I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.

The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).

For example one easy case, it's when $a>0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3} = - \root 3 \of 3 $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a<0$ , for example $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?

$\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple.

  • 1
    Do you mean the polynmial $x^n-a \in \mathbb{Q}[x]$? If not then your "polynomial" is a fixed number. – Fly by Night Sep 29 '12 at 20:38
up vote 1 down vote accepted

The minimal polynomial of $\zeta_n$ and $\sqrt[n]a$ are almost always different. In fact, every root in $\mathbb C$ of the $n$th cyclotomic polynomial has absolute value $1$, whereas $\sqrt[n]a$ usually has not. The only exceptions are $a=1$ and $a=-1$. For $a=1$, the splitting field is simply $\mathbb Q[\zeta_n]$. For $a=-1$, a solution to $x^n+1=0$ is also a solution to $x^{2n}-1=0$, hence the splitting field is simple $\mathbb Q[\zeta_{2n}]$.

Let $K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{a})$. Then the splitting field of $X^n-a$ is given by $F=KL$ and has degree $$ [L:\mathbb{Q}]=[KL:\mathbb{Q}]=\frac{[K:\mathbb{Q}][L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}=\frac{\phi(n)n}{2^s}, $$ where $s\ge 0$ satisfies $2^s\mid \phi(n)$ and $K\cap L=\mathbb{Q}(\sqrt[2^s]{a})$. So the point is that we need to determine the intersection $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a})$. This intersection is often not just $\mathbb{Q}$; note for example that $\sqrt{p}\in \mathbb{Q}(\zeta_p)$ for all primes $p\equiv 1 \bmod 4$. A complete answer for this (and the whole question) has been published by Jacobson and Velez in "The Galois group of a radical extension of the rationals", in 1990. Let $G$ be the Galois group of $X^n-a$. It is obvious that $G$ embedds naturally into the semidirect product of the Galois group $(\mathbb{Z}/n)^{\times}$ of $X^n-1$ and the automorphism group $Aut_{\mathbb{Q}}(\mathbb{Q}(\sqrt[n]{a})\cong \mathbb{Z}/n$.

Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

Sepcial attention is needed for the case $n=2^k$; see also this MO-question.

A few remarks on this specific galois group: Consider the field extension $\mathbb{Q}(\zeta)|\mathbb{Q}$ and $\zeta$ a primitive root of $X^{n} - 1 \in \mathbb{Q}[X]$. A general fact of cyclotomic field extensions is that in this case: $\mathrm{Gal}(\mathbb{Q}(\zeta)|\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^{*}$ whereas $|(\mathbb{Z}/n\mathbb{Z})^{*}|=\varphi(n)$. Let $a \in \mathbb{Q}$ such that $\sqrt[n]{a} \notin \mathbb{Q} $ then the splitting field of $X^n - a$ is as mentioned above given by $$ \mathrm{SF}(X^n - a)=\mathbb{Q}(\zeta, \sqrt[n]{a}).$$ Furthermore the automorphism group of $ \mathbb{Q}(\sqrt[n]{a})$ is cyclic of order $[\mathbb{Q}(\sqrt[n]{a}):\mathbb{Q}]=n$ hence $\mathrm{Aut}_{\mathbb{Q}}(\mathbb{Q}(\sqrt[n]{a})) \cong \mathbb{Z}/n\mathbb{Z}$. Note that both groups embed into the full galois group and moreover $\mathbb{Q}(\sqrt[n]{a}) \cap \mathbb{Q}(\zeta)= \mathbb{Q} $. One could then suggest (following theorems of galois theory) that the full galois groups is given by their product but this fails if we consider $\mathrm{Gal}(X^4 - 5) \cong \mathrm{D}_{4} $ which is not a direct product of cyclic groups. However if $ n=p $ is prime then $(\mathbb{Z}/p\mathbb{Z})^{*}=\mathbb{Z}/(p-1)\mathbb{Z}\triangleleft \mathrm{G}$ and since $|\mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$ $$ \mathrm{gcd}(|(\mathbb{Z}/p\mathbb{Z})^{*}|, [G:(\mathbb{Z}/p\mathbb{Z})^{*}]) =\mathrm{gcd}(p-1,p)=1. $$ This yields a subgroup $\mathrm{U} \subseteq \mathrm{G}$ such that $\mathrm{G}\cong (\mathbb{Z}/p\mathbb{Z})^{*}\ltimes \mathrm{U} $ and $|\mathrm{U}|=p $ hence $\mathrm{U}\cong \mathbb{Z}/p\mathbb{Z}$ therefore $$\mathrm{Gal}(X^p-a) \cong \mathbb{Z}/(p-1)\mathbb{Z} \ltimes \mathbb{Z}/p\mathbb{Z}. $$

  • 1
    The claim that "moreover $\mathbb{Q}(\sqrt[n]{a}) \cap \mathbb{Q}(\zeta)= \mathbb{Q} $" is *not* true in general. Take $X^{10}-5$, so $n=10$ and $a=5$. Then the intersection is $\mathbb{Q}(\sqrt{5})$, because $\sqrt{p}\in \mathbb{Q}(\zeta_p)$ for all primes $p\equiv 1\bmod 4$, and this holds for $p=5$. – Dietrich Burde Dec 23 '16 at 14:37
  • See also math.stackexchange.com/a/2529029/300700 – nguyen quang do Nov 20 '17 at 13:15

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.