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Math Help - Primitive root in quadratic field

  1. #1
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    Primitive root in quadratic field

    Let \omega _{m} = e^{ \frac {2 \pi i} {m}} , then  \omega _{m} is one of the primitive m th roots of unity.

    a) Show that  \mathbb {Q} [ \omega _{m} ] \subseteq \mathbb {Q} [ \omega _n ] if m|n.

    Proof.

    Let mt = n for some integer t.

    Pick an element a+b \omega _m \in \mathbb {Q} [ \omega _m ] , with a and b rational numbers.

    Then it equals to  a + b e^ { \frac {2 \pi i } {m} } = a + be^{ \frac {2 \pi i (t) } {n} } = a + be^t + be ^ { \frac {2 \pi i } {n} } Now, is this element in  \mathbb {Q} [ \omega _n ] ? I just don't know if  e^t is a rational number.

    b) Show that  \sqrt {p} \in \mathbb {Q} [ \omega _{4p} ] .

    Proof.

    There is a theorem that state: Let  \omega be a primitive pth root of unity, p an odd prime.  \sqrt {p} \in \mathbb {Q} [ \omega ] if  p \equiv 1 \ (mod \ 4 )

    Now, is  \omega _{4p} a primitive pth roof of unity?

    Thank you.
    Last edited by tttcomrader; April 28th 2008 at 08:02 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    a) Show that  \mathbb {Q} [ \omega _{m} ] \subseteq \mathbb {Q} [ \omega _n ] if m|n.
    Remember the definition of \mathbb{Q}(\alpha) (here \alpha is algebraic number), it is defined to be the smallest field containing \mathbb{Q} and \alpha*.

    Thus, \mathbb{Q}(\omega_m) is the smallest field containing \mathbb{Q} and \omega_m. Thus, if you can show that \omega_m \in \mathbb{Q}(\omega_n) then by definition \mathbb{Q}(\omega_m)\subseteq \mathbb{Q}(\omega_n). Can you show this last step?

    *)Meaning if F is any other field containing \alpha and \mathbb{Q} then \mathbb{Q}(\alpha)\subseteq F.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    b) Show that  \sqrt {p} \in \mathbb {Q} [ \omega _{4p} ] .
    I do not know this theorem and I only see one way of deriving it.
    If p is an odd prime, and \omega = e^{2\pi i/p} then the quadradic Gauss sum, i.e. g=\sum_{k=1}^{p-1} (k/p)\omega^k  = \left\{ \begin{array}{c}\sqrt{p} \mbox{ if }p\equiv 1(\bmod 4) \\ i\sqrt{p} \mbox{ if }p\equiv 3(\bmod 4) \end{array} \right.

    Note that (k/p) = \pm 1 because it is the Legendre symbol. This means if p\equiv 1(\bmod 4) then \sqrt{p} = g \in \mathbb{Q} (\omega_p). Thus, then \sqrt{p} \in \mathbb{Q}(\omega_{4p}) by excercise #1.

    If, however, p\equiv 3(\bmod 4) then g = i\sqrt{p} \in \mathbb{Q}( \omega_p ) \subseteq     \mathbb{Q} ( \omega_{4p} ). But i \in \mathbb{Q}(\omega_{4p}) because i = \omega_{4p}^p. Thus, by closure g/i = \sqrt{p} \in \mathbb{Q}(\omega_{4p}).
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  4. #4
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    Please don't laugh for my understanding in this material is very weak.

    for a), I have  \omega _{m} = e^{ \frac {2 \pi i }{m} } = e^{ \frac {2 \pi i (t) }{n} }= \omega _n ^t, this is in  \mathbb {Q} [ \omega _n] since it is cyclic, right?

    And there is also one more question in this problem:

    c) Let m be square-free, show that  \mathbb {Q} [ \sqrt {m} ] \subseteq \mathbb {Q} [ \omega _{4m} ]

    So should I try to get   \sqrt {m} \  |  \ \omega _{4m}

    Thanks.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    Please don't laugh for my understanding in this material is very weak.

    for a), I have  \omega _{m} = e^{ \frac {2 \pi i }{m} } = e^{ \frac {2 \pi i (t) }{n} }= \omega _n ^t, this is in  \mathbb {Q} [ \omega _n] since it is cyclic, right?
    That works.

    c) Let m be square-free, show that  \mathbb {Q} [ \sqrt {m} ] \subseteq \mathbb {Q} [ \omega _{4m} ]

    So should I try to get   \sqrt {m} \  |  \ \omega _{4m}
    If m is square-free then m=p_1...p_n for distinct primes. By other excerise it means \sqrt{p_k}\in \mathbb{Q}(\omega_{4p_k}) for k=1,...,n. But \mathbb{Q}(\omega_{4m}) because p_k|m. Thus, \sqrt{p_k}\in \mathbb{Q}(\omega_{4m}) for k=1,...,n. Since fields are closed under multiplication it means \sqrt{p_1}\cdot ... \cdot \sqrt{p_m} \in \mathbb{Q}(\omega_{4m}) thus \sqrt{m} = \sqrt{p_1...p_m} \in \mathbb{Q}(\omega_{4m}). Why does this complete the proof?
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  6. #6
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    The proof is complete because like in the first exercise,  \mathbb {Q} [ \sqrt {m} ] is the smallest field that contains both  \mathbb {Q} and  \sqrt {m} , if  \sqrt {m} \in \mathbb {Q} [ \omega _{4m} ], it completes the proof.

    Thank you very much!!!!!!
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