Herstein's Abstract Algebra: Ch 4. Sect 3. # 5.
If I is an ideal of R and A is a subring of R, show that I ∩ A is an ideal of A
So where exactly are you stuck?
To show a set is an ideal you need to varify two things.
1. The set is a sub group under "+"
2. That the set "absorbs" elements under multipication.
For any element r in the ring and any elment x in the ideal that
xr is in the ideal and rx is in the ideal.
Try to show the above and post back if you get stuck.
(1) [ I ∩ A, +] is a group because [I, +] and [A, +] are subgroups of R and the intersection of two subgroups is a subgroup.
(2) Let x ∈ I ∩ A and let r ∈ A. Then x ∈ I and x ∈ A. Since I is an ideal of R, xr ∈ I and rx ∈ I. Since x ∈ A, xr ∈ A and rx ∈ A. So xr ∈ I ∩ A and rx ∈ I ∩ A.
From (1) and (2) I ∩ A is an ideal of A.