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Math Help - Invertible ideal

  1. #1
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    Invertible ideal

    Hi there

    I have the ring \mathbb{Z}[\sqrt{-5}] with field of fractions K and am to decide whether any of the ideals (2), (1+\sqrt{-5}), (2, 1+\sqrt{-5}) are invertible or not.

    Well, if the ideal I is invertible, we have that II^*=R where I^*=\{q\in K | qI \subset R\}
    so q is on the form \frac{a+b\sqrt{-5}}{c+d\sqrt{-5}}, correct?

    Then I would say that the ideal (2, 1+\sqrt{-5}) is invertible, because whatever the denominator is in the above fraction, it disappaers when multiplying with (2, 1+\sqrt{-5}).

    Did I get that right?
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  2. #2
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    Quote Originally Posted by Carl View Post
    Hi there

    I have the ring \mathbb{Z}[\sqrt{-5}] with field of fractions K and am to decide whether any of the ideals (2), (1+\sqrt{-5}), (2, 1+\sqrt{-5}) are invertible or not.

    Well, if the ideal I is invertible, we have that II^*=R where I^*=\{q\in K | qI \subset R\}
    so q is on the form \frac{a+b\sqrt{-5}}{c+d\sqrt{-5}}, correct?

    Then I would say that the ideal (2, 1+\sqrt{-5}) is invertible, because whatever the denominator is in the above fraction, it disappaers when multiplying with (2, 1+\sqrt{-5}).

    Did I get that right?
    if I=(a), \ a \neq 0, then I^{-1}=Ra^{-1}. so, in general, every non-zero principal idea of an integral domain is invertible.

    if I=(2,1+\sqrt{-5}), then I^2=(2) and therefore I is invertible. (why?)
    Last edited by NonCommAlg; January 19th 2010 at 10:55 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    if I=(2,1+\sqrt{-5}), then I^2=(2) and therefore I is invertible. (why?)
    Well, I find it hard to see why this property will make the ideal invertible. I would say that it has something to do with the fact that 2 and 1+\sqrt{-5} are both irreducible elements in the ring?
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  4. #4
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    Quote Originally Posted by Carl View Post
    Well, I find it hard to see why this property will make the ideal invertible. I would say that it has something to do with the fact that 2 and 1+\sqrt{-5} are both irreducible elements in the ring?
    sorry for the late response. from I^2=(2), you get \frac{1}{2}II=R, which means that I^{-1}=\frac{1}{2}I.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    sorry for the late response.[/tex]
    No problem, the exercise is first for next week.
    Ok, I got it now, thanks!
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