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Thread: Algebraic Numbers

  1. #1
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    Algebraic Numbers

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

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

    Thanks
    Last edited by bram kierkels; Apr 1st 2010 at 02:51 AM.
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  2. #2
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    Quote Originally Posted by bram kierkels View Post
    Let $\displaystyle z=e^{\frac{2\pi i}{6}}$
    I want to show $\displaystyle \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3})$
    Any hints?

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

    Thanks

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

    Now, I'm guessing you may have wanted to prove $\displaystyle \mathbb{Q}(z)=\mathbb{Q}(\sqrt{-3})\,,\,\,\sqrt{-3}=\sqrt{3}\,i$ , 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 $\displaystyle z=e^{\frac{2\pi i}{6}}$
    I want to show $\displaystyle \mathbb{Q}(z)=\mathbb{Q}(\sqrt{3}i)$
    $\displaystyle w\,\in\,\mathbb Q(z)$

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

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

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

    Conversely,

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

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

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

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

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