Proof of the third isomorphism theorem for rings

Proof

1. Let $R$ be a ring and let $I \subseteq J$ be two-sided ideals of $R$. Put $J/I = \set{a + I : a \in J}$, a subset of the quotient ring $R/I$.
2. For an ideal $L$ and $a, b \in R$, $a + L = b + L$ if and only if $a - b \in L$. Indeed, if the cosets are equal, then $a \in b + L$. Conversely, if $a - b = x \in L$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for each $y \in L$.
3. The set $J/I$ contains $0 + I$, since $0 \in J$. For $a, b \in J$, $(a + I) - (b + I) = (a - b) + I$ lies in $J/I$, since $a - b \in J$. So $J/I$ is an additive subgroup of $R/I$.
4. For $r \in R$ and $a \in J$, $(r + I)(a + I) = r a + I$ and $(a + I)(r + I) = a r + I$. Both lie in $J/I$, since $r a, a r \in J$. So $J/I$ is a two-sided ideal of $R/I$.
5. Let $\rho \colon R \to R/J$, $\rho(a) = a + J$. By the definition of the operations of $R/J$, $\rho$ is a ring homomorphism. By step 2, its kernel is $J$.
6. Since $I \subseteq J = \ker \rho$, the universal property of the quotient ring gives a ring homomorphism $\psi \colon R/I \to R/J$ with $\psi(a + I) = a + J$.
7. The map $\psi$ is surjective, because each $a + J$ equals $\psi(a + I)$.
8. The kernel of $\psi$ is $J/I$. First, if $a \in J$, then $\psi(a + I) = a + J = 0 + J$ by step 2.
9. Conversely, let $\psi(a + I) = 0 + J$. Then $a \in J$ by step 2, so $a + I \in J/I$.
10. By the first isomorphism theorem applied to $\psi$, with steps 7 to 9, the map $(R/I)/(J/I) \to R/J$, $(a + I) + J/I \mapsto \psi(a + I) = a + J$, is a ring isomorphism.