{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\n1. Let $(B, \\wedge, \\vee, \\neg, 0, 1)$ be a [[Boolean algebra|0XADJKM699]].\nPut $a + b = (a \\wedge \\neg b) \\vee (\\neg a \\wedge b)$ and $a \\cdot b = a \\wedge b$.\nThen $(B, +, \\cdot, 0, 1)$ is a [[Boolean ring|0RCV1ZFP8G]].\n2. Let $(R, +, \\cdot, 0, 1)$ be a Boolean ring.\nPut $a \\wedge b = ab$, $a \\vee b = a + b + ab$ and $\\neg a = 1 + a$.\nThen $(R, \\wedge, \\vee, \\neg, 0, 1)$ is a Boolean algebra.\n3. The two constructions are inverse to each other.\n4. A map between Boolean algebras is a homomorphism if and only if it is a unital ring homomorphism of the matching rings.\n\n## Remarks\n\nThe proof is a direct check of axioms.\nIn a Boolean ring, $(a + b)^2 = a + b$ gives $ab + ba = 0$, and $a = -a$ follows from $(a + a)^2 = a + a$, so the ring is commutative of characteristic $2$; this makes the lattice laws routine.\nIn the other direction, the key point is that $+$ is associative, which reduces to the distributive laws and [[De Morgan's laws|0XYVDWGQX2]].\nFor the power set of a set $X$, the ring addition is the [[symmetric difference|0ME21QN125]] $A \\mathbin{\\triangle} B$ and the product is $A \\cap B$.\nIn categorical terms, the theorem gives an isomorphism between the category of Boolean algebras and the category of Boolean rings.\n\n## Explanations\n\n### L1 human\n\nBoolean algebras use \"and\", \"or\" and \"not\"; Boolean rings use \"exclusive or\" and \"and\" as plus and times.\nThe theorem says that these are two notations for the same thing.\nOne can always translate between them without losing anything.\n\n### L2 human\n\nThe ring view brings tools of commutative algebra to Boolean algebras.\nIdeals of the ring are exactly the ideals of the Boolean algebra, and prime ideals are maximal, with quotient the field $\\F_2$.\nSo ultrafilters, which are complements of maximal ideals, correspond to ring homomorphisms onto $\\F_2$, and the Stone space of the algebra is the prime spectrum of the ring.\nThis is the bridge used in one standard proof of [[Stone's representation theorem|S4QJ395PA5]].\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "about": [
      "0XADJKM699",
      "0RCV1ZFP8G"
    ],
    "kind": "statement",
    "msc": [
      "06E20",
      "06E05"
    ],
    "name": "Equivalence of Boolean rings and Boolean algebras",
    "role": "theorem",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Boolean_ring"
      }
    ],
    "tag": "PQEPV7F2RV",
    "uses": [
      "0XADJKM699",
      "0RCV1ZFP8G",
      "0ME21QN125"
    ]
  },
  "kind": "statement",
  "link": "https://tamnd.github.io/math2/node/PQEPV7F2RV/",
  "links": [
    {
      "field": "about",
      "target": "0XADJKM699"
    },
    {
      "field": "about",
      "target": "0RCV1ZFP8G"
    },
    {
      "field": "uses",
      "target": "0XADJKM699"
    },
    {
      "field": "uses",
      "target": "0RCV1ZFP8G"
    },
    {
      "field": "uses",
      "target": "0ME21QN125"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/PQEPV7F2RV.md",
  "msc": [
    "06E20",
    "06E05"
  ],
  "name": "Equivalence of Boolean rings and Boolean algebras",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/PQE/PQEPV7F2RV.md",
  "status": null,
  "tag": "PQEPV7F2RV",
  "uri": "m2://PQEPV7F2RV",
  "url": "https://tamnd.github.io/math2/wiki/equivalence-of-boolean-rings-and-boolean-algebras/",
  "writing": {
    "tag": "PQEPV7F2RV",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}