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Show that the splitting field of $X^n-a$ over a field $K$ is $K(\alpha, \zeta_n)$, where $\alpha$ is a $n$-th root of $a$ and $\zeta_n$ is a primitive $n$-root of unity.

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    $\begingroup$ I would expect that your lecture notes or book describe the splitting field of $X^3-2$ over the rationals as an example. Which step do you have problems generalizing? Mind you, you should have problems at some points. If for example $K$ has characteristic two, then there aren't any $n$th primitive roots of unity for even $n$. $\endgroup$ – Jyrki Lahtonen Dec 17 '13 at 19:28
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Notice that $\alpha \zeta_n^k$, $k=0,\ldots,n-1$ are the roots of the polynomial given. So, the splitting field will contain both $\alpha$, (k=0) case and $\zeta_n$. This means that the splitting field contains $K(\alpha,\zeta_n)$. Also, if the splitting field contains $\alpha$ and $\zeta_n$, it also contains all the roots of the polynomial(products of $\alpha$ and $\zeta_n$) which means that the splitting field is exactly $K(\alpha,\zeta_n)$.

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