
050612, 08:54 PM #1
 Posts
 3,742
Ideals question
This is another past exam question.
Define .
For I,J,K ideals of R, where R is a commutative ring, prove that:
(a) (I:J) is an ideal of R
(b) I∩J is an ideal of R
(c) (I: (J+K)) = (I:J)∩(I:K)
Well, you have to figure out what the elements of I, J, and K look like; say r_{1} ∈ R > r_{1}j ∈ I, when j ∈ J. Then show that r_{1}j  r_{2}j, say, is in (I:J), and that r_{1}j r_{2}j is in (I:J). That's for the (a) question.
Have I got the idea here, or what?

050712, 02:41 AM #2
 Posts
 3,742
The definition says that for elements r_{1}, r_{2} ∈ R, j ∈ J, r_{1} and r_{2} are in (I:J) when r_{1}j, r_{2}j ∈ I, right?
So I need to show that r_{1}  r_{2} is also in (I:J) when (r_{1}  r_{2})j ∈ I, and that (r_{1}r_{2})j ∈ I, so r_{1}r_{2} is in (I:J)?

050712, 08:39 PM #3
 Posts
 3,742
If (I:J) is an ideal of R, R/A (where A = (I:J)) is the ring of cosets {r + A  r ∈ R} such that for s,t ∈ R, (s + A)(t + A) = st + A = 0 + A when s ∈ A or t ∈ A, i.e. when s = aj or t = aj for some a in R and j in J.
I still can't see how any of that helps to prove A = (I:J) is an ideal. Do I try to prove that R/A is a coset ring and so A is an ideal (if multiplication is 'well defined')? Just a hint might be all I need.

050912, 11:52 PM #4
 Posts
 3,742
Damn, I nearly got it.
The answer for part (a) is as follows:
Let a,b ∈ (I:J) such that aj, bj ∈ I, ∀j ∈ J.
So aj  bj = (ab)j ∈ I, and hence a  b ∈ (I:J) ........(1)
Let r ∈ R, such that (ra)j ∈ I, ∀j ∈ J.
So r(aj) ∈ I since aj ∈ I (and because I is an ideal of R).
So that ra ∈ (I:J) ...............................................(2)
Therefore, by (1) and (2), (I:J) is an ideal of R.
Similar Threads

By ScaryMonster in forum Free ThoughtsLast Post: 022712, 02:54 AMReplies: 6

By audric in forum General Science & TechnologyLast Post: 010412, 08:43 PMReplies: 1

By S.A.M. in forum PoliticsLast Post: 110911, 04:16 PMReplies: 103

By Michael in forum Religion ArchivesLast Post: 100608, 01:50 PMReplies: 44
Bookmarks