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Can we finitely extend a field with Galois group $S_n$ to a a field with a given permutation group
Clash Royale CLAN TAG#URR8PPP
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$newcommandQmathbb Q$Suppose I have a finite Galois field extension $K/Q$ with Galois group $S_n$ (originally I thought of any normal subgroup of $S_n$ for minimal $n$ but this might be too ambitious) and suppose that $mgeq n$ and I have a transitive subgroup $G$ of $S_m$ (acting on $m$ elements) and we have the following containment $S_nunlhd G$. Can I now (constructively) find a Galois field extension $L/K$ such that $mathrmGal(L/Q) = G$? By "constructively", I mean I want to adjoin $Q$ with finite elements that I can also determine by some construction (roots of some polynomials). In fact, I prefer to obtain an irreducible polynomial over $Q$ with degree $m$ such that $L$ is the splitting field of this polynomial.
If this is not possible for a general $G$, for what $G$ do we know this to be possible (maybe taking a group that is the semidirect product of say $C_2^k$ with $S_n$ or something similar?)
Edit : I thought I add the following condition because the above could still be hard to solve. In my case anyway, I am only dealing with $G$ solvable and I actually know that there is a number field Galois over $Q$ and with Galois group $G$.
galois-theory algebraic-number-theory symmetric-groups
asked Jul 24 at 12:44
quantum
40429
add a comment |Â
up vote
1
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$newcommandQmathbb Q$Suppose I have a finite Galois field extension $K/Q$ with Galois group $S_n$ (originally I thought of any normal subgroup of $S_n$ for minimal $n$ but this might be too ambitious) and suppose that $mgeq n$ and I have a transitive subgroup $G$ of $S_m$ (acting on $m$ elements) and we have the following containment $S_nunlhd G$. Can I now (constructively) find a Galois field extension $L/K$ such that $mathrmGal(L/Q) = G$? By "constructively", I mean I want to adjoin $Q$ with finite elements that I can also determine by some construction (roots of some polynomials). In fact, I prefer to obtain an irreducible polynomial over $Q$ with degree $m$ such that $L$ is the splitting field of this polynomial.
If this is not possible for a general $G$, for what $G$ do we know this to be possible (maybe taking a group that is the semidirect product of say $C_2^k$ with $S_n$ or something similar?)
Edit : I thought I add the following condition because the above could still be hard to solve. In my case anyway, I am only dealing with $G$ solvable and I actually know that there is a number field Galois over $Q$ and with Galois group $G$.
galois-theory algebraic-number-theory symmetric-groups
asked Jul 24 at 12:44
quantum
40429
1
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$newcommandQmathbb Q$Suppose I have a finite Galois field extension $K/Q$ with Galois group $S_n$ (originally I thought of any normal subgroup of $S_n$ for minimal $n$ but this might be too ambitious) and suppose that $mgeq n$ and I have a transitive subgroup $G$ of $S_m$ (acting on $m$ elements) and we have the following containment $S_nunlhd G$. Can I now (constructively) find a Galois field extension $L/K$ such that $mathrmGal(L/Q) = G$? By "constructively", I mean I want to adjoin $Q$ with finite elements that I can also determine by some construction (roots of some polynomials). In fact, I prefer to obtain an irreducible polynomial over $Q$ with degree $m$ such that $L$ is the splitting field of this polynomial.
If this is not possible for a general $G$, for what $G$ do we know this to be possible (maybe taking a group that is the semidirect product of say $C_2^k$ with $S_n$ or something similar?)
Edit : I thought I add the following condition because the above could still be hard to solve. In my case anyway, I am only dealing with $G$ solvable and I actually know that there is a number field Galois over $Q$ and with Galois group $G$.
galois-theory algebraic-number-theory symmetric-groups
asked Jul 24 at 12:44
quantum
40429
$newcommandQmathbb Q$Suppose I have a finite Galois field extension $K/Q$ with Galois group $S_n$ (originally I thought of any normal subgroup of $S_n$ for minimal $n$ but this might be too ambitious) and suppose that $mgeq n$ and I have a transitive subgroup $G$ of $S_m$ (acting on $m$ elements) and we have the following containment $S_nunlhd G$. Can I now (constructively) find a Galois field extension $L/K$ such that $mathrmGal(L/Q) = G$? By "constructively", I mean I want to adjoin $Q$ with finite elements that I can also determine by some construction (roots of some polynomials). In fact, I prefer to obtain an irreducible polynomial over $Q$ with degree $m$ such that $L$ is the splitting field of this polynomial.
If this is not possible for a general $G$, for what $G$ do we know this to be possible (maybe taking a group that is the semidirect product of say $C_2^k$ with $S_n$ or something similar?)
Edit : I thought I add the following condition because the above could still be hard to solve. In my case anyway, I am only dealing with $G$ solvable and I actually know that there is a number field Galois over $Q$ and with Galois group $G$.
galois-theory algebraic-number-theory symmetric-groups
asked Jul 24 at 12:44
quantum
40429
edited Jul 24 at 14:25
asked Jul 24 at 12:44
quantum
40429
asked Jul 24 at 12:44
quantum
40429
asked Jul 24 at 12:44
quantum
40429
40429
1
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24
add a comment |Â
1
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24
1
1
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24
add a comment |Â
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1
This is open in general -- if $G$ is a finite group, we do not know that there is a finite extension $L/mathbb Q$ with Galois group $L$ (this is the Inverse Galois Problem). So we certainly can't construct one.
– Mathmo123
Jul 24 at 12:50
There might be some hope if $G/S_n$ is solvable, as extensions can be constructed by adding radicals.
– Mathmo123
Jul 24 at 13:19
For my particular case I know that $G$ is a galois group of a field, but I do not know if it can come from the Galois group of the field extension of $K$. I could add that, does that make a significant difference to the difficulty of the problem? Actually I even know $G$ is solvable. Could you provide your reason (maybe in an answer post) why there is some "hope" here?
– quantum
Jul 24 at 13:21
The hope is that you can reduce the problem to the case of constructing cyclic extensions. Although even then, it's not immediately clear to me how to proceed, since a Galois extension of a Galois extension need not be Galois. For example, I know how to construct an extension $L/K$ with Galois group $C_2$, but I don't know how to ensure that this field has Galois group, say, $S_nrtimes C_2$ over $mathbb Q$ rather than $S_ntimes C_2$.
– Mathmo123
Jul 24 at 13:56
ah ok. Thanks! I'm editing the post and adding I assume $G$ is solvable and I know that $G$ is Galois-groupable..
– quantum
Jul 24 at 14:24