Second isomorphism theorem for rings - Imported Knowlpedia Content

Second isomorphism theorem (rings): Let $R$ be a ring, let $A\subseteq R$ be a subring, and let $I\triangleleft R$ be an ideal. Then $A\cap I\triangleleft A$ , $A+I:=\{a+i:a\in A,\ i\in I\}$ is a subring of $R$ , and there is a natural isomorphism

A/(A\cap I)\ \cong\ (A+I)/I

of quotient rings.

Remarks

This is most efficiently proved by restricting the quotient map $R\to R/I$ to $A$ and applying the first isomorphism theorem.