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I believe that over $\mathbb{Q}[x]$, the splitting field of the polynomial $x^p-2$ is finite and separable, and so it must also be simple. However, I am unable to find, for a general $p$, what the element of the simple extension must be.

blanchey
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1 Answers1

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The splitting field is $L=\Bbb Q(\zeta_p,\sqrt[p]{2})$. A primitive element for the extension $L\mid \Bbb Q$ is $\alpha=\zeta_p+\sqrt[p]{2}$. We have $[\Bbb Q(\alpha):\Bbb Q]=p(p-1)$, which is also the degree of $[L:\Bbb Q]$.

Dietrich Burde
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  • Assuming $p$ was meant to be prime in the problem. More generally, $[L:\mathbb Q]\leq p\cdot\phi(p).$ With $p=8$ then $\sqrt{2}\in \mathbb Q[\zeta_8]$ and then $L:Q]=4\cdot 4=16<8\cdot\phi(8).$ – Thomas Andrews May 09 '19 at 16:59
  • @ThomasAndrews Yes, you are right. I was assuming that $p$ is prime. We had discussed the general case for example here. – Dietrich Burde May 09 '19 at 17:45