# Algebraic Numbers

• Mar 31st 2010, 03:12 AM
bram kierkels
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
• Mar 31st 2010, 03:55 AM
tonio
Quote:

Originally Posted by bram kierkels
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
• Apr 1st 2010, 09:21 AM
proscientia
Quote:

Originally Posted by bram kierkels
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 $\displaystyle \mathbb Q(z)=\mathbb Q(\sqrt3\,i).$