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It is not hard to check that the three roots of $x^3-2=0$ is $\sqrt[3]{2}, \sqrt[3]{2}\zeta_3, \sqrt[3]{2}\zeta_3^{2}$, hence the splitting field for $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}[\sqrt[3]{2}, \sqrt[3]{2}\zeta_3, \sqrt[3]{2}\zeta_3^{2}]$. However, since $\sqrt[3]{2}\zeta_3^{2}$ can be compute through $\sqrt[3]{2}, \sqrt[3]{2}\zeta_3$ then the splitting field is $\mathbb{Q}[\sqrt[3]{2}, \sqrt[3]{2}\zeta_3]$.

In the case $x^5-2=0$, in the book Galois theory by J.S.Milne, the author said that the splitting field is $\mathbb{Q}[\sqrt[5]{2}, \zeta_5]$.

My question is :

  1. How can the other roots of $x^5-2$ be represented in term of $\sqrt[5]{2}, \zeta_5$, so that he can write the splitting field is$\mathbb{Q}[\sqrt[5]{2}, \zeta_5] $ ?
  2. Is the splitting field for $x^n -a$ over $\mathbb{Q}$ is $\mathbb{Q}[\alpha,\zeta_n]$, where $\alpha$ is the real $n$-th root of $a$ ?
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Try $\zeta_5^j\root5\of 2$ with $j=0,1,2,3,4$? –  Jyrki Lahtonen Jan 1 '13 at 16:20
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Note that the roots of $x^n-a$ are $\alpha ,\alpha\zeta_n,\alpha\zeta_n^2,\ldots,\alpha\zeta_n^{n-1}$ where $\alpha$ is as you said. Then the splitting field of $x^n-a$ is $F:=\mathbb{Q}(\alpha ,\alpha\zeta_n,\ldots,\alpha\zeta_n^{n-1})$. Now $\alpha\in F$ and $\zeta_n=\alpha^{-1}(\alpha\zeta_n)\in F$, hence $\mathbb{Q(\alpha ,\zeta_n})\subseteq F$. Now $\alpha\zeta_n^i\in \mathbb{Q}(\alpha ,\alpha\zeta_n)$ for each $i$, hence $F\subseteq \mathbb{Q}(\alpha ,\alpha\zeta_n)$. Therefore $F=\mathbb{Q}(\alpha ,\alpha\zeta_n).$

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Note that $(\alpha\zeta_n^k)^n = \alpha^n\zeta_n^{nk}=\alpha^n=a$, $0\le k<n$.

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  1. Assuming that by $\zeta_5$ you mean the fifth primitive root of unity, the roots of $x^5-2$ are simply $\sqrt[5]{2}, \sqrt[5]{2} \zeta_5, \sqrt[5]{2} \zeta_5^2, \sqrt[5]{2} \zeta_5^3, \sqrt[5]{2} \zeta_5^4$.

  2. The roots of $x^n - a$ are given when $x^n=a$. Let $x=re^{i\theta}$ be such a root, then $\left( re^{i\theta} \right)^n = a$, meaning $r^n=a$ and $n\theta = 2\pi k$. This means that the roots are always $\sqrt[n]{a},\sqrt[n]{a}\zeta_n^1,...,\sqrt[n]{a}\zeta_n^{n-1} \in \mathbb{Q}[\sqrt[n]{a},\zeta_n]$.

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