "field": "uses",
      "target": "051VYKA1EX"
    },
    {
      "field": "uses",
      "target": "0CHQKZFK5S"
    },
    {
      "field": "uses",
      "target": "0YVRNEDSVG"
    },
    {
      "field": "uses",
      "target": "072D5G4SK6"
    },
    {
      "field": "uses",
      "target": "0TPW25CBN6"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/3SMMPJ6DVN.md",
  "msc": [
    "11A15"
  ],
  "name": "Proof of Eisenstein's lemma",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/3SM/3SMMPJ6DVN.md",
  "status": null,
  "tag": "3SMMPJ6DVN",
  "uri": "m2://3SMMPJ6DVN",
  "url": "https://math2.tamnd.com/wiki/proof-of-eisensteins-lemma/",
  "writing": {
    "tag": "3SMMPJ6DVN",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}