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Compute the Galois group of $x^n-p$ over $\Bbb Q$. Here, $n\in\Bbb N$ and $p$ is a prime number.

Let $\eta$ be the primitive $n$th root of unity. Then $\Bbb Q(\sqrt[n]{p},\eta)$ is a splitting field of $x^n-p$ over $\Bbb Q$. Let $H$ be a fixed field of $\Bbb Q(\eta)$ under the Galois correspondence. Since $\Bbb Q(\eta)/\Bbb Q$ is Galois, $H$ is normal in $G=\operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q)$. Hence from the s.e.s. which splits $$0\to H = \operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q(\eta))\to G = \operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q)\xrightarrow{\pi} G/H = \operatorname{Gal}(\Bbb Q(\eta)/\Bbb Q)\to 0$$ since we can define an inverse $p:G/H\to G$ by extension $\sigma\in \operatorname{Gal} (\Bbb Q(\eta)/\Bbb Q)$ by $\sigma(\sqrt[n]{p}) =\sqrt[n]{p}$, we can conclude that $G\simeq H\rtimes (G/H)$.

It's a naive generalization of this answer which may require more conditions. Does the proof make sense? What extra conditions are necessary to make this proof is valid?

one potato two potato
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1 Answers1

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In the end it depends what you allow yourself to know about semi-direct products, in particular what equivalent conditions there are to achieve the structure of a semi-direct product. If you say a split exact sequence defines a semi-direct product then your proof is perfectly valid and correct.

There are, however, different, more explicit approaches possible too. For example, one can show that for a group $G$ with subgroups $N,H$ such that

  1. $N\triangleleft G$
  2. $NH=G$
  3. $N\cap H=\{1\}$

we have $G=N\rtimes H$ with $H$ acting on $N$ by conjugation (the general $H\to\operatorname{Aut}(N)$ being yet another equivalent definition for semi-direct products).

For $n=p$ this gives a particular simple proof. The conditions are this case readily checked using the subgroups generated by $\sigma$ and $\tau$, with

$$ \sigma\colon\begin{cases}\sqrt[p]{p}&\mapsto\sqrt[p]{p}\zeta_p\\\zeta_p&\mapsto\zeta_p\end{cases}\quad\text{and}\quad\tau\colon\begin{cases}\sqrt[p]{p}&\mapsto\sqrt[p]{p}\\\zeta_p&\mapsto\zeta_p^c\end{cases} $$

and $c$ a generator of $(\mathbb Z/p\mathbb Z)^\ast$. It might be possible to generalize this strategy for general $n$ but you might have to make a more elaborate argument.

mrtaurho
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  • So my argument holds? I mean from the above argument, can I conclude $G\simeq H\rtimes G/H$ with no further conditions or something to make the argument works? – one potato two potato Sep 23 '21 at 23:18
  • @love_sodam As I said: it depents on how you define semi-direct products in the first place. Having a split exact sequence is one of many possible but equivalent definitions. – mrtaurho Sep 24 '21 at 00:33
  • Yes my definition of semidirect product is same as your definition. And I know if the sequence splits then it's a semidirect product of ~~. – one potato two potato Sep 24 '21 at 00:44
  • @love_sodam This should answer your question completely, does it? Then you could accept the answer, or is there anything you want me to add before? – mrtaurho Oct 13 '21 at 16:34
  • Oh sorry. I forgot this post. So my proof is valid. If we drop the the condition $p$ (constant factor), then the proof is no longer valid since we don't know $\Bbb Q(\sqrt[n]{p},\eta)$ is a splitting field of $x^n-p$ over $\Bbb Q$ right? – one potato two potato Oct 13 '21 at 16:49
  • @love_sodam If you mean by "the condition $p$" $x^n-p$ with $p$ prime, then more or less yes. $\mathbb Q(\sqrt[n]{p},\eta)$ would still contain the splitting field (after all, $x^n-p$ splits there) but it might be too large. Of the top of my head I think it should still work for squarefree $p$ but will get tricky if $p$ contains powers of primes. – mrtaurho Oct 13 '21 at 16:56