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Thread: Ideals question

  1. #1

    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 r1 ∈ R -> r1j ∈ I, when j ∈ J. Then show that r1j - r2j, say, is in (I:J), and that r1j r2j is in (I:J). That's for the (a) question.

    Have I got the idea here, or what?

  2. #2
    The definition says that for elements r1, r2 ∈ R, j ∈ J, r1 and r2 are in (I:J) when r1j, r2j ∈ I, right?

    So I need to show that r1 - r2 is also in (I:J) when (r1 - r2)j ∈ I, and that (r1r2)j ∈ I, so r1r2 is in (I:J)?

  3. #3
    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.

  4. #4
    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 = (a-b)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.

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