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Math Help - Ring of integers

  1. #1
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    Ring of integers

    Hello,

    let K=\mathbb Q(\sqrt d) be a number field.

    It is well known, that the ring of integers \mathcal O_K of K is given by \mathbb Z[\xi] with

     \xi = \frac{1}{2}+\frac{1}{2}\sqrt d  \ \ \quad  \text{ if } \quad d \equiv 1 \text{ mod }4
     \xi = \sqrt d \quad \quad  \ \ \quad  \text{ if } \quad d \equiv 2,3 \text{ mod }4.

    How to prove that this is the same as the ring \mathbb Z[ \theta] with \theta = \frac{1}{2}d + \frac{1}{2} \sqrt{d} ?

    Thanks for your help in advance.

    Bye,
    Alexander
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Ring of integers

    Quote Originally Posted by AlexanderW View Post
    Hello,

    let K=\mathbb Q(\sqrt d) be a number field.

    It is well known, that the ring of integers \mathcal O_K of K is given by \mathbb Z[\xi] with

     \xi = \frac{1}{2}+\frac{1}{2}\sqrt d  \ \ \quad  \text{ if } \quad d \equiv 1 \text{ mod }4
     \xi = \sqrt d \quad \quad  \ \ \quad  \text{ if } \quad d \equiv 2,3 \text{ mod }4.

    How to prove that this is the same as the ring \mathbb Z[ \theta] with \theta = \frac{1}{2}d + \frac{1}{2} \sqrt{d} ?

    Thanks for your help in advance.

    Bye,
    Alexander
    Alexander,

    Just check that \xi\in\mathbb{Z}[\theta] and \theta\in\mathbb{Z}[\xi]. Where are you having trouble?

    Best,
    Alex
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  3. #3
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    Re: Ring of integers

    Maybe the exercise is incorrect, because:

    I would like to check that \theta \in \mathbb Z[\xi] ... then the case  d \equiv 2,3 \text{ mod 4} yields a contradiction:

    \frac{1}{2}d+\frac{1}{2}\sqrt{d} = a+b \cdot \sqrt{d} \in \mathbb Z[\xi] is not possible for a,b \in \mathbb Z.

    Quote Originally Posted by Drexel28 View Post
    Alexander,

    Just check that \xi\in\mathbb{Z}[\theta] and \theta\in\mathbb{Z}[\xi]. Where are you having trouble?

    Best,
    Alex
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Ring of integers

    Quote Originally Posted by AlexanderW View Post
    Maybe the exercise is incorrect, because:

    I would like to check that \theta \in \mathbb Z[\xi] ... then the case  d \equiv 2,3 \text{ mod 4} yields a contradiction:

    \frac{1}{2}d+\frac{1}{2}\sqrt{d} = a+b \cdot \sqrt{d} \in \mathbb Z[\xi] is not possible for a,b \in \mathbb Z.
    You seem to be correct. The minimal polynomial for \frac{1}{2}d+\frac{1}{2}\sqrt{d} when d=2 is 2x^2-4x+1, and thus this element isn't even integral over \mathbb{Z}. Are you sure the exercise is written down correctly? Can you give more context (book its in, etc.)?

    Best,
    Alex
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