Second isomorphism theorem (rings): Let RR be a ring, let A⊆RA\subseteq R be a , and let I◃RI\triangleleft R be a two-sided . Then A∩I◃AA\cap I\triangleleft A, A+I:={a+i:a∈A, i∈I}A+I:=\{a+i:a\in A,\ i\in I\} is a subring of RR, and there is a natural isomorphism

A/(A∩I) ≅ (A+I)/IA/(A\cap I)\ \cong\ (A+I)/I

of .

Remarks

This is most efficiently proved by restricting the quotient map R→R/IR\to R/I to AA and applying the .