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Math Help - Proving (I:a) is an ideal of a ring

  1. #1
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    Proving (I:a) is an ideal of a ring

    Suppose R is a commutative ring and I is an ideal of R and a∈R.

    Define (I:a) = {r∈R : ar∈ I}


    I want to prove that (I:a) is also an ideal of R.

    I know i need to prove:
    1) 0∈(I:a)

    2) If b,c∈ (I:a) , then b + c ∈ (I:a)

    3) If b∈ (I:a) and s∈ R, then bs∈(I:a) and sb∈(I:a)

    I started by trying to prove 1):

    I know 0∈I (by definition of an ideal) so ar=0 for some a, some r.
    Hence, 0∈(I:a)

    I am not sure if this is correct, or how to proceed with the rest of the proof.

    Any help is appreciated.
    Thanks
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  2. #2
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    Quote Originally Posted by Siknature View Post
    Suppose R is a commutative ring and I is an ideal of R and a∈R.

    Define (I:a) = {r∈R : ar∈ I}


    I want to prove that (I:a) is also an ideal of R.

    I know i need to prove:
    1) 0∈(I:a)

    2) If b,c∈ (I:a) , then b + c ∈ (I:a)

    3) If b∈ (I:a) and s∈ R, then bs∈(I:a) and sb∈(I:a)

    I started by trying to prove 1):

    I know 0∈I (by definition of an ideal) so ar=0 for some a, some r.
    Hence, 0∈(I:a)

    I am not sure if this is correct, or how to proceed with the rest of the proof.

    Any help is appreciated.
    Thanks
    1) Is it true that 0\in (I:a) , i.e. is it true that  a\cdot 0=0\in I ?

    2) is it true that ab\in I\,,\,\,ac\in I\Longrightarrow a(b+c)=ab+ac\in I ?

    3) Is it true that b\in (I:a)\,,\,\,r\in R\Longrightarrow br\in (I:a) , i.e.  a(br)=abr\in I ?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    1) Is it true that 0\in (I:a) , i.e. is it true that  a\cdot 0=0\in I ?

    2) is it true that ab\in I\,,\,\,ac\in I\Longrightarrow a(b+c)=ab+ac\in I ?

    3) Is it true that b\in (I:a)\,,\,\,r\in R\Longrightarrow br\in (I:a) , i.e.  a(br)=abr\in I ?

    Tonio
    Yes i think it is true in all 3 cases because we know I is an ideal and by the definition of ideal i think that all these cases must hold.

    Thanks for your help in showing me this.

    A further question i would appreciate your help with is this:

    suppose bR is an associate of a.

    How do i prove that (I:b) = (I:a) ?

    I started off by saying that a,bR are associates if b=ua for some unit uR.

    I tried substituting this in to (I:b)

    So (I:b) = {rR : br R}
    = {rR : uar R} for some unit uR

    but i cant see how to proceed for here to try to make it equal (I:a).

    Do you have any suggestions?
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  4. #4
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    Quote Originally Posted by Siknature View Post
    Yes i think it is true in all 3 cases because we know I is an ideal and by the definition of ideal i think that all these cases must hold.

    Thanks for your help in showing me this.

    A further question i would appreciate your help with is this:

    suppose bR is an associate of a.

    How do i prove that (I:b) = (I:a) ?

    I started off by saying that a,bR are associates if b=ua for some unit uR.

    I tried substituting this in to (I:b)

    So (I:b) = {rR : br R}
    = {rR : uar R} for some unit uR

    but i cant see how to proceed for here to try to make it equal (I:a).

    Do you have any suggestions?

    Let x\in(I:b)\Longrightarrow bx=i\in I , but  i=bx=uax\Longrightarrow if uv=1 , then  vi=vuax=ax\in I\Longrightarrow x\in (I:a)\Longrightarrow (I:b)\subset (I:a).

    In a similar way can the other direction be proved.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Let x\in(I:b)\Longrightarrow bx=i\in I , but  i=bx=uax\Longrightarrow if uv=1 , then  vi=vuax=ax\in I\Longrightarrow x\in (I:a)\Longrightarrow (I:b)\subset (I:a).

    In a similar way can the other direction be proved.

    Tonio
    Is I⊆(I:a) ?

    Is a∈(I:a) ?
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  6. #6
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    Quote Originally Posted by Siknature View Post
    Is I⊆(I:a) ?


    Obviously (why?) yes...!

    Is a∈(I:a) ?

    Not necessarily. For example, 3\notin (4\mathbb{Z}:3)\,,\,\,R=\mathbb{Z}

    Tonio
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  7. #7
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    Quote Originally Posted by tonio View Post
    Is I⊆(I:a) ?


    Obviously (why?) yes...!

    Is a∈(I:a) ?
    I am not sure why this is obvious.

    Is it because every element of (I:a) must belong to I by definition of (I:a) ?
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  8. #8
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    Quote Originally Posted by Siknature View Post
    I am not sure why this is obvious.

    Is it because every element of (I:a) must belong to I by definition of (I:a) ?


    What happens if you take any element of I and multiply it by ANY element of R? So what happens if you multiply any element of I by a?!

    Tonio
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  9. #9
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    Quote Originally Posted by tonio View Post
    What happens if you take any element of I and multiply it by ANY element of R? So what happens if you multiply any element of I by a?!

    Tonio
    Oh, I see....

    If you take any element of I and multiply it by any element of R,the result is in I (by definition of an ideal).


    So if you multiply any element by aR, the result will still be in I.

    Hence, I⊆(I:a)
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  10. #10
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    Quote Originally Posted by Siknature View Post
    Oh, I see....

    If you take any element of I and multiply it by any element of R,the result is in I (by definition of an ideal).


    So if you multiply any element of I by aR, the result will still be in I.

    Hence, I⊆(I:a)

    Check the red addition to your post above!

    Tonio
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  11. #11
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    Quote Originally Posted by tonio View Post
    Check the red addition to your post above!

    Tonio
    Oh yes!

    Thankyou, I very much appreciate your help in this series of questions.
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