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  "body": "\n## Proof\n\n1. Let $R$ be a [[ring|0894J980RK]], let $S$ be a [[subring|0859D58WPD]] of $R$, and let $I$ be a [[two-sided ideal|0DDA6CY7GJ]] of $R$. Put $T = S + I = \\set{s + x : s \\in S, x \\in I}$.\n2. The set $T$ contains $1 = 1 + 0$, since $1 \\in S$ and $0 \\in I$.\n3. Let $s + x$ and $s' + x'$ be in $T$, with $s, s' \\in S$ and $x, x' \\in I$. Then $(s + x) - (s' + x') = (s - s') + (x - x')$, with $s - s' \\in S$ and $x - x' \\in I$. So the difference lies in $T$.\n4. The product is $(s + x)(s' + x') = s s' + (s x' + x s' + x x')$. Here $s s' \\in S$. Since $I$ is a two-sided ideal of $R$, the three terms $s x'$, $x s'$, $x x'$ lie in $I$. So the product lies in $T$.\n5. By steps 2 to 4, $T$ is a subring of $R$.\n6. Each $x \\in I$ equals $0 + x$ with $0 = 1 - 1 \\in S$. So $I \\subseteq T$. Also $I$ is an additive subgroup, and for $t \\in T$ and $x \\in I$, $t x \\in I$ and $x t \\in I$, because $I$ is a two-sided ideal of $R$. So $I$ is a two-sided ideal of $T$.\n7. So the [[quotient ring|0VFW4Y69BT]] $T/I$ exists. Define $\\varphi \\colon S \\to T/I$ by $\\varphi(s) = s + I$.\n8. The map $\\varphi$ is a [[ring homomorphism|0GGE0Z0JQ1]]. By the definition of the operations of $T/I$, $\\varphi(s + s') = \\varphi(s) + \\varphi(s')$ and $\\varphi(s s') = \\varphi(s) \\varphi(s')$. Also $\\varphi(1) = 1 + I$, the identity of $T/I$.\n9. For $a, b \\in T$, $a + I = b + I$ if and only if $a - b \\in I$. Indeed, if the cosets are equal, then $a \\in b + I$. Conversely, if $a - b = x \\in I$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for each $y \\in I$.\n10. The map $\\varphi$ is surjective. Indeed, an element of $T/I$ is $(s + x) + I$ with $s \\in S$, $x \\in I$. Since $(s + x) - s = x \\in I$, step 9 gives $(s + x) + I = s + I = \\varphi(s)$.\n11. The [[kernel|0AAHE7GEW6]] of $\\varphi$ is $S \\cap I$. Indeed, by step 9, $\\varphi(s) = 0 + I$ if and only if $s \\in I$.\n12. By the [[first isomorphism theorem|0ZMYX7GMH0]] applied to $\\varphi$, the kernel $S \\cap I$ is a two-sided ideal of $S$.\n13. The same theorem, with step 10, shows that $S/(S \\cap I) \\to T/I$, $s + (S \\cap I) \\mapsto \\varphi(s) = s + I$, is a [[ring isomorphism|0B6QC0C0G1]].\n\n## Idea\n\nSend each $s \\in S$ to its coset modulo $I$ in the bigger ring $S + I$. Every coset of $S + I$ already has a representative in $S$, so this map is onto. It kills exactly the elements of $S$ that lie in $I$, and the first isomorphism theorem does the rest.\n",
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