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

For finding the 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 $\sqrt[p]{2} \rightarrow \sqrt[p]{2}\zeta$.



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 $\zeta \rightarrow \zeta^2$



For similar reasons, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}$, as for every $a\in \mathbb{F}_p^\times$, 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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – Andy
    Aug 9, 2013 at 14:51
  • $\begingroup$ 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 $$galois(x^7-2)$$ $$"7T4", \{"7:6", "F_42(7)"\}, "-", 42, \{"(1 2 3 4 5 6 7)", "(1 3 2 6 4 5)"\} .$$ $\endgroup$
    – user64494
    Aug 9, 2013 at 15:39

3 Answers 3


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...

  • 2
    $\begingroup$ 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 $G=G_1\ltimes G_2$ $\endgroup$
    – user87543
    Aug 10, 2013 at 4:55
  • 1
    $\begingroup$ I get only $17$ automorphism $\sigma$ and $\tau$: $Id,\sigma,\sigma^{2},\sigma^3,\sigma^4,\sigma\circ\tau,\sigma^2\circ\tau,\sigma^3\circ\tau,\sigma^4\circ\tau,\sigma\circ\tau^2,\sigma^2\circ\tau^2,\sigma^3\circ\tau^2,\sigma^4\circ\tau^2,\sigma\circ\tau^3,\sigma^2\circ\tau^3,\sigma^3\circ\tau^3,\sigma^4\circ\tau^3$ So I am not sure how to find the three others, because $\tau^{j}\circ\sigma\circ\tau^{-1}(\alpha=\sigma^{2^j}(\alpha)$ and $\tau^{j}\circ\sigma\circ\tau^{-1}(\zeta)=\sigma^{2^j}(\zeta).$ $\endgroup$
    – JeSuis
    Nov 26, 2016 at 17:02
  • 1
    $\begingroup$ @JeSuis Presumably you are asking about the case $p=5$? All the 20 automorphisms are $\sigma^i\circ\tau^j$ with $i=0,1,2,3,4$, $j=0,1,2,3$. $\endgroup$ Nov 26, 2016 at 18:46
  • 1
    $\begingroup$ @JyrkiLahtonen arf I forgot $\tau,\tau^2,\tau^3$. Thanks! $\endgroup$
    – JeSuis
    Nov 26, 2016 at 18:51

I thought I'd add a realization of this Galois group that can be analysed easily with direct computation, since it turns out that the Galois group is actually quite easy to describe.

Let $G = Gal(X^p-2)$ (or indeed $Gal(X^p-q)$ for any prime $q$). Then, as observed, the splitting field is $L = \mathbb{Q}(\zeta,\sqrt[p]{2})$. So any element $\varphi \in G$ is determined by $\varphi(\zeta)$ and $\varphi(\sqrt[p]{2})$. But $\varphi(\zeta) = \zeta^a$ for some $1 \leq i \leq p-1$ and $\varphi(\sqrt[p]{2}) = \zeta^b \sqrt[p]{2}$ for some $0 \leq b \leq p-1$ We thus get a map:

$$\chi:G \rightarrow G' :=\bigg\{\begin{pmatrix} a & b \\ 0 & 1\\ \end{pmatrix} : a \in \mathbb{F}_p^{\times}, b \in \mathbb{F}_p\bigg\} \subset \mathrm{GL}(2, \mathbb{F}_p) $$

$$\chi(\varphi)= \begin{pmatrix} a & b \\ 0 & 1\\ \end{pmatrix}$$

This is quickly seen to be both bijective and a homomorphism, so an isomorphism, explicitly identifying $G$ with the easy to understand group $G'$.

The nicest thing about this isomorphism, I think, is that it is extremly explicit. You can see the subgroups generated by $\sigma$ and $\tau$, you can easily tell which subgroups are normal and you can quickly compute the composition of given elements. This also gives you for free that the size of the extension is $p(p-1)$, although that was actually easy to compute.


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.

  • $\begingroup$ 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 $\big<\sigma \big> \cap \big< \tau \big>=e$ $\endgroup$
    – user87543
    Aug 9, 2013 at 16:48
  • $\begingroup$ @PraphullaKoushik If $H$, $K$ are finite subgroups of a group $G$, then $$|HK|=\frac{|H||K|}{|H\cap K|}$$ $\endgroup$ Aug 9, 2013 at 16:56
  • $\begingroup$ @PraphullaKoushik So $|\big<\sigma \big> \big< \tau \big>|=|G|$, so they are equal. $\endgroup$ Aug 9, 2013 at 17:03
  • $\begingroup$ :( 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 $\endgroup$
    – user87543
    Aug 10, 2013 at 4:48

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