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Math Help - 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 \zeta_5 be a primitive fifth root of unity and \mathcal O_K the ring of integers in K=\mathbb Q(\zeta_5).

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

    I know that \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 \mathbb Q(\sqrt m) is \mathbb Z \cdot 1 \oplus  \mathbb Z \cdot \sqrt{m} if m \equiv 2,3 \text{ mod }4 and \mathbb Z\cdot 1  \oplus \mathbb Z\cdot \left ( \frac{1}{2}+\frac{1}{2}\sqrt{m} \right ) if m \equiv 1 \text{ mod }4.

    In the case of \zeta_3 it is easy because of

    \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 -3 \equiv 1 \text{ mod }4.

    But \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 \mathcal O_K looks like if 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 \zeta_5 be a primitive fifth root of unity and \mathcal O_K the ring of integers in K=\mathbb Q(\zeta_5).

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

    I know that \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 \mathbb Q(\sqrt m) is \mathbb Z \cdot 1 \oplus  \mathbb Z \cdot \sqrt{m} if m \equiv 2,3 \text{ mod }4 and \mathbb Z\cdot 1  \oplus \mathbb Z\cdot \left ( \frac{1}{2}+\frac{1}{2}\sqrt{m} \right ) if m \equiv 1 \text{ mod }4.

    In the case of \zeta_3 it is easy because of

    \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 -3 \equiv 1 \text{ mod }4.

    But \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 \mathcal O_K looks like if K=\mathbb Q(\zeta_5).

    Please help me.

    Bye,
    Alexander
    Hint:  \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 \mathcal O_K = \mathbb Z[\zeta_5] for K=\mathbb Q(\zeta_5)

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

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

    So, i have to add the zeros of f_{\zeta_5} to \mathbb Q to get \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 \mathcal O_K = \mathbb Z[\zeta_5] for K=\mathbb Q(\zeta_5)

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

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

    So, i have to add the zeros of f_{\zeta_5} to \mathbb Q to get \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 f_{\zeta_5}(x) satisfy the equation x^5 = 1. However, if your textbook doesn't contain a proof that \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 \mathcal O_K dividing the ideal generated by 3. Do you know anything about the structure of the residue field \mathcal O_K/P?
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