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Math Help - Algebraic Numbers

  1. #1
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    Algebraic Numbers

    Let z=e^{\frac{2\pi i}{6}}
    I want to show \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3}i)
    Any hints?

    (For z\in\mathbb{C}, the field generated by z over \mathbb{Q} is defined to be the smallest subfield of \mathbb{C} which contains both \mathbb{Q} and z, denoted by \mathbb{Q}(z) )

    Thanks
    Last edited by bram kierkels; April 1st 2010 at 03:51 AM.
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  2. #2
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    Quote Originally Posted by bram kierkels View Post
    Let z=e^{\frac{2\pi i}{6}}
    I want to show \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3})
    Any hints?

    (For z\in\mathbb{C}, the field generated by z over \mathbb{Q} is defined to be the smallest subfield of \mathbb{C} which contains both \mathbb{Q} and z, denoted by \mathbb{Q}(z) )

    Thanks

    z=e^{\pi i\slash 3}=\frac{1}{2}+\frac{\sqrt{3}}{2}\,i\notin\mathbb{  R} . Since clearly \mathbb{Q}(\sqrt{3})\subset\mathbb{R} , it is impossible that \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3}) ...

    Now, I'm guessing you may have wanted to prove \mathbb{Q}(z)=\mathbb{Q}(\sqrt{-3})\,,\,\,\sqrt{-3}=\sqrt{3}\,i , but then the first part must be enough to show the way to prove what you want.

    Tonio
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  3. #3
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    Quote Originally Posted by bram kierkels View Post
    Let z=e^{\frac{2\pi i}{6}}
    I want to show \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3}i)
    w\,\in\,\mathbb Q(z)

    \Rightarrow\quad w\ =\ r_0+r_1z+r_2z^2+r_3z^3+r_4z^4+r_5z^5 for some r_0,\ldots,r_5\in\mathbb Q

    \Rightarrow\quad w\ =\ \left[r_0+\frac{r_1}2-\frac{r_2}2-r_3-\frac{r_4}2+\frac{r_5}2\right]+\left[\frac{r_1}2+\frac{r_2}2-\frac{r_4}2-\frac{r_5}2\right]\sqrt3\,i

    \Rightarrow\quad w\,\in\,\mathbb Q(\sqrt3\,i)

    Conversely,

    w\,\in\,\mathbb Q(\sqrt3\,i)

    \Rightarrow\quad w\ =\ s_1+s_2\sqrt3\,i for some s_1,s_2\in\mathbb Q

    \Rightarrow\quad w\ =\ (s_1-s_2)+2s_2\,z

    \Rightarrow\quad w\,\in\,\mathbb Q(z)

    Hence \mathbb Q(z)=\mathbb Q(\sqrt3\,i).
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