---
tag: 1NM7SH936M
kind: proof
name: Proof of the second isomorphism theorem for rings
msc: [16D25]
proves: 0HJ4DTW6FN
uses: [0894J980RK, 0859D58WPD, 0DDA6CY7GJ, 0VFW4Y69BT, 0GGE0Z0JQ1, 0AAHE7GEW6, 0ZMYX7GMH0, 0B6QC0C0G1]
sources:
  - url: https://proofwiki.org/wiki/Second_Isomorphism_Theorem_for_Rings
---

## Proof

1. 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 $T = S + I = \set{s + x : s \in S, x \in I}$.
2. The set $T$ contains $1 = 1 + 0$, since $1 \in S$ and $0 \in I$.
3. 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 $T$.
4. 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 $T$.
5. By steps 2 to 4, $T$ is a subring of $R$.
6. 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 $T$.
7. So the [[quotient ring|0VFW4Y69BT]] $T/I$ exists. Define $\varphi \colon S \to T/I$ by $\varphi(s) = s + I$.
8. 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 $T/I$.
9. 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 $y \in I$.
10. 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 $(s + x) + I = s + I = \varphi(s)$.
11. The [[kernel|0AAHE7GEW6]] of $\varphi$ is $S \cap I$. Indeed, by step 9, $\varphi(s) = 0 + I$ if and only if $s \in I$.
12. By the [[first isomorphism theorem|0ZMYX7GMH0]] applied to $\varphi$, the kernel $S \cap I$ is a two-sided ideal of $S$.
13. 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]].

## Idea

Send 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.