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Math Help - Product of non principal ideals is principal.

  1. #1
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    Product of non principal ideals is principal.

    Hello, Im trying to prove that the product of two non principal ideals is a principal ideal.

    I consider the non principals ideals $\left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$

    The book saids the result is (3). But I don`t know how to do the product.

    Regards...
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  2. #2
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    Quote Originally Posted by orbit View Post
    Hello, Im trying to prove that the product of two non principal ideals is a principal ideal.

    I consider the non principals ideals $\left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$

    The book saids the result is (3). But I don`t know how to do the product.

    Regards...

    Well, SOMETIMES the prod. of two non principal ideals is principal, not always...surely you meant this.

    Now, \left( {3,1 - \sqrt { - 5} } \right)\left( {3,1 + \sqrt { - 5} } \right)=(9,6,3(1\pm \sqrt{-5})) \in (3)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Well, SOMETIMES the prod. of two non principal ideals is principal, not always...surely you meant this.

    Now, \left( {3,1 - \sqrt { - 5} } \right)\left( {3,1 + \sqrt { - 5} } \right)=(9,6,3(1\pm \sqrt{-5})) \in (3)

    Tonio
    Thanks for aswering. But I donīt know how to do the product... wich are the steps to get the result. That`s my question.

    Regards
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  4. #4
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    Quote Originally Posted by orbit View Post
    Thanks for aswering. But I donīt know how to do the product... wich are the steps to get the result. That`s my question.

    Regards

    Just multiply the ideals's generators with each other.

    Tonio
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  5. #5
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    the elements of IJ are sums of the form Σij, where i is in I, and j is in J. what are the possible products ij?

    (3)(3) = 9

    (3)(1 + √(-5))

    (1 - √(-5))(3)

    (1 - √(-5)(1 + √(-5)) = 1 - (-5) = 6.

    3 divides all of these products, so certainly IJ is contained in (3).

    on the other hand 3 = 9 - 6 which is an element of IJ, so (3) is contained in IJ.
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  6. #6
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    Thanks

    Quote Originally Posted by Deveno View Post
    the elements of IJ are sums of the form Σij, where i is in I, and j is in J. what are the possible products ij?

    (3)(3) = 9

    (3)(1 + √(-5))

    (1 - √(-5))(3)

    (1 - √(-5)(1 + √(-5)) = 1 - (-5) = 6.

    3 divides all of these products, so certainly IJ is contained in (3).

    on the other hand 3 = 9 - 6 which is an element of IJ, so (3) is contained in IJ.
    Thank you very much!
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