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X in ideal if x^n is in ideal?

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data
    In a ring with multiplicative identity, If x^n is in an ideal then is x also in the ideal? with n a natural number.


    3. The attempt at a solution
    I can't find a proof. Which is likely to mean the statement is false?
     
  2. jcsd
  3. Mar 29, 2007 #2

    radou

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    Let x^n be in the ideal I of a ring R. It can be represented as x^n = x^(n-1) x. Now, either both x^(n-1) and x are in the ideal (in that case the statement if proved), or one of them is in the ideal. If x is in the ideal, the proof is finished. If x^(n-1) is in the ideal, then you can apply the same argument again.

    I hope this works, since I'm a bit new to rings.
     
  4. Mar 29, 2007 #3

    Dick

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    The statement that a*b in an ideal implies a and/or b in the ideal is only true for prime ideals. Not any ideal. The statement is false.
     
  5. Mar 29, 2007 #4

    Dick

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    Eg. consider the ideal generated by 4 over the integers.
     
  6. Mar 29, 2007 #5

    radou

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    [This is a sign I should stop learning from various lecture notes, and turn to books. :uhh:]
     
  7. Mar 29, 2007 #6

    matt grime

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    No. It is a sign you should think about what you're reading.
     
  8. Mar 29, 2007 #7

    matt grime

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    Just think about an example for a second. Like the simplest ring there is, the integers. Why didn't you do some examples to see if it was true?
     
  9. Mar 29, 2007 #8
    Right, I should have done that. So the ideal generated by <x^2> in a ring certainly wouldn't have x in it. So the statement is false.
     
  10. Mar 29, 2007 #9

    matt grime

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    It is perfectly possible for the ideal generated by x^2 to have x in it. It is just not *certain* to have x in it.
     
  11. Mar 29, 2007 #10
    Good point. An example might be the ideal <2^2> in N which contains no element 2.
     
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