{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Let $R$ be a [[ring|0894J980RK]] and let $I \\subseteq J$ be [[two-sided ideals|0DDA6CY7GJ]] of $R$. Put $J/I = \\set{a + I : a \\in J}$, a subset of the [[quotient ring|0VFW4Y69BT]] $R/I$.\n2. For an ideal $L$ and $a, b \\in R$, $a + L = b + L$ if and only if $a - b \\in L$. Indeed, if the cosets are equal, then $a \\in b + L$. Conversely, if $a - b = x \\in L$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for each $y \\in L$.\n3. The set $J/I$ contains $0 + I$, since $0 \\in J$. For $a, b \\in J$, $(a + I) - (b + I) = (a - b) + I$ lies in $J/I$, since $a - b \\in J$. So $J/I$ is an additive subgroup of $R/I$.\n4. For $r \\in R$ and $a \\in J$, $(r + I)(a + I) = r a + I$ and $(a + I)(r + I) = a r + I$. Both lie in $J/I$, since $r a, a r \\in J$. So $J/I$ is a two-sided ideal of $R/I$.\n5. Let $\\rho \\colon R \\to R/J$, $\\rho(a) = a + J$. By the definition of the operations of $R/J$, $\\rho$ is a [[ring homomorphism|0GGE0Z0JQ1]]. By step 2, its [[kernel|0AAHE7GEW6]] is $J$.\n6. Since $I \\subseteq J = \\ker \\rho$, the [[universal property of the quotient ring|0R74MZYNRB]] gives a ring homomorphism $\\psi \\colon R/I \\to R/J$ with $\\psi(a + I) = a + J$.\n7. The map $\\psi$ is surjective, because each $a + J$ equals $\\psi(a + I)$.\n8. The kernel of $\\psi$ is $J/I$. First, if $a \\in J$, then $\\psi(a + I) = a + J = 0 + J$ by step 2.\n9. Conversely, let $\\psi(a + I) = 0 + J$. Then $a \\in J$ by step 2, so $a + I \\in J/I$.\n10. By the [[first isomorphism theorem|0ZMYX7GMH0]] applied to $\\psi$, with steps 7 to 9, the map $(R/I)/(J/I) \\to R/J$, $(a + I) + J/I \\mapsto \\psi(a + I) = a + J$, is a [[ring isomorphism|0B6QC0C0G1]].\n\n## Idea\n\nReducing modulo $J$ kills everything in $I$, so it factors through $R/I$. The resulting map $R/I \\to R/J$ is onto, and it kills exactly the cosets of elements of $J$. The first isomorphism theorem turns this into the isomorphism.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "16D25"
    ],
    "name": "Proof of the third isomorphism theorem for rings",
    "proves": "073C80EC63",
    "sources": [
      {
        "url": "https://proofwiki.org/wiki/Third_Isomorphism_Theorem_for_Rings"
      }
    ],
    "tag": "NHSEJ8KGTA",
    "uses": [
      "0894J980RK",
      "0DDA6CY7GJ",
      "0VFW4Y69BT",
      "0GGE0Z0JQ1",
      "0AAHE7GEW6",
      "0R74MZYNRB",
      "0ZMYX7GMH0",
      "0B6QC0C0G1"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/NHSEJ8KGTA/",
  "links": [
    {
      "field": "proves",
      "target": "073C80EC63"
    },
    {
      "field": "uses",
      "target": "0894J980RK"
    },
    {
      "field": "uses",
      "target": "0DDA6CY7GJ"
    },
    {
      "field": "uses",
      "target": "0VFW4Y69BT"
    },
    {
      "field": "uses",
      "target": "0GGE0Z0JQ1"
    },
    {
      "field": "uses",
      "target": "0AAHE7GEW6"
    },
    {
      "field": "uses",
      "target": "0R74MZYNRB"
    },
    {
      "field": "uses",
      "target": "0ZMYX7GMH0"
    },
    {
      "field": "uses",
      "target": "0B6QC0C0G1"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/NHSEJ8KGTA.md",
  "msc": [
    "16D25"
  ],
  "name": "Proof of the third isomorphism theorem for rings",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/NHS/NHSEJ8KGTA.md",
  "status": null,
  "tag": "NHSEJ8KGTA",
  "uri": "m2://NHSEJ8KGTA",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-third-isomorphism-theorem-for-rings/",
  "writing": {
    "tag": "NHSEJ8KGTA",
    "kind": "proof",
    "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
      }
    ]
  }
}