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

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- November 14th 2011, 01:35 PMThatPinkSockHerstein: Subrings and Ideals
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** - November 14th 2011, 03:34 PMTheEmptySetRe: Herstein: Subrings and Ideals
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. - November 14th 2011, 05:00 PMThatPinkSockRe: Herstein: Subrings and Ideals
(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.