Mixing
Computing the Galois group of polynomials $x^n-a in mathbb{Q}[x]$
$begingroup$
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.
galois-theory
asked Sep 29 '12 at 20:25
AndyAndy
6024
$endgroup$
add a comment |
$begingroup$
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.
galois-theory
asked Sep 29 '12 at 20:25
AndyAndy
6024
$endgroup$
1
$begingroup$
Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38
add a comment |
$begingroup$
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.
galois-theory
asked Sep 29 '12 at 20:25
AndyAndy
6024
$endgroup$
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.
galois-theory
galois-theory
asked Sep 29 '12 at 20:25
AndyAndy
6024
asked Sep 29 '12 at 20:25
AndyAndy
6024
asked Sep 29 '12 at 20:25
AndyAndy
6024
asked Sep 29 '12 at 20:25
AndyAndy
6024
asked Sep 29 '12 at 20:25
AndyAndy
6024
6024
1
$begingroup$
Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38
add a comment |
1
$begingroup$
Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38
1
1
$begingroup$
Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38
$begingroup$
Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $mathbb Q[zeta_{2n}]$.
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
$begingroup$
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
$$
[F:mathbb{Q}]=[KL:mathbb{Q}]=frac{[K:mathbb{Q}][L:mathbb{Q}]}{[Kcap L:mathbb{Q}]}=frac{phi(n)n}{2^s},
$$
where $sge 0$ satisfies $2^smid phi(n)$ and $Kcap L=mathbb{Q}(sqrt[2^s]{a})$.
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 $pequiv 1 bmod 4$. 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 $Gcong mathbb{Z}/n rtimes (mathbb{Z}/n)^{times}$ if and only if $n$ is odd, or $n$ is even and $sqrt{a}notin mathbb{Q}(zeta_n)$.
Sepcial attention is needed for the case $n=2^k$; see also this MO-question.
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
$endgroup$
add a comment |
$begingroup$
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: $mathrm{Gal}(mathbb{Q}(zeta)|mathbb{Q})cong (mathbb{Z}/nmathbb{Z})^{*}$ whereas $|(mathbb{Z}/nmathbb{Z})^{*}|=varphi(n)$. 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 $mathrm{Aut}_{mathbb{Q}}(mathbb{Q}(sqrt[n]{a})) cong mathbb{Z}/nmathbb{Z}$. 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 $(mathbb{Z}/pmathbb{Z})^{*}=mathbb{Z}/(p-1)mathbb{Z}triangleleft mathrm{G}$ and since $|mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$
$$ mathrm{gcd}(|(mathbb{Z}/pmathbb{Z})^{*}|, [G:(mathbb{Z}/pmathbb{Z})^{*}]) =mathrm{gcd}(p-1,p)=1. $$
This yields a subgroup $mathrm{U} subseteq mathrm{G}$ such that $mathrm{G}cong (mathbb{Z}/pmathbb{Z})^{*}ltimes mathrm{U} $ and $|mathrm{U}|=p $ hence $mathrm{U}cong mathbb{Z}/pmathbb{Z}$ therefore
$$mathrm{Gal}(X^p-a) cong mathbb{Z}/(p-1)mathbb{Z} ltimes mathbb{Z}/pmathbb{Z}. $$
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
$endgroup$
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
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– nguyen quang do
Nov 20 '17 at 13:15
add a comment |
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3 Answers
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3 Answers
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$begingroup$
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 $mathbb Q[zeta_{2n}]$.
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
$begingroup$
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 $mathbb Q[zeta_{2n}]$.
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
add a comment |
$begingroup$
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 $mathbb Q[zeta_{2n}]$.
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
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 $mathbb Q[zeta_{2n}]$.
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
answered Sep 29 '12 at 20:47
Hagen von EitzenHagen von Eitzen
277k22269496
277k22269496
add a comment |
add a comment |
$begingroup$
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
$$
[F:mathbb{Q}]=[KL:mathbb{Q}]=frac{[K:mathbb{Q}][L:mathbb{Q}]}{[Kcap L:mathbb{Q}]}=frac{phi(n)n}{2^s},
$$
where $sge 0$ satisfies $2^smid phi(n)$ and $Kcap L=mathbb{Q}(sqrt[2^s]{a})$.
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 $pequiv 1 bmod 4$. 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 $Gcong mathbb{Z}/n rtimes (mathbb{Z}/n)^{times}$ if and only if $n$ is odd, or $n$ is even and $sqrt{a}notin mathbb{Q}(zeta_n)$.
Sepcial attention is needed for the case $n=2^k$; see also this MO-question.
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
$endgroup$
add a comment |
$begingroup$
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
$$
[F:mathbb{Q}]=[KL:mathbb{Q}]=frac{[K:mathbb{Q}][L:mathbb{Q}]}{[Kcap L:mathbb{Q}]}=frac{phi(n)n}{2^s},
$$
where $sge 0$ satisfies $2^smid phi(n)$ and $Kcap L=mathbb{Q}(sqrt[2^s]{a})$.
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 $pequiv 1 bmod 4$. 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 $Gcong mathbb{Z}/n rtimes (mathbb{Z}/n)^{times}$ if and only if $n$ is odd, or $n$ is even and $sqrt{a}notin mathbb{Q}(zeta_n)$.
Sepcial attention is needed for the case $n=2^k$; see also this MO-question.
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
$endgroup$
add a comment |
$begingroup$
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
$$
[F:mathbb{Q}]=[KL:mathbb{Q}]=frac{[K:mathbb{Q}][L:mathbb{Q}]}{[Kcap L:mathbb{Q}]}=frac{phi(n)n}{2^s},
$$
where $sge 0$ satisfies $2^smid phi(n)$ and $Kcap L=mathbb{Q}(sqrt[2^s]{a})$.
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 $pequiv 1 bmod 4$. 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 $Gcong mathbb{Z}/n rtimes (mathbb{Z}/n)^{times}$ if and only if $n$ is odd, or $n$ is even and $sqrt{a}notin mathbb{Q}(zeta_n)$.
Sepcial attention is needed for the case $n=2^k$; see also this MO-question.
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
$endgroup$
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
$$
[F:mathbb{Q}]=[KL:mathbb{Q}]=frac{[K:mathbb{Q}][L:mathbb{Q}]}{[Kcap L:mathbb{Q}]}=frac{phi(n)n}{2^s},
$$
where $sge 0$ satisfies $2^smid phi(n)$ and $Kcap L=mathbb{Q}(sqrt[2^s]{a})$.
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 $pequiv 1 bmod 4$. 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 $Gcong mathbb{Z}/n rtimes (mathbb{Z}/n)^{times}$ if and only if $n$ is odd, or $n$ is even and $sqrt{a}notin mathbb{Q}(zeta_n)$.
Sepcial attention is needed for the case $n=2^k$; see also this MO-question.
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
edited Jan 3 at 13:06
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
answered Dec 23 '16 at 14:56
Dietrich BurdeDietrich Burde
78.3k64386
78.3k64386
add a comment |
add a comment |
$begingroup$
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: $mathrm{Gal}(mathbb{Q}(zeta)|mathbb{Q})cong (mathbb{Z}/nmathbb{Z})^{*}$ whereas $|(mathbb{Z}/nmathbb{Z})^{*}|=varphi(n)$. 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 $mathrm{Aut}_{mathbb{Q}}(mathbb{Q}(sqrt[n]{a})) cong mathbb{Z}/nmathbb{Z}$. 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 $(mathbb{Z}/pmathbb{Z})^{*}=mathbb{Z}/(p-1)mathbb{Z}triangleleft mathrm{G}$ and since $|mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$
$$ mathrm{gcd}(|(mathbb{Z}/pmathbb{Z})^{*}|, [G:(mathbb{Z}/pmathbb{Z})^{*}]) =mathrm{gcd}(p-1,p)=1. $$
This yields a subgroup $mathrm{U} subseteq mathrm{G}$ such that $mathrm{G}cong (mathbb{Z}/pmathbb{Z})^{*}ltimes mathrm{U} $ and $|mathrm{U}|=p $ hence $mathrm{U}cong mathbb{Z}/pmathbb{Z}$ therefore
$$mathrm{Gal}(X^p-a) cong mathbb{Z}/(p-1)mathbb{Z} ltimes mathbb{Z}/pmathbb{Z}. $$
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
$endgroup$
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
add a comment |
$begingroup$
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: $mathrm{Gal}(mathbb{Q}(zeta)|mathbb{Q})cong (mathbb{Z}/nmathbb{Z})^{*}$ whereas $|(mathbb{Z}/nmathbb{Z})^{*}|=varphi(n)$. 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 $mathrm{Aut}_{mathbb{Q}}(mathbb{Q}(sqrt[n]{a})) cong mathbb{Z}/nmathbb{Z}$. 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 $(mathbb{Z}/pmathbb{Z})^{*}=mathbb{Z}/(p-1)mathbb{Z}triangleleft mathrm{G}$ and since $|mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$
$$ mathrm{gcd}(|(mathbb{Z}/pmathbb{Z})^{*}|, [G:(mathbb{Z}/pmathbb{Z})^{*}]) =mathrm{gcd}(p-1,p)=1. $$
This yields a subgroup $mathrm{U} subseteq mathrm{G}$ such that $mathrm{G}cong (mathbb{Z}/pmathbb{Z})^{*}ltimes mathrm{U} $ and $|mathrm{U}|=p $ hence $mathrm{U}cong mathbb{Z}/pmathbb{Z}$ therefore
$$mathrm{Gal}(X^p-a) cong mathbb{Z}/(p-1)mathbb{Z} ltimes mathbb{Z}/pmathbb{Z}. $$
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
$endgroup$
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
add a comment |
$begingroup$
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: $mathrm{Gal}(mathbb{Q}(zeta)|mathbb{Q})cong (mathbb{Z}/nmathbb{Z})^{*}$ whereas $|(mathbb{Z}/nmathbb{Z})^{*}|=varphi(n)$. 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 $mathrm{Aut}_{mathbb{Q}}(mathbb{Q}(sqrt[n]{a})) cong mathbb{Z}/nmathbb{Z}$. 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 $(mathbb{Z}/pmathbb{Z})^{*}=mathbb{Z}/(p-1)mathbb{Z}triangleleft mathrm{G}$ and since $|mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$
$$ mathrm{gcd}(|(mathbb{Z}/pmathbb{Z})^{*}|, [G:(mathbb{Z}/pmathbb{Z})^{*}]) =mathrm{gcd}(p-1,p)=1. $$
This yields a subgroup $mathrm{U} subseteq mathrm{G}$ such that $mathrm{G}cong (mathbb{Z}/pmathbb{Z})^{*}ltimes mathrm{U} $ and $|mathrm{U}|=p $ hence $mathrm{U}cong mathbb{Z}/pmathbb{Z}$ therefore
$$mathrm{Gal}(X^p-a) cong mathbb{Z}/(p-1)mathbb{Z} ltimes mathbb{Z}/pmathbb{Z}. $$
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
$endgroup$
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: $mathrm{Gal}(mathbb{Q}(zeta)|mathbb{Q})cong (mathbb{Z}/nmathbb{Z})^{*}$ whereas $|(mathbb{Z}/nmathbb{Z})^{*}|=varphi(n)$. 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 $mathrm{Aut}_{mathbb{Q}}(mathbb{Q}(sqrt[n]{a})) cong mathbb{Z}/nmathbb{Z}$. 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 $(mathbb{Z}/pmathbb{Z})^{*}=mathbb{Z}/(p-1)mathbb{Z}triangleleft mathrm{G}$ and since $|mathrm{G}|=p(p-1)$ we can apply Schur-Zassenhaus because for all primes $p$
$$ mathrm{gcd}(|(mathbb{Z}/pmathbb{Z})^{*}|, [G:(mathbb{Z}/pmathbb{Z})^{*}]) =mathrm{gcd}(p-1,p)=1. $$
This yields a subgroup $mathrm{U} subseteq mathrm{G}$ such that $mathrm{G}cong (mathbb{Z}/pmathbb{Z})^{*}ltimes mathrm{U} $ and $|mathrm{U}|=p $ hence $mathrm{U}cong mathbb{Z}/pmathbb{Z}$ therefore
$$mathrm{Gal}(X^p-a) cong mathbb{Z}/(p-1)mathbb{Z} ltimes mathbb{Z}/pmathbb{Z}. $$
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
edited Feb 8 '15 at 15:02
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
answered Feb 7 '15 at 18:41
einfachEriceinfachEric
193
193
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
add a comment |
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
1
1
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
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 $pequiv 1bmod 4$, and this holds for $p=5$.
$endgroup$
– Dietrich Burde
Dec 23 '16 at 14:37
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
$begingroup$
See also math.stackexchange.com/a/2529029/300700
$endgroup$
– nguyen quang do
Nov 20 '17 at 13:15
add a comment |
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Do you mean the polynmial $x^n-a in mathbb{Q}[x]$? If not then your "polynomial" is a fixed number.
$endgroup$
– Fly by Night
Sep 29 '12 at 20:38