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Thread: 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 $\displaystyle A\cap B =0$ show that $\displaystyle ab=0=ba$.
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  2. #2
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    By definition of ideal if $\displaystyle a\in A$ and $\displaystyle b\in B$ then $\displaystyle ab,ba\in A\cap B=0$ (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, $\displaystyle a$ multiplied by any element of $\displaystyle R$ will be in A and will also be in B. So, $\displaystyle b$ multiplied by any element $\displaystyle a$ in $\displaystyle R$ will be $\displaystyle \in B$ and same for $\displaystyle A$. Because the problem tells us that the union of A and B is 0, then because $\displaystyle ab\in A,B$, $\displaystyle ab=0$.


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


    Hows this sound?
    Not quite. By definition of ideal if $\displaystyle a\in A$ and $\displaystyle b\in R$ then $\displaystyle ab,ba\in A$, if in addition $\displaystyle b \in B$, since $\displaystyle A\subset R$ we have $\displaystyle ab,ba\in B$ so all this implies that if $\displaystyle a\in A$ and $\displaystyle b\in B$ then $\displaystyle ab,ba \in A\cap B=0$ (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 $\displaystyle b \in B$, since $\displaystyle A\subset R$ we have $\displaystyle ab,ba\in B$
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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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