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Thread: Prove, that an ideal is prime in O_K

  1. #1
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    Prove, that an ideal is prime in O_K

    Hello,

    let $\displaystyle \zeta_5$ be a primitive fifth root of unity and $\displaystyle \mathcal O_K$ the ring of integers in $\displaystyle K=\mathbb Q(\zeta_5)$.

    How to prove, that $\displaystyle 3 \cdot \mathcal O_K$ is a prime ideal in $\displaystyle \mathcal O_K$?

    I know that $\displaystyle \mathcal O_K$ is a dedekind ring and therefore each ideal has a factorization in prime ideals and furthermore i know that the ring of integers of $\displaystyle \mathbb Q(\sqrt m)$ is $\displaystyle \mathbb Z \cdot 1 \oplus \mathbb Z \cdot \sqrt{m}$ if $\displaystyle m \equiv 2,3 \text{ mod }4$ and $\displaystyle \mathbb Z\cdot 1 \oplus \mathbb Z\cdot \left ( \frac{1}{2}+\frac{1}{2}\sqrt{m} \right ) $ if $\displaystyle m \equiv 1 \text{ mod }4$.

    In the case of $\displaystyle \zeta_3$ it is easy because of

    $\displaystyle \mathbb Q(\zeta_3) = \mathbb Q \left ( \frac{-1}{2}+\frac{1}{2}i\sqrt{3} \right )=\mathbb Q(i\sqrt{3}) = \mathbb Q(\sqrt{-3})$ and $\displaystyle -3 \equiv 1 \text{ mod }4$.

    But $\displaystyle \zeta_5={{\rm e}^{ \frac{2 \pi i}{5} }} = \cos \left ( \frac{2 \pi }{5} \right ) + i \sin \left ( \frac{2 \pi }{5} \right )$ and now i can't simplify the term and i don't know how the ring $\displaystyle \mathcal O_K$ looks like if $\displaystyle K=\mathbb Q(\zeta_5)$.

    Please help me.

    Bye,
    Alexander
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by AlexanderW View Post
    Hello,

    let $\displaystyle \zeta_5$ be a primitive fifth root of unity and $\displaystyle \mathcal O_K$ the ring of integers in $\displaystyle K=\mathbb Q(\zeta_5)$.

    How to prove, that $\displaystyle 3 \cdot \mathcal O_K$ is a prime ideal in $\displaystyle \mathcal O_K$?

    I know that $\displaystyle \mathcal O_K$ is a dedekind ring and therefore each ideal has a factorization in prime ideals and furthermore i know that the ring of integers of $\displaystyle \mathbb Q(\sqrt m)$ is $\displaystyle \mathbb Z \cdot 1 \oplus \mathbb Z \cdot \sqrt{m}$ if $\displaystyle m \equiv 2,3 \text{ mod }4$ and $\displaystyle \mathbb Z\cdot 1 \oplus \mathbb Z\cdot \left ( \frac{1}{2}+\frac{1}{2}\sqrt{m} \right ) $ if $\displaystyle m \equiv 1 \text{ mod }4$.

    In the case of $\displaystyle \zeta_3$ it is easy because of

    $\displaystyle \mathbb Q(\zeta_3) = \mathbb Q \left ( \frac{-1}{2}+\frac{1}{2}i\sqrt{3} \right )=\mathbb Q(i\sqrt{3}) = \mathbb Q(\sqrt{-3})$ and $\displaystyle -3 \equiv 1 \text{ mod }4$.

    But $\displaystyle \zeta_5={{\rm e}^{ \frac{2 \pi i}{5} }} = \cos \left ( \frac{2 \pi }{5} \right ) + i \sin \left ( \frac{2 \pi }{5} \right )$ and now i can't simplify the term and i don't know how the ring $\displaystyle \mathcal O_K$ looks like if $\displaystyle K=\mathbb Q(\zeta_5)$.

    Please help me.

    Bye,
    Alexander
    Hint: $\displaystyle \mathcal O_K = \mathbb{Z}[\zeta_5] $.
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  3. #3
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    Hello

    thanks for the hint.

    So i have to proof at first, that $\displaystyle \mathcal O_K = \mathbb Z[\zeta_5]$ for $\displaystyle K=\mathbb Q(\zeta_5)$

    The minimal polynomial of $\displaystyle \zeta_5$ is (since 5 is a prime number) the cyclotomic polynomial:

    $\displaystyle f_{\zeta_5}(x)=x^4+x^3+x^2+x+1$

    So, i have to add the zeros of $\displaystyle f_{\zeta_5}$ to $\displaystyle \mathbb Q$ to get $\displaystyle \mathbb Q(\zeta_5)$.

    But how does one find out the zeros of a polynomial of degree 5??

    Bye,
    Alexander
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  4. #4
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    Quote Originally Posted by AlexanderW View Post
    Hello

    thanks for the hint.

    So i have to proof at first, that $\displaystyle \mathcal O_K = \mathbb Z[\zeta_5]$ for $\displaystyle K=\mathbb Q(\zeta_5)$

    The minimal polynomial of $\displaystyle \zeta_5$ is (since 5 is a prime number) the cyclotomic polynomial:

    $\displaystyle f_{\zeta_5}(x)=x^4+x^3+x^2+x+1$

    So, i have to add the zeros of $\displaystyle f_{\zeta_5}$ to $\displaystyle \mathbb Q$ to get $\displaystyle \mathbb Q(\zeta_5)$.

    But how does one find out the zeros of a polynomial of degree 5??

    Bye,
    Alexander
    Recall that the roots of $\displaystyle f_{\zeta_5}(x)$ satisfy the equation $\displaystyle x^5 = 1$. However, if your textbook doesn't contain a proof that $\displaystyle \mathcal O_K = \mathbb Z[\zeta_5]$, I doubt that you should be proving that fact to solve the problem. Suppose that P is a prime ideal in $\displaystyle \mathcal O_K$ dividing the ideal generated by 3. Do you know anything about the structure of the residue field $\displaystyle \mathcal O_K/P$?
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