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A Rings and Ideals Problem

  1. Jul 2, 2008 #1
    1. The problem statement, all variables and given/known data
    If A and B are ideals of a commutative ring R with unity and A + B = R, show that [itex]A \cap B[/itex] = AB.

    3. The attempt at a solution
    Showing [itex]AB \subseteq A \cap B[/itex] is easy. I'm having trouble with containment in the other direction:

    Let [itex]x \in A \cap B[/itex]. Then x is in A and x is in B. To show that x belongs to AB, it suffices to show that 1 belongs to either A or B and so 1x or x1 belongs to AB. It seems to me that 1 isn't necessarily in A or B so this approach is unfruitful. Is there another decomposition of x into ab where a belongs to A and B belongs to B?
     
  2. jcsd
  3. Jul 2, 2008 #2

    matt grime

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    If you show 1 is in A, then that means A=R. Clearly you don't want to do that. What information haven't you used? That A+B=R, ie there exist elements x in A and y in B such that x+y=1.
     
  4. Jul 2, 2008 #3
    I have thought about that but it lead nowhere: Write 1 as a + b. Then x1 = xa + xb. Hmm...I see it know. Since R is commutative, xa = ax and so ax is in AB since x is in B. Similarly, xb is in AB. If AB is closed under addition, then surely xa + xb is in AB and so x is in AB. All that remains is to show that AB is closed under addition. Right?
     
  5. Jul 2, 2008 #4

    matt grime

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    That is part of the *definition* of AB.
     
  6. Jul 2, 2008 #5
    AB = {ab : a in A and b in B}. How does that definition demonstrate that AB is closed under addition?
     
  7. Jul 2, 2008 #6

    matt grime

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    I think you should go and check your definition of the *ideal* AB.
     
  8. Jul 2, 2008 #7
    Aha! There is an exercise where AB is defined. I had overlooked it. Thanks a lot.
     
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