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This was a problem in a test:

Prove that the Galois group of $x^p-2$ (where $p$ is a prime) over $\mathbb Q$ isomorphic to the following matrix group under multiplication:

$$\begin{pmatrix} a & b \\ 0 & 1\\ \end{pmatrix}$$

where $a \in \mathbb F_{p}$ and $b \in \mathbb F_{p} \backslash \{0\}.$

I don't know why the Galois group is not simply the order $p$ cyclic group (since the roots of the polynomial over $\mathbb C$ are $\sqrt[p]{2}\epsilon_{i}$ where the $\epsilon_{i}$ are the $p$-th roots of unity),

Can someone help me?

Nesa
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    Complex conjugation will also be an automorphism, which fixes some roots but not others. On the other hand, the $\sqrt[p]{2} \mapsto \sqrt[p]{2}\zeta$ automorphism, which would generate that cyclic group, moves all the roots. – Kaj Hansen Jun 06 '17 at 23:30
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    The splitting field over $\Bbb{Q}$ will need to contain the $p^{th}$ roots of unity. Hence the $\Bbb{F}_p\backslash {0}$ – sharding4 Jun 06 '17 at 23:53
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    What we want to prove is $\sigma \in Gal(\mathbb Q(\sqrt[p]2, \zeta)/\mathbb Q)$ iff $\sigma(\zeta^k\sqrt[p]2) = \sigma(\zeta^k)\sigma(\sqrt[p]2)=\zeta^{nk}\zeta^m\sqrt[p]{2}$ for some $n \in \mathbb{F}_p^\times, m \in \mathbb{F}_p$.

    Thus $$\sigma_{n',m'} \circ \sigma_{n,m}(\zeta^k\sqrt[p]2) = \sigma_{n',m'} (\zeta^{nk}\zeta^m\sqrt[p]2) = \zeta^{n'(nk+m) + m'}\sqrt[p]2 = \sigma_{n'n,n'm+m'}(\zeta^k\sqrt[p]2)$$

     – reuns Jun 07 '17 at 04:48
  • Hint: Map that matrix to the automorphism that is determined by $\zeta\mapsto \zeta^a$, $\root p\of2\mapsto \zeta^b\root p\of 2$. – Jyrki Lahtonen Jun 07 '17 at 06:38
  • Related. – Jyrki Lahtonen Feb 22 '22 at 07:27

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Let $\zeta = e^{\frac{2\pi i}p}$, a primitive $p$-th root of unity and let $K$ be the splitting field of $x^p-2$ over $\mathbb Q$. Then claim that $K = \mathbb Q(\sqrt[p]2, \zeta)$ and $[K:\mathbb Q] = p(p-1)$. Since $x^p -2$ is an separable polynomial over $\mathbb Q$ (why?) it follows that $K/\mathbb Q$ is Galois and so the Galois group has order $p(p-1)$. Following the comments one can show that $Gal(K/\mathbb Q)$ is isomorphic to the matrix group you state above, which is also isomorphic to the nonabelian group $C_p \rtimes C_{p-1}$

Proof of claim: As you said $\sqrt[p]2\zeta^k$ for $k = 0, 1, \dots, p-1$ are the roots of $x^p - 2$; so clearly $K \subset \mathbb Q(\sqrt[p]2, \zeta)$. Now note that $\sqrt[p]2 \in K$ and also $\zeta = \frac{(\sqrt[p]2\zeta)^{p+1}}{2\sqrt[p]2} \in K$; hence $\mathbb Q(\sqrt[p]2, \zeta) \subset K$. Now note that $K$ contains the subfields $\mathbb Q(\sqrt[p]2)$ and $\mathbb Q(\zeta)$, which have degree $p$ and $p-1$ resp. over $\mathbb Q$. Since $\gcd(p, p-1) = 1$, by a basic field theory result it follows that the compositum field $\mathbb Q(\sqrt[p]2)\mathbb Q(\zeta) = K$ has degree $p(p-1)$ over $\mathbb Q$. $\square$

cat
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