Third isomorphism theorem for rings

Statement

Let $R$ be a ring and let $I \subseteq J$ be two-sided ideals of $R$. Then $J/I = \{a + I : a \in J\}$ is a two-sided ideal of the quotient ring $R/I$. The map $(R/I)/(J/I) \to R/J$, $(a + I) + J/I \mapsto a + J$, is a ring isomorphism.

Remarks

The proof applies the first isomorphism theorem to the map $R/I \to R/J$, $a + I \mapsto a + J$. It is surjective, and its kernel is $J/I$. For example, $6\mathbb{Z} \subseteq 2\mathbb{Z}$, and $(\mathbb{Z}/6\mathbb{Z})/(2\mathbb{Z}/6\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.

Explanations

L1 human

You can make a quotient in two steps or in one step. Dividing first by $I$ and then by the image of $J$ gives the same ring as dividing directly by $J$.