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X in ideal if x^n is in ideal?

  • Thread starter pivoxa15
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  • #1
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Homework Statement


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.


The Attempt at a Solution


I can't find a proof. Which is likely to mean the statement is false?
 

Answers and Replies

  • #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.
 
  • #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.
 
  • #4
Dick
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Eg. consider the ideal generated by 4 over the integers.
 
  • #5
radou
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[This is a sign I should stop learning from various lecture notes, and turn to books. :uhh:]
 
  • #6
matt grime
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No. It is a sign you should think about what you're reading.
 
  • #7
matt grime
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I can't find a proof. Which is likely to mean the statement is false?

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?
 
  • #8
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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.
 
  • #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.
 
  • #10
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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.
Good point. An example might be the ideal <2^2> in N which contains no element 2.
 

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