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Algebra problem

  1. Sep 19, 2005 #1

    Galileo

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    I want to show that if I and J are coprime ideals of a ring R, so I+J=R, then for any positive numbers m and n we also have [itex]I^n+I^m=R[/itex].

    I thought the easiest way to do it was to show that [itex]1 \in I^n+J^m[/itex] given that there exist [itex]i\in I[/itex] and [itex]j\in J[/itex] such that [itex]i+j=1[/itex]. But I haven't had much luck yet. Any hint would be appreciated.
     
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  3. Sep 19, 2005 #2

    matt grime

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    I'm going to hope that R is commutative, for then consider (i+j) raised to some power to do with m and n but larger than both.
     
  4. Sep 19, 2005 #3

    Galileo

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    Yes, R is commutative. I forgot to mention that.
    I already had tried expanding (i+j) to some power. For example m+n or mn:
    [tex]1=(i+j)^{m+n}=(i+j)^m(i+j)^n=\sum_{k=0}^m {m \choose k}i^{m-k}j^k\sum_{k=0}^{n}{n \choose k}i^{n-k}j^k=\sum_{k=0}^{m+n}i^{m+n-k}j^k[/tex]

    [tex]1=(i+j)^{mn}=\sum_{k=0}^{mn}{mn \choose k}i^{mn-k}j^k=\left(\sum_{k=0}^m {m \choose k}i^{m-k}j^k\right)^n[/tex]
    I can't see where that leads me. I understand that if I can write 1 as [itex]i^n x+j^m y[/itex] whatever x and y are, then I`m done.
     
  5. Sep 19, 2005 #4

    Galileo

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    I think I've got it.

    [tex]1=(i+j)^{m+n}=\sum_{k=0}^{m+n}{m+n \choose k}i^{m+n-k}j^k[/tex]
    [tex]\sum_{k=0}^{m}{m+n \choose k}i^{m+n-k}j^k +\sum_{k=m+1}^{m+n}{m+n \choose k}i^{m+n-k}j^k=[/tex]
    [tex]i^n\left(\sum_{k=0}^{m}{m+n \choose k}i^{m-k}j^k\right) +j^m\left(\sum_{k=0}^{n-1}{m+n \choose m+1+k}i^{n-1-k}j^{1+k}\right)[/tex]
     
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