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Question is to Determine the Galois group of $x^p-2$ for an odd prime p.

For finding Galois group, we look for the splitting field of $x^p-2$ which can be seen as $\mathbb{Q}(\sqrt[p]{2},\zeta)$ where $\zeta$ is a primitive $p^{th}$ root of unity.

Consider $\mathbb{Q}\subset \mathbb{Q}(\zeta) \subset \mathbb{Q}(\sqrt[p]{2},\zeta)$.we know that $\mathbb{Q}(\sqrt[p]{2},\zeta)$ is Galois over $\mathbb{Q}(\zeta)$, we find Corresponding Galois Group say $G_1$.

Consider $\mathbb{Q}\subset \mathbb{Q}(\sqrt[p]{2}) \subset \mathbb{Q}(\sqrt[p]{2},\zeta)$. we know that $\mathbb{Q}(\sqrt[p]{2},\zeta)$ is galois over $ \mathbb{Q}(\sqrt[p]{2})$, we find Corresponding Galois Group say $G_2$.

Then Galois Group of $\mathbb{Q}(\sqrt[p]{2},\zeta)$ would possibly be Product of these two subgroups $G_1$ and $G_2$ with some relation between the generators.

For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta))$, consider $\tau: \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow \mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\zeta$ and sending $\sqrt[p]{2} \rightarrow \sqrt[p]{2}\zeta$.

$\tau(\sqrt[p]{2})=\sqrt[p]{2}\zeta$,

$\tau^2(\sqrt[p]{2})=\tau(\tau(\sqrt[p]{2}))=\tau(\sqrt[p]{2}\zeta)=\tau(\sqrt[p]{2})\tau(\zeta)=\sqrt[p]{2}\zeta^2$,

For similar Reasons, $\tau^{p}(\sqrt[p]{2})=\sqrt[p]{2}\zeta^p=\sqrt[p]{2}$.

No power of $\tau$ less than $p$ gives identity as no power of $\zeta$ less than $p$ gives identity.

So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta)) \cong \mathbb{Z}_p \cong \big< \tau \big>$.

For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2}))$, consider $\sigma : \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow \mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\sqrt[p]{2}$ and sending $\zeta \rightarrow \zeta^2$

$\sigma(\zeta)=\zeta^2$

$\sigma^2(\zeta)=\sigma(\sigma(\zeta))=\sigma(\zeta^2)=\zeta^{(2^2)}$

For similar reasons, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}$, as for every $a\in \mathbb{F}_p$, we have $a^{p-1}=1$ we have in particular $2^{p-1} \equiv~1~mod~p$.

So, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}=\zeta$ (as $\zeta$ is a $p^{th}$ root of unity).

No power of $\sigma$ less than $p-1$ gives identity as $2\in \mathbb{F}_p$ generates Multiplicative group, no power of $2$ less than $p-1$ can be equal to $1~mod~p$.

So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2})) \cong \mathbb{Z}_{p-1} \cong \big< \sigma\big>$.

As $[\mathbb{Q}(\sqrt[p]{2},\zeta):\mathbb{Q}]=p(p-1)$ and $|\sigma|=p-1$ and $|\tau|=p$ i strongly feel Galois group should be possibly generated by $\sigma$ and $\tau$ with "Some extra related conditions"But not very sure to confirm this.

I am not able to go any further, I can see that $\sigma$ and $\tau$ do not commute with each other. I am unable to produce a know group which contain isomorphic copies of $\mathbb{Z}_{p-1}$ and $\mathbb{Z}_p$ as subgroups.

I would be thankful if some one can help me out in this case.

Thank You.

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Since $\sigma$ and $\tau$ don't commute, you could be looking for relations of the form $\sigma^k \tau^m \sigma^{-k} = \tau^l$ for some $m, k, l \in \mathbb{Z}$. My suggestion it to try with $m = k = 1$ and see what you get out of that. Also, do you recognize what kind of "product" this is? –  Andy Aug 9 '13 at 14:51
    
Maple produces by $$galois(x^3-2) $$ $$"3T2", \{"S(3)"\}, "-", 6, \{"(1 3)", "(2 3)"\} ,$$galois(x^5-2); $$"5T3", \{"5:4", "F(5)"\}, "-", 20, \{"(1 2 3 4 5)", "(1 2 4 3)"\} ,$$ and $$galois(x^7-2)$$ $$"7T4", \{"7:6", "F_42(7)"\}, "-", 42, \{"(1 2 3 4 5 6 7)", "(1 3 2 6 4 5)"\} .$$ –  user64494 Aug 9 '13 at 15:39
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2 Answers 2

up vote 1 down vote accepted

Here it might be best to realize the Galois group as a group of permutations of the roots of the polynomial of interest (as in user64494's comment under the OP). You have already observed that the roots of $x^p-2$ are $x_i=\zeta_p^{i-1}\root p\of 2, i=1,2,\ldots,p.$ You also know that the splitting field is of degree $p(p-1)$, so that is also the order of the Galois group.

Let us consider the action of your automorphism $\tau$ defined by $\tau(\zeta_p)=\zeta_p$ and $\tau(\root p\of 2)=\zeta_p\root p\of 2$. So we see that $\tau(x_i)=x_{i+1}$, if $i<p$, and $\tau(x_p)=x_1$. The action of $\tau$ on the chosen indexing of the roots thus corresponds to the $p$-cycle $\tau=(123\cdots p).$

On the other hand the automorphism $\sigma_a:x_1\mapsto x_1, \zeta_p\mapsto \zeta_p^a,1\le a<p,$ keeps the real root $x_1$ fixed, and permutes the others according to the rule $x_i=x_1\zeta_p^{i-1}\mapsto x_1\zeta_p^{a(i-1)}=x_{1+a(i-1)}$, where the subscript is calculated modulo $p$. You see that all these share $x_1$ as a fixed point (this was also clear from your construction of $\sigma$:s as elements of the Galois group $G_2=Gal(\mathbb{Q}(x_1,\zeta_p)/\mathbb{Q}(\zeta_p))$.

You can either look at all these automorphisms as elements of $S_p$. This works beautifully, once you have found a generator of $G_2$. This is equivalent to finding a generator of the multiplicative group $\mathbb{Z}_p^*$, i.e. a primitive root. There is no general formula for such a generator, so I won't say much about that (this may be a cause of your difficulties). We simply know that one exists! But if $p$ is fixed, say $p=5$ or another smallish prime, then I recommend this way, as you can easily calculate with permutations.

PVAL is strongly hinting at the possibility that you may get a semi-direct product of $G_1$ and $G_2$. Indeed, you will see that one of the two subgroups is stable under conjugation by elements of the other. Which way does it work? I'm a mean dude and won't tell you! But you do remember that the fixed field of a normal subgroup is itself Galois over the base field. So which of the fields $\mathbb{Q}(\zeta_p)$ or $\mathbb{Q}(\root p\of2)$ is Galois over the rationals? The group of automorphisms associated to that field should be a normal subgroup of the big Galois group. After figuring that out, you can start studying the effect of either $\sigma_a\tau\sigma_a^{-1}$ or $\tau\sigma_a\tau^{-1}$ all according to which feels more interesting...

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Yes, Yes, As $\mathbb{Q}(\zeta_p)$ is Galois, Corresponding Group $G_2$ is Normal subgroup, $G_1\cap G_2=\{e\}$ and $|G_1||G_2|=|G|$ So, we have $G=G_1\ltimes G_2$ –  Praphulla Koushik Aug 10 '13 at 4:55
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Hint: Clearly $\big<\sigma \big> \big< \tau \big>=G$ (count the order of each), and $\big<\sigma \big> \cap \big< \tau \big>=e$. If one of these is normal, then it's easy to write as a semi-direct product of groups.

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I could not understand why $\big<\sigma \big> \big< \tau \big>=G$ as i do not know what $G$ is at this point. only thing i know is $|G|$ and $|\big<\sigma \big>|| \big< \tau \big>|=|G|$, I agree with $\big<\sigma \big> \cap \big< \tau \big>=e$ –  Praphulla Koushik Aug 9 '13 at 16:48
    
@PraphullaKoushik If $H$, $K$ are finite subgroups of a group $G$, then $$|HK|=\frac{|H||K|}{|H\cap K|}$$ –  PVAL Aug 9 '13 at 16:56
    
@PraphullaKoushik So $|\big<\sigma \big> \big< \tau \big>|=|G|$, so they are equal. –  PVAL Aug 9 '13 at 17:03
    
:( I got it... I convinced myself saying the same even before posting the Question... I was not so sure of the way i did:) :) Thank YOu –  Praphulla Koushik Aug 10 '13 at 4:48
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