---
tag: D97V0DPD37
kind: proof
name: Proof of the correspondence theorem for rings
msc: [16D25]
proves: 0D1R4SVMBQ
uses: [0894J980RK, 0DDA6CY7GJ, 0VFW4Y69BT, 0MYSA6MYNA, 0FGCFY1AZX, 0Q3F20XB0A, 0SNQEF7Z53]
sources:
  - url: https://en.wikipedia.org/wiki/Ideal_(ring_theory)
---

## Proof

1. 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.
2. 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 $a \in R$.
3. 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 $y \in I$.
4. 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 $R/I$.
5. 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 $\mathcal{B}$.
6. 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 $I \subseteq \pi^{-1}(K)$.
7. 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, $\pi^{-1}(K) \in \mathcal{A}$.
8. 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 $b \in J$.
9. By step 3, $a - b \in I \subseteq J$. So $a = b + (a - b) \in J$. Thus $\pi^{-1}(\pi(J)) = J$.
10. 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 $\pi(\pi^{-1}(K)) = K$.
11. 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 $K \mapsto \pi^{-1}(K)$.
12. 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.

## Idea

Push 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.