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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 $\\mathcal{A}$ and $\\mathcal{B}$ be as in the statement.\n2. By the definition of the operations of $R/I$, $\\pi(a + b) = \\pi(a) + \\pi(b)$, $\\pi(a - b) = \\pi(a) - \\pi(b)$ and $\\pi(r a) = \\pi(r) \\pi(a)$. Every element of $R/I$ is $\\pi(a)$ for some $a \\in R$.\n3. For $a, b \\in R$, $\\pi(a) = \\pi(b)$ if and only if $a - b \\in I$. Indeed, if $a + I = b + I$, 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 $y \\in I$.\n4. Let $J \\in \\mathcal{A}$. The [[image|0Q3F20XB0A]] $\\pi(J)$ contains $\\pi(0) = 0 + I$. For $a, b \\in J$, $\\pi(a) - \\pi(b) = \\pi(a - b) \\in \\pi(J)$. So $\\pi(J)$ is an additive subgroup of $R/I$.\n5. For $r \\in R$ and $a \\in J$, $\\pi(r) \\pi(a) = \\pi(r a)$, and $r a \\in J$ because $J$ is a left [[ideal|0MYSA6MYNA]]. With step 2, $\\pi(J)$ is a left ideal of $R/I$. So $J \\mapsto \\pi(J)$ maps $\\mathcal{A}$ to $\\mathcal{B}$.\n6. Let $K \\in \\mathcal{B}$, and let $\\pi^{-1}(K) = \\set{a \\in R : \\pi(a) \\in K}$ be its [[preimage|0FGCFY1AZX]]. For $x \\in I$, $\\pi(x) = 0 + I \\in K$ by step 3. So $I \\subseteq \\pi^{-1}(K)$.\n7. For $a, b \\in \\pi^{-1}(K)$, $\\pi(a - b) = \\pi(a) - \\pi(b) \\in K$. For $r \\in R$, $\\pi(r a) = \\pi(r) \\pi(a) \\in K$, since $K$ is a left ideal. So $\\pi^{-1}(K)$ is a left ideal of $R$ that contains $I$, that is, $\\pi^{-1}(K) \\in \\mathcal{A}$.\n8. Let $J \\in \\mathcal{A}$. Each $a \\in J$ has $\\pi(a) \\in \\pi(J)$, so $J \\subseteq \\pi^{-1}(\\pi(J))$. Conversely, let $a \\in \\pi^{-1}(\\pi(J))$. Then $\\pi(a) = \\pi(b)$ for some $b \\in J$.\n9. By step 3, $a - b \\in I \\subseteq J$. So $a = b + (a - b) \\in J$. Thus $\\pi^{-1}(\\pi(J)) = J$.\n10. Let $K \\in \\mathcal{B}$. Every element of $\\pi(\\pi^{-1}(K))$ is $\\pi(a)$ with $\\pi(a) \\in K$. Conversely, each $k \\in K$ is $\\pi(a)$ for some $a \\in R$ by step 2, and then $a \\in \\pi^{-1}(K)$. So $\\pi(\\pi^{-1}(K)) = K$.\n11. By steps 9 and 10, the maps $J \\mapsto \\pi(J)$ and $K \\mapsto \\pi^{-1}(K)$ are inverse to each other. So $J \\mapsto \\pi(J) = J/I$ is a [[bijection|0SNQEF7Z53]] from $\\mathcal{A}$ to $\\mathcal{B}$, with inverse $K \\mapsto \\pi^{-1}(K)$.\n12. If $J_1 \\subseteq J_2$, then $\\pi(J_1) \\subseteq \\pi(J_2)$. If $K_1 \\subseteq K_2$, then $\\pi^{-1}(K_1) \\subseteq \\pi^{-1}(K_2)$. So the bijection and its inverse preserve inclusion.\n\n## Idea\n\nPush an ideal forward with $\\pi$, or pull an ideal back with $\\pi^{-1}$. Pulling back after pushing forward adds all of $I$, which changes nothing when the ideal already contains $I$. Pushing forward after pulling back changes nothing because $\\pi$ is onto.\n",
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