Mixing
Why we can represent automorphisms in $text{Gal}(Bbb Q(sqrt[4]{2},i)/Bbb Q)$ as permutations in $S_4$ but not...
The splitting field of $x^4-2$ over $Bbb Q$ is $G=text{Gal}(Bbb Q(sqrt[4]{2},i)/Bbb Q)$. By primitive element theorem, $K=Bbb Q(alpha)$ for some $alpha$ and $[K:Bbb Q]=8$. So I know that the minimal polynomial of $alpha$ over $Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $sigma,~sigma'in G$ such that $forall rin{sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i},~sigma(r)=sigma'(r)$.
Picture:
abstract-algebra galois-theory automorphism-group
asked Dec 31 '18 at 14:27
SharaShara
725
|
show 19 more comments
The splitting field of $x^4-2$ over $Bbb Q$ is $G=text{Gal}(Bbb Q(sqrt[4]{2},i)/Bbb Q)$. By primitive element theorem, $K=Bbb Q(alpha)$ for some $alpha$ and $[K:Bbb Q]=8$. So I know that the minimal polynomial of $alpha$ over $Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $sigma,~sigma'in G$ such that $forall rin{sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i},~sigma(r)=sigma'(r)$.
Picture:
abstract-algebra galois-theory automorphism-group
asked Dec 31 '18 at 14:27
SharaShara
725
2
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
2
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
1
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
2
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
2
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30
|
show 19 more comments
The splitting field of $x^4-2$ over $Bbb Q$ is $G=text{Gal}(Bbb Q(sqrt[4]{2},i)/Bbb Q)$. By primitive element theorem, $K=Bbb Q(alpha)$ for some $alpha$ and $[K:Bbb Q]=8$. So I know that the minimal polynomial of $alpha$ over $Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $sigma,~sigma'in G$ such that $forall rin{sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i},~sigma(r)=sigma'(r)$.
Picture:
abstract-algebra galois-theory automorphism-group
asked Dec 31 '18 at 14:27
SharaShara
725
The splitting field of $x^4-2$ over $Bbb Q$ is $G=text{Gal}(Bbb Q(sqrt[4]{2},i)/Bbb Q)$. By primitive element theorem, $K=Bbb Q(alpha)$ for some $alpha$ and $[K:Bbb Q]=8$. So I know that the minimal polynomial of $alpha$ over $Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $sigma,~sigma'in G$ such that $forall rin{sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i},~sigma(r)=sigma'(r)$.
Picture:
abstract-algebra galois-theory automorphism-group
abstract-algebra galois-theory automorphism-group
asked Dec 31 '18 at 14:27
SharaShara
725
asked Dec 31 '18 at 14:27
SharaShara
725
edited Dec 31 '18 at 14:50
Shara
asked Dec 31 '18 at 14:27
SharaShara
725
asked Dec 31 '18 at 14:27
SharaShara
725
asked Dec 31 '18 at 14:27
SharaShara
725
725
2
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
2
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
1
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
2
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
2
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30
|
show 19 more comments
2
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
2
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
1
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
2
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
2
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30
2
2
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
2
2
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
1
1
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
2
2
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
2
2
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30
|
show 19 more comments
1 Answer
1
active
oldest
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Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $ngeq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $sqrt[4]{2}$, $-sqrt[4]{2}$, $isqrt[4]{2}$, $-isqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $mathbb{Q}left(sqrt[4]{2},iright)supsetmathbb{Q}left(sqrt[4]{2}right)supset mathbb{Q}$, since $mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $sigma$ in $mbox{Gal}left(mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)right)$ such that send $imapsto -i$ (acts on the roots of $x^2+1$) and leave $mathbb{Q}left(sqrt[4]{2}right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/mathbb{Q}$, so $sigma$ lie to $mbox{Gal}left(K/mathbb{Q}right)$
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
add a comment |
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Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $ngeq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $sqrt[4]{2}$, $-sqrt[4]{2}$, $isqrt[4]{2}$, $-isqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $mathbb{Q}left(sqrt[4]{2},iright)supsetmathbb{Q}left(sqrt[4]{2}right)supset mathbb{Q}$, since $mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $sigma$ in $mbox{Gal}left(mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)right)$ such that send $imapsto -i$ (acts on the roots of $x^2+1$) and leave $mathbb{Q}left(sqrt[4]{2}right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/mathbb{Q}$, so $sigma$ lie to $mbox{Gal}left(K/mathbb{Q}right)$
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
add a comment |
Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $ngeq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $sqrt[4]{2}$, $-sqrt[4]{2}$, $isqrt[4]{2}$, $-isqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $mathbb{Q}left(sqrt[4]{2},iright)supsetmathbb{Q}left(sqrt[4]{2}right)supset mathbb{Q}$, since $mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $sigma$ in $mbox{Gal}left(mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)right)$ such that send $imapsto -i$ (acts on the roots of $x^2+1$) and leave $mathbb{Q}left(sqrt[4]{2}right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/mathbb{Q}$, so $sigma$ lie to $mbox{Gal}left(K/mathbb{Q}right)$
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
add a comment |
Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $ngeq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $sqrt[4]{2}$, $-sqrt[4]{2}$, $isqrt[4]{2}$, $-isqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $mathbb{Q}left(sqrt[4]{2},iright)supsetmathbb{Q}left(sqrt[4]{2}right)supset mathbb{Q}$, since $mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $sigma$ in $mbox{Gal}left(mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)right)$ such that send $imapsto -i$ (acts on the roots of $x^2+1$) and leave $mathbb{Q}left(sqrt[4]{2}right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/mathbb{Q}$, so $sigma$ lie to $mbox{Gal}left(K/mathbb{Q}right)$
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $ngeq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $sqrt[4]{2}$, $-sqrt[4]{2}$, $isqrt[4]{2}$, $-isqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $mathbb{Q}left(sqrt[4]{2},iright)supsetmathbb{Q}left(sqrt[4]{2}right)supset mathbb{Q}$, since $mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $sigma$ in $mbox{Gal}left(mathbb{Q}left(sqrt[4]{2},iright)/mathbb{Q}left(sqrt[4]{2}right)right)$ such that send $imapsto -i$ (acts on the roots of $x^2+1$) and leave $mathbb{Q}left(sqrt[4]{2}right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/mathbb{Q}$, so $sigma$ lie to $mbox{Gal}left(K/mathbb{Q}right)$
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
edited Jan 1 at 2:13
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
answered Jan 1 at 2:00
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
807110
807110
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
add a comment |
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
I did an incomplete answer at the end, because I thought that is was enough for you. But, if you need more help I can add more detailed information of comment here
– José Alejandro Aburto Araneda
Jan 1 at 2:17
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
But generally the analysis is always done first in the group-world, and then pull back to see its related fixed field right?
– Shara
Jan 1 at 2:52
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
Hm, depends of the facts you know about the group or the extension. I will give you two examples. First, if the extension is given by a splitting field of a polynomial of degree $n$, and the extension have degree $n!$, then the galois group is the symmetric group of order $n!$ (the galois group order equals the order of the galois extension). Second, the extension obtained from a cyclotomic polynomial, for instance, the splitting field of $x^p-1$ where $p$ is a prime number, you will get easily that the if we consider $zeta$ one generator (a finite subset of a field is cylic)
– José Alejandro Aburto Araneda
Jan 1 at 8:53
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
...of the groups of roots of that polynomials, then an element of the galois group and their powers acting on $zeta$ will leave to that the galois group embedds on the set of units of $mathbb{Z}/pmathbb{Z}$ (the argument works for any positive integer in fact)
– José Alejandro Aburto Araneda
Jan 1 at 8:56
add a comment |
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2
$K$ is the splitting field of $X^4-2$.
– Lord Shark the Unknown
Dec 31 '18 at 14:29
2
@mouthetics Wait, $[Bbb Q(sqrt[4]{2},i):Bbb Q]=8$, how can ${sqrt[4]{2},~sqrt[4]{2}i,~-sqrt[4]{2},~-sqrt[4]{2}i}$ which consists of only four elements be a basis of it?
– Shara
Dec 31 '18 at 15:33
1
The four roots do not form a vector space basis over Q, such a basis has eight elements, the simplest being the four powers of fourth root of 2 together with each of those times $i$.
– Ned
Dec 31 '18 at 15:37
2
The fact that a homomorphism is uniquely determined once we know how it maps a set of generators is everywhere in algebra. Students probably first meet this principle in linear algebra in the form: a linear transformation is fully described when we know how it maps the elements of a basis. Unlike in linear algebra in other structures we cannot choose the images of the generators any which way we please, but that's besides the point.
– Jyrki Lahtonen
Dec 31 '18 at 16:27
2
(cont'd) The four zeros generate the splitting field as an extension field of $Bbb{Q}$, and therefore knowing their images determines an automorphism uniquely if such an automorphism exists. The latter is a separate question. Anyway, nothing stops you from finding a single generating element for this field extension, locating its eight conjugates, and rewriting the Galois group as a group of permutations of that octet.
– Jyrki Lahtonen
Dec 31 '18 at 16:30