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  "body": "\n## Proof\n\n1. Let $f \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]], and write $K = \\ker f$ for its [[kernel|0AAHE7GEW6]].\n2. As for any additive map, $f(0) = 0$ and $f(a - b) = f(a) - f(b)$ for $a, b \\in R$. Indeed, $f(0) = f(0) + f(0)$ gives $f(0) = 0$, and then $f(a - b) + f(b) = f(a)$.\n3. The set $K$ is an additive subgroup of $R$. Indeed, $0 \\in K$, and if $a, b \\in K$, then $f(a - b) = 0 - 0 = 0$.\n4. For $r \\in R$ and $a \\in K$, $f(r a) = f(r) \\cdot 0 = 0$ and $f(a r) = 0 \\cdot f(r) = 0$. So $r a, a r \\in K$, and $K$ is a [[two-sided ideal|0DDA6CY7GJ]] of $R$.\n5. The image $f(R)$ is a [[subring|0859D58WPD]] of $S$. Indeed, $1 = f(1) \\in f(R)$, $f(a) - f(b) = f(a - b)$ and $f(a) f(b) = f(a b)$.\n6. Since $K \\subseteq \\ker f$, the [[universal property of the quotient ring|0R74MZYNRB]] gives a ring homomorphism $\\bar{f} \\colon R/K \\to S$ with $\\bar{f}(a + K) = f(a)$. Here $R/K$ is the [[quotient ring|0VFW4Y69BT]].\n7. The values of $\\bar{f}$ are the values $f(a)$, so $\\bar{f}$ takes values in $f(R)$. As a map $R/K \\to f(R)$, it is still a ring homomorphism, because $f(R)$ is a subring with the operations of $S$.\n8. This map is surjective onto $f(R)$, because each $f(a)$ equals $\\bar{f}(a + K)$.\n9. Its kernel is $\\set{0 + K}$. Indeed, $\\bar{f}(a + K) = 0$ means $f(a) = 0$, that is, $a \\in K$. Then $a - 0 \\in K$, so $a + K = 0 + K$.\n10. A ring homomorphism with zero kernel [[is injective|0MMCB8RXGA]]. So $\\bar{f} \\colon R/K \\to f(R)$ is bijective, and it is a [[ring isomorphism|0B6QC0C0G1]].\n11. If $f$ is surjective, then $f(R) = S$. So $\\bar{f}$ is an isomorphism from $R/\\ker f$ to $S$, and $R/\\ker f \\cong S$.\n\n## Idea\n\nThe kernel is an ideal, so one can form $R/\\ker f$, and $f$ passes to a map on cosets. This map hits the whole image. Its kernel is only the zero coset, because dividing by $\\ker f$ removed exactly what $f$ sends to zero.\n",
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