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  "body": "\n## Proof\n\n1. Let $R$ be a [[ring|0894J980RK]], let $I$ be a [[two-sided ideal|0DDA6CY7GJ]] of $R$, and let $\\pi \\colon R \\to R/I$, $\\pi(a) = a + I$, be the quotient map to the [[quotient ring|0VFW4Y69BT]]. Let $f \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]] with $I \\subseteq \\ker f$, where $\\ker f$ is the [[kernel|0AAHE7GEW6]].\n2. For $a, b \\in R$, $a + I = b + I$ if and only if $a - b \\in I$. Indeed, if $a + I = b + I$, then $a = a + 0 \\in b + I$, so $a = b + x$ with $x \\in I$. Conversely, if $a - b = x \\in I$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for $y \\in I$, so the two cosets are equal.\n3. The map $f$ satisfies $f(0) = 0$, since $f(0) = f(0 + 0) = f(0) + f(0)$. It also satisfies $f(-a) = -f(a)$, since $f(a) + f(-a) = f(0) = 0$. So $f(a - b) = f(a) - f(b)$.\n4. Define $\\bar{f}(a + I) = f(a)$. This does not depend on the choice of $a$. Indeed, if $a + I = b + I$, then $a - b \\in I \\subseteq \\ker f$ by step 2. So $f(a) - f(b) = f(a - b) = 0$ by step 3.\n5. The map $\\bar{f}$ preserves addition: $\\bar{f}((a + I) + (b + I)) = \\bar{f}((a + b) + I) = f(a + b) = f(a) + f(b)$.\n6. The map $\\bar{f}$ preserves multiplication: $\\bar{f}((a + I)(b + I)) = \\bar{f}(a b + I) = f(a b) = f(a) f(b)$.\n7. The identity of $R/I$ is $1 + I$, and $\\bar{f}(1 + I) = f(1) = 1$. So $\\bar{f}$ is a ring homomorphism.\n8. For $a \\in R$, $\\bar{f}(\\pi(a)) = \\bar{f}(a + I) = f(a)$. So $\\bar{f} \\circ \\pi = f$. This proves existence.\n9. For uniqueness, let $g \\colon R/I \\to S$ be a ring homomorphism with $g \\circ \\pi = f$. Every element of $R/I$ is $a + I = \\pi(a)$ for some $a \\in R$.\n10. Then $g(a + I) = g(\\pi(a)) = f(a) = \\bar{f}(a + I)$. So $g = \\bar{f}$.\n11. Hence there is exactly one ring homomorphism $\\bar{f} \\colon R/I \\to S$ with $\\bar{f} \\circ \\pi = f$, and it is given by $\\bar{f}(a + I) = f(a)$.\n\n## Idea\n\nA homomorphism that kills $I$ takes the same value at all elements of a coset $a + I$. So it gives a well defined map on cosets. That map inherits the homomorphism rules from $f$, and it is forced on every coset because $\\pi$ is surjective.\n",
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