Let be a ring homomorphism.
- Prove that if is an ideal, then is an ideal. Apply this to the special case when is a subring of and is the inclusion map to deduce that if is an ideal and a subring, then is an ideal.
- Prove that if is surjective and an ideal then is an ideal. Give an example where this fails if is not surjective.
- Let . Now and , so that . By hypothesis, , so that is closed under subtraction. Now let . We have since is an ideal of ; thus . Similarly, . Thus is an ideal of .
The final deduction is immediate since .
- Note that , so that . Now let . Now and for some . Since , is closed under subtraction. Now let . Since is surjective, there exists with . Now since is an ideal; thus . Similarly, . Thus is an ideal of .
For the counterexample, consider the inclusion . is an ideal of itself but not of since is not an integer.