How many extension dimension of $\Bbb Q(\omega_3,\omega_5)$, $\Bbb Q(2^{1/3},i)$and $\Bbb Q(2^{1/3},\omega_3)$ where $\omega_n=e^{2\pi i/n}$
the first i think 6 extension the second is three third is 4
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Let me know if the edit is what you intended. I am not sure if you were thinking about $e$; since $e$ is transcendental. – Pedro May 01 '14 at 18:12
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no i want 3,and 5 under e – user146264 May 01 '14 at 18:25
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The usual notation I have seen for that is $\omega_n$. – Pedro May 01 '14 at 18:28
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If $n$ and $m$ are coprime, then $\mathbb{Q}(\omega_n,\omega_m)=\mathbb{Q}(\omega_{nm})$ and the degree is $$ [\mathbb{Q}(\omega_{nm}):\mathbb{Q}]=\phi(nm)=\phi(n)\phi(m). $$ With $n=3, m=5$ we obtain that $[\mathbb{Q}(\omega_3,\omega_5):\mathbb{Q}]=2\cdot 4=8$. For the second one we have $$ [\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})]\cdot [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=2\cdot 3=6. $$ The field is the splitting field of $x^3-2$, as you know - see can we have split field of the real number and how i get the extension dimension?. The last one is for you.
Dietrich Burde
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can u explain please y u said 2*4 since w3 has three element and w5 has 5 element? – user146264 May 03 '14 at 02:33
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The degree of $\mathbb{Q}(\omega_3)$ is $\phi(3)=2$, so it is a quadratic number field. Note that $1+\omega_3+\omega_3^2=0$, so we do not have "three elements" linear independent. – Dietrich Burde May 05 '14 at 08:30