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$\Bbb Q(\omega)= \Bbb Q(\sqrt3,\iota)$ This is written in my text book that i am following. But I think this is a typo. Since $\Bbb Q(\sqrt3,\iota)$ is a larger field. in which $\Bbb Q(\omega)$ is contained.

According to me, It should be:
$\Bbb Q(\omega)= \Bbb Q(\sqrt3.\iota)$

Further, here minimal polynomial of $\sqrt3.\iota$ is $x^2+3$. So degree of extension $\Bbb Q(\omega)$ over $\Bbb Q$ is $2$, and basis is $\{1,\sqrt3.\iota\}$, right!

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Is $\;\iota=i=\sqrt{-1}\;$ ? And is $\;\omega=e^{\frac{2\pi i}3}\;$ – DonAntonio May 2 '14 at 19:33
yes its iota and omega is the root of unity. – Pitt HarmanN - FreshmaN May 2 '14 at 22:10
up vote 1 down vote accepted

You are right, the field is $\mathbb{Q}(\sqrt{-3})$. It is not $\mathbb{Q}(\sqrt{3}.i)$. That is a larger field. We show that $\sqrt{3}\not\in \mathbb{Q}(\omega)$.

Note that $\omega^2=-1-\omega$. So if $\sqrt{3}$ were in $\mathbb{Q}(\omega)$, were, we would have $\sqrt{3}=a+b\omega$ for some rationals $a$ and $b$. But since $\sqrt{3}$ is real, we must have $b=0$. That implies that $\sqrt{3}$ is rational, which is not the case.

Remark: To prove that $\mathbb{Q}(\omega)=\mathbb{Q}(\sqrt{-3})$ we only need to show that $\sqrt{-3}\in \mathbb{Q}(\omega)$ and that $\omega\in \mathbb{Q}(\sqrt{-3})$. Since $\omega=\frac{-1+\sqrt{-3}}{2}$, this is straightforward.

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Thanks, fixed. I do have trouble with minus signs. – André Nicolas May 2 '14 at 19:53
Thank you Andre for clearing it. – Pitt HarmanN - FreshmaN May 2 '14 at 22:14
You are welcome. You knew that the answer in the book was not right. – André Nicolas May 2 '14 at 22:19

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