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Thread: Ring of integers

  1. #1
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    Ring of integers

    Hello,

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

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

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

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

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

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

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

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

    Thanks for your help in advance.

    Bye,
    Alexander
    Alexander,

    Just check that $\displaystyle \xi\in\mathbb{Z}[\theta]$ and $\displaystyle \theta\in\mathbb{Z}[\xi]$. 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 $\displaystyle \theta \in \mathbb Z[\xi]$ ... then the case $\displaystyle d \equiv 2,3 \text{ mod 4} $ yields a contradiction:

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

    Quote Originally Posted by Drexel28 View Post
    Alexander,

    Just check that $\displaystyle \xi\in\mathbb{Z}[\theta]$ and $\displaystyle \theta\in\mathbb{Z}[\xi]$. 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 $\displaystyle \theta \in \mathbb Z[\xi]$ ... then the case $\displaystyle d \equiv 2,3 \text{ mod 4} $ yields a contradiction:

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

    Best,
    Alex
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