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Math Help - Not Principal Ideal

  1. #1
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    Not Principal Ideal

    How would you prove it's not a P.I.D. by using I?

    Let be an ideal of . Prove that I is not a principal ideal of .

    I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by gummy_ratz View Post
    How would you prove it's not a P.I.D. by using I?

    Let be an ideal of . Prove that I is not a principal ideal of .

    I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!
    Think about norms. N(2)=4, N(1+\sqrt{-5})=1+25 = 26 (in general, N(a+b\sqrt{-5})=a^2+25b^2).

    You should therefore notice that if I is a principle ideal, the principle element (or what ever it is called - lets denote it by x) has norm 2 (why?). You should also notice that there exists some a and b such that x(a+b\sqrt{-5}) = 1+\sqrt{-5} and N(a+b\sqrt{-5})=13. So...can you find a contradiction in what I have just written?
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    hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
    So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

    But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

    Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by gummy_ratz View Post
    hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
    So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

    But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

    Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?
    Not quite...have you come across norms before? The point is that N(xy)=N(x)N(y) for all x, y in your ring. So, specifically, N(x(a+b\sqrt{-5})=N(x)N(a+b\sqrt{-5}). Use this.
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  5. #5
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    Why would the norm be ?

    Wouldn't it be a^2 + 5b^2?

    Since (a+bsqrt(-5))*(a-bsqrt(-5)) = a^2 + 5b^2?
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  6. #6
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    Quote Originally Posted by gummy_ratz View Post
    How would you prove it's not a P.I.D. by using I?

    Let be an ideal of . Prove that I is not a principal ideal of .

    I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!


    Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

    typo, surely) , prove that 6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}) are two fundamentally

    different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

    it cannot be a PID.

    Tonio
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

    typo, surely) , prove that 6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}) are two fundamentally

    different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

    it cannot be a PID.

    Tonio
    Yes, well, more of a mental-o (I was just squaring everything in sight!)
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