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Let $\zeta =e^{\pi i/12}$.

  • Find the extension degree of $\mathbb{Q}\leq \mathbb{Q}(\zeta)$
  • Show that $\mathbb{Q}(\zeta)=\mathbb{Q}(\sqrt{2} , \sqrt{3} , i)$

$\zeta$ is a root of $x^{24}-1$

$Irr(\zeta, \mathbb{Q})=\Phi_{24}(x)$ which has degree $\phi(24)=8$

So the extension degree is $8$.

Is this ccorrect??

For the second question are we looking a primitive element??

Mary Star
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    Yes, since $\zeta=\zeta_{24}$, the degree is $[\mathbb{Q}(\zeta):\mathbb{Q}]=\phi(24)=8$ with Euler's totient function. Of course, also $[\mathbb{Q}(\sqrt{2},\sqrt{3},i):\mathbb{Q}]=8$. – Dietrich Burde Jan 08 '15 at 16:16
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    Note that $\mathbb{Q}(\zeta^2)=\mathbb{Q}(\sqrt{3},i)$ and $\mathbb{Q}(\zeta^3)=\mathbb{Q}(\sqrt{2},i)$. – Dietrich Burde Jan 08 '15 at 16:25

1 Answers1

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The degree is equal to $8$ in both cases, i.e., $[\mathbb{Q}(\zeta):\mathbb{Q}]=8=[\mathbb{Q}(\sqrt{2},\sqrt{3},i):\mathbb{Q}]$; and one field is contained in the other, see here for details, i.e., the relation between cyclotomic fields and quadratic fields.

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Dietrich Burde
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  • To show that $\mathbb{Q}(\zeta)=\mathbb{Q}(\sqrt{2}, \sqrt{3}, i)$ could we also do it as followed??

    The primitive element that we are looking for is $a$, which is a linear combination of the elements $\sqrt{2}$, $\sqrt{3}$, $i$.

    $$\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \ \frac{1}{2}+i\frac{\sqrt{3}}{2}$$

     – Mary Star Jan 08 '15 at 19:01
  • Can we take as primitive element the following:

    $$a= \left ( \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right ) \cdot \left ( \frac{1}{2}+i\frac{\sqrt{3}}{2} \right ) =e^{\frac{\pi i}{4}} \cdot e^{\frac{\pi i}{3}}=e^{\frac{7 \pi i}{12}}=\left ( e^{\frac{\pi i}{12}} \right )^7=\zeta^7$$

    So, $$\mathbb{Q}(\sqrt{2}, \sqrt{3}, i)=\mathbb{Q}(\zeta^7)=\mathbb{Q}(\zeta)$$

    Or is this wrong??

     – Mary Star Jan 08 '15 at 19:02