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Thread: Invertible ideal

  1. #1
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    Invertible ideal

    Hi there

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

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

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

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

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

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

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

    if $\displaystyle I=(2,1+\sqrt{-5}),$ then $\displaystyle I^2=(2)$ and therefore $\displaystyle I$ is invertible. (why?)
    Last edited by NonCommAlg; Jan 19th 2010 at 09:55 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    if $\displaystyle I=(2,1+\sqrt{-5}),$ then $\displaystyle I^2=(2)$ and therefore $\displaystyle I$ 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 $\displaystyle 1+\sqrt{-5}$ 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 $\displaystyle 1+\sqrt{-5}$ are both irreducible elements in the ring?
    sorry for the late response. from $\displaystyle I^2=(2),$ you get $\displaystyle \frac{1}{2}II=R,$ which means that $\displaystyle I^{-1}=\frac{1}{2}I.$
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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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