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I have for example the cyclotomic field extension $\mathbb{Q}(\zeta_3)$ . And I think it's possible: $\mathbb{Q}(\sqrt{\zeta_3})$ . Then it will be possible the field extension of the cyclotomic extension: $\mathbb{Q}(\zeta_3,\sqrt{\zeta_3})$ .

How $\mathbb{Q}(\zeta_3)$ and $\mathbb{Q}(\sqrt{\zeta_3})$ are linearly disjoint over $\mathbb{Q}$ , I have to multiply its two respective bases to get the basis of $K=\mathbb{Q}(\zeta_3,\sqrt{\zeta_3})$ .

Then: $K=\{1,\zeta_3\}\,\,x\,\, \{1,\sqrt{\zeta_3}\}\,=\,\{1,\,\zeta_3,\,\sqrt{\zeta_3},\,\,\zeta_3\sqrt{\zeta_3}\}$

Is it correct?

Is it truly the basis $k_1=\mathbb{Q}(\sqrt{\zeta_3})=\{1,\sqrt{\zeta_3}\}$ , how I'm thinking?

Thanks in advance

mref
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    Which extension do you finally consider? $\mathbb{Q}(\zeta_3,\sqrt{\zeta_3})$ over $\Bbb Q$? – Dietrich Burde Mar 02 '20 at 15:27
  • @Dietrich Burde Yes – mref Mar 02 '20 at 15:37
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    It would be "more interesting" to consider the fields obtained by adding successively $3^n$-roots of $\zeta_3$. – nguyen quang do Mar 03 '20 at 08:43
  • @nguyenquangdo True. $ \left( \dfrac{1+\sqrt{-3}}{2} \right)^2=\dfrac{-1+\sqrt{-3}}{2}$ . What would X of: $X^3=\dfrac{-1+\sqrt{-3}}{2}$ ? – mref Mar 04 '20 at 14:51
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    You don't need to do any computation if you know about cyclotomic fields. If $p$ is an odd prime, let $\zeta_n$ denote a primitive $p^n$-th root of unity. Then $\mathbf Q(\zeta_n)/\mathbf Q$ is a cyclic extension of degree $p^{n-1}(p-1)$. In particular, for $n \ge 2, \mathbf Q(\zeta_n)/\mathbf Q(\zeta_1)$ is cyclic of degree $p^{n-1}$ – nguyen quang do Mar 04 '20 at 17:50
  • @nguyenquangdo Ok – mref Mar 04 '20 at 22:20

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Let $a=\sqrt{\zeta_3}$. Since $a^2=\zeta_3$ we have $a^6=1$ and $a$ is a $6$-th root of unity. However, we have $$ \Bbb Q(\zeta_6)=\Bbb Q(\zeta_3), $$ see here: Is $\mathbb Q(\zeta_6)=\mathbb {Q}(\zeta_3)$?

So the degree is indeed $2$.

Dietrich Burde
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