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Math Help - ring/ideal

  1. #1
    Senior Member sfspitfire23's Avatar
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    ring/ideal

    Any input?

    If A and B are ideals of R such that A\cap B =0 show that ab=0=ba.
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  2. #2
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    By definition of ideal if a\in A and b\in B then ab,ba\in A\cap B=0 (I'm assuming your ring commutative, but I believe if A,B are two sided ideals the proof still works)
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  3. #3
    Senior Member sfspitfire23's Avatar
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    How does my reasoning sound?

    By definition of ideal, a multiplied by any element of R will be in A and will also be in B. So, b multiplied by any element a in R will be \in B and same for A. Because the problem tells us that the union of A and B is 0, then because ab\in A,B, ab=0.


    Hows this sound?
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  4. #4
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    Quote Originally Posted by sfspitfire23 View Post
    How does my reasoning sound?

    By definition of ideal, a multiplied by any element of R will be in A and will also be in B. So, b multiplied by any element a in R will be \in B and same for A. Because the problem tells us that the union of A and B is 0, then because ab\in A,B, ab=0.


    Hows this sound?
    Not quite. By definition of ideal if a\in A and b\in R then ab,ba\in A, if in addition b \in B, since A\subset R we have ab,ba\in B so all this implies that if a\in A and b\in B then ab,ba \in A\cap B=0 (not the union!)
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  5. #5
    Senior Member sfspitfire23's Avatar
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    Hm. I guess I get it all but this part here.


    Quote Originally Posted by Jose27 View Post
    if in addition b \in B, since A\subset R we have ab,ba\in B
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  6. #6
    Member Focus's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Hm. I guess I get it all but this part here.
    It is the exact same reason why it is in A.
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