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  "body": "\n## Statement\n\nLet $f \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]].\nThen the [[kernel|0AAHE7GEW6]] $\\ker f$ is a two-sided ideal of $R$, and the image $f(R)$ is a subring of $S$.\nThe map $R/\\ker f \\to f(R)$, $a + \\ker f \\mapsto f(a)$, is a [[ring isomorphism|0B6QC0C0G1]].\nIn particular, if $f$ is surjective, then $R/\\ker f \\cong S$.\n\n## Remarks\n\nThe reduction map $\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ has kernel $n\\mathbb{Z}$.\nThe evaluation map $\\mathbb{R}[x] \\to \\mathbb{C}$, $p \\mapsto p(i)$, is surjective with kernel $(x^2 + 1)$. So $\\mathbb{R}[x]/(x^2 + 1) \\cong \\mathbb{C}$.\nThe theorem has the same form for groups and for modules.\n\n## Explanations\n\n### L1 human\n\nA homomorphism forgets exactly the information in its kernel.\nIf you divide out the kernel, the homomorphism becomes a bijection onto its image.\nSo every image of a ring is a quotient of that ring.\n",
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