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  "body": "\n## Statement\n\nLet $R$ be a [[ring|0894J980RK]] and let $I \\subseteq J$ be two-sided ideals of $R$.\nThen $J/I = \\{a + I : a \\in J\\}$ is a two-sided ideal of the [[quotient ring|0VFW4Y69BT]] $R/I$.\nThe map $(R/I)/(J/I) \\to R/J$, $(a + I) + J/I \\mapsto a + J$, is a [[ring isomorphism|0B6QC0C0G1]].\n\n## Remarks\n\nThe proof applies the [[first isomorphism theorem|0ZMYX7GMH0]] to the map $R/I \\to R/J$, $a + I \\mapsto a + J$. It is surjective, and its kernel is $J/I$.\nFor example, $6\\mathbb{Z} \\subseteq 2\\mathbb{Z}$, and $(\\mathbb{Z}/6\\mathbb{Z})/(2\\mathbb{Z}/6\\mathbb{Z}) \\cong \\mathbb{Z}/2\\mathbb{Z}$.\n\n## Explanations\n\n### L1 human\n\nYou can make a quotient in two steps or in one step.\nDividing first by $I$ and then by the image of $J$ gives the same ring as dividing directly by $J$.\n",
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  "name": "Third isomorphism theorem for rings",
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