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Elements of $Gal(K(alpha_1, cdots, alpha_n): K)$ which send a specific subset of $alpha_1, cdots, alpha_n$ to another specific subset

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0

$begingroup$

Suppose $L:K$ is a splitting field extension for a monic, separable polynomial $f in K[t]$ which is irreducible over $K$. Let $alpha_1, cdots, alpha_n$ be the distinct roots of $f$ in $L$. Let $k < n/2$ and choose a subset $beta_1, cdots, beta_k, delta_1, cdots, delta_k subset alpha_1, cdots, alpha_n$ of the roots (all the elements are distinct.)

I know that the maps in $Gal(L:K)$ permute the roots $alpha_1, cdots, alpha_n$. Can I always find an element $tau in Gal(L:K)$ such that $tau(beta_1) = delta_1, cdots, tau(beta_k) = delta_k$?

If not, how do I know which subsets of $alpha_1, cdots, alpha_n$ can be mapped to which other subsets? (Is there a way of knowing this, in general?)

abstract-algebra galois-theory

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asked Mar 10 at 20:59

38917803891780

856

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0

$begingroup$

Suppose $L:K$ is a splitting field extension for a monic, separable polynomial $f in K[t]$ which is irreducible over $K$. Let $alpha_1, cdots, alpha_n$ be the distinct roots of $f$ in $L$. Let $k < n/2$ and choose a subset $beta_1, cdots, beta_k, delta_1, cdots, delta_k subset alpha_1, cdots, alpha_n$ of the roots (all the elements are distinct.)

I know that the maps in $Gal(L:K)$ permute the roots $alpha_1, cdots, alpha_n$. Can I always find an element $tau in Gal(L:K)$ such that $tau(beta_1) = delta_1, cdots, tau(beta_k) = delta_k$?

If not, how do I know which subsets of $alpha_1, cdots, alpha_n$ can be mapped to which other subsets? (Is there a way of knowing this, in general?)

abstract-algebra galois-theory

share|cite|improve this question

asked Mar 10 at 20:59

38917803891780

856

$endgroup$

add a comment | 

$begingroup$

Suppose $L:K$ is a splitting field extension for a monic, separable polynomial $f in K[t]$ which is irreducible over $K$. Let $alpha_1, cdots, alpha_n$ be the distinct roots of $f$ in $L$. Let $k < n/2$ and choose a subset $beta_1, cdots, beta_k, delta_1, cdots, delta_k subset alpha_1, cdots, alpha_n$ of the roots (all the elements are distinct.)

I know that the maps in $Gal(L:K)$ permute the roots $alpha_1, cdots, alpha_n$. Can I always find an element $tau in Gal(L:K)$ such that $tau(beta_1) = delta_1, cdots, tau(beta_k) = delta_k$?

If not, how do I know which subsets of $alpha_1, cdots, alpha_n$ can be mapped to which other subsets? (Is there a way of knowing this, in general?)

abstract-algebra galois-theory

share|cite|improve this question

asked Mar 10 at 20:59

38917803891780

856

$endgroup$

share|cite|improve this question

asked Mar 10 at 20:59

38917803891780

856

share|cite|improve this question

asked Mar 10 at 20:59

38917803891780

856

asked Mar 10 at 20:59

38917803891780

856

asked Mar 10 at 20:59

38917803891780

856

1 Answer
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The short answer is no. Consider the cylotomic field extension $mathbbQ(zeta_p)$ over $mathbbQ$ with Galois group $mathbbZ_p^times$, and basis $zeta_p, ..., zeta_p^p-1$. Then there is no automorphism that will send $zeta_p$ to $zeta_p^2$ and $zeta_p^3$ to $zeta_p^4$.

In general this depends on the structure of the Galois group. A permutation group on $n$ elements is called $k$-transitive if any ordered $k$-tuple can be mapped to any other ordered $k$-tuple (which is exactly what you have asked). Multiply transitive groups (specifically doubly transitive groups) are rare - all doubly transitive groups are known. There is a list of classifications here http://mathworld.wolfram.com/TransitiveGroup.html

There are extensions for which this is possible. For instance, Hilbert showed that $S_n$ and $A_n$ can represented as Galois groups of Galois extensions of $mathbbQ$. (It is a nice exercise to check that $S_n$ is $n$-transitive and $A_n$ is $n-2$-transitive).

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answered Mar 10 at 21:16

nammienammie

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2

$begingroup$

The short answer is no. Consider the cylotomic field extension $mathbbQ(zeta_p)$ over $mathbbQ$ with Galois group $mathbbZ_p^times$, and basis $zeta_p, ..., zeta_p^p-1$. Then there is no automorphism that will send $zeta_p$ to $zeta_p^2$ and $zeta_p^3$ to $zeta_p^4$.

In general this depends on the structure of the Galois group. A permutation group on $n$ elements is called $k$-transitive if any ordered $k$-tuple can be mapped to any other ordered $k$-tuple (which is exactly what you have asked). Multiply transitive groups (specifically doubly transitive groups) are rare - all doubly transitive groups are known. There is a list of classifications here http://mathworld.wolfram.com/TransitiveGroup.html

There are extensions for which this is possible. For instance, Hilbert showed that $S_n$ and $A_n$ can represented as Galois groups of Galois extensions of $mathbbQ$. (It is a nice exercise to check that $S_n$ is $n$-transitive and $A_n$ is $n-2$-transitive).

share|cite|improve this answer

answered Mar 10 at 21:16

nammienammie

1449

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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2

$begingroup$

The short answer is no. Consider the cylotomic field extension $mathbbQ(zeta_p)$ over $mathbbQ$ with Galois group $mathbbZ_p^times$, and basis $zeta_p, ..., zeta_p^p-1$. Then there is no automorphism that will send $zeta_p$ to $zeta_p^2$ and $zeta_p^3$ to $zeta_p^4$.

In general this depends on the structure of the Galois group. A permutation group on $n$ elements is called $k$-transitive if any ordered $k$-tuple can be mapped to any other ordered $k$-tuple (which is exactly what you have asked). Multiply transitive groups (specifically doubly transitive groups) are rare - all doubly transitive groups are known. There is a list of classifications here http://mathworld.wolfram.com/TransitiveGroup.html

There are extensions for which this is possible. For instance, Hilbert showed that $S_n$ and $A_n$ can represented as Galois groups of Galois extensions of $mathbbQ$. (It is a nice exercise to check that $S_n$ is $n$-transitive and $A_n$ is $n-2$-transitive).

share|cite|improve this answer

answered Mar 10 at 21:16

nammienammie

1449

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

The short answer is no. Consider the cylotomic field extension $mathbbQ(zeta_p)$ over $mathbbQ$ with Galois group $mathbbZ_p^times$, and basis $zeta_p, ..., zeta_p^p-1$. Then there is no automorphism that will send $zeta_p$ to $zeta_p^2$ and $zeta_p^3$ to $zeta_p^4$.

In general this depends on the structure of the Galois group. A permutation group on $n$ elements is called $k$-transitive if any ordered $k$-tuple can be mapped to any other ordered $k$-tuple (which is exactly what you have asked). Multiply transitive groups (specifically doubly transitive groups) are rare - all doubly transitive groups are known. There is a list of classifications here http://mathworld.wolfram.com/TransitiveGroup.html

There are extensions for which this is possible. For instance, Hilbert showed that $S_n$ and $A_n$ can represented as Galois groups of Galois extensions of $mathbbQ$. (It is a nice exercise to check that $S_n$ is $n$-transitive and $A_n$ is $n-2$-transitive).

share|cite|improve this answer

answered Mar 10 at 21:16

nammienammie

1449

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 10 at 21:16

nammienammie

1449

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nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Mar 10 at 21:16

nammienammie

1449

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 10 at 21:16

nammienammie

1449