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Must a degree $6$ field extension with Galois group $S_3$ be the splitting field of a cubic? [duplicate]
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Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic?
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Prove (or disprove by example) the following: If $F$ is a field extension of $K$, not necessarily a Galois extension, with $[F:K]=6$ and $mathrmAut_K F simeq S_3$, then $F$ must be the splitting field of an irreducible cubic in $K[x]$.
This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if $F$ doesn't have to be Galois over $K$, then there must be an example where $F$ isn't a splitting field at all, right? I've yet to come up with such an example though.
abstract-algebra field-theory galois-theory
asked Jul 29 at 6:57
Mike Pierce
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marked as duplicate by Mike Pierce, max_zorn, Xander Henderson, John Ma, Jyrki Lahtonen abstract-algebra
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Jul 31 at 4:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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This question already has an answer here:
Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic?
1 answer
Prove (or disprove by example) the following: If $F$ is a field extension of $K$, not necessarily a Galois extension, with $[F:K]=6$ and $mathrmAut_K F simeq S_3$, then $F$ must be the splitting field of an irreducible cubic in $K[x]$.
This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if $F$ doesn't have to be Galois over $K$, then there must be an example where $F$ isn't a splitting field at all, right? I've yet to come up with such an example though.
abstract-algebra field-theory galois-theory
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
marked as duplicate by Mike Pierce, max_zorn, Xander Henderson, John Ma, Jyrki Lahtonen abstract-algebra
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Jul 31 at 4:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
1
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47
add a comment |Â
up vote
2
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up vote
2
down vote
favorite
This question already has an answer here:
Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic?
1 answer
Prove (or disprove by example) the following: If $F$ is a field extension of $K$, not necessarily a Galois extension, with $[F:K]=6$ and $mathrmAut_K F simeq S_3$, then $F$ must be the splitting field of an irreducible cubic in $K[x]$.
This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if $F$ doesn't have to be Galois over $K$, then there must be an example where $F$ isn't a splitting field at all, right? I've yet to come up with such an example though.
abstract-algebra field-theory galois-theory
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
This question already has an answer here:
Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic?
1 answer
Prove (or disprove by example) the following: If $F$ is a field extension of $K$, not necessarily a Galois extension, with $[F:K]=6$ and $mathrmAut_K F simeq S_3$, then $F$ must be the splitting field of an irreducible cubic in $K[x]$.
This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if $F$ doesn't have to be Galois over $K$, then there must be an example where $F$ isn't a splitting field at all, right? I've yet to come up with such an example though.
This question already has an answer here:
Does a Galois group being $S_3$ correspond to the extension being the splitting field of a cubic?
1 answer
abstract-algebra field-theory galois-theory
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
edited Jul 30 at 22:38
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
asked Jul 29 at 6:57
Mike Pierce
10.9k93573
10.9k93573
marked as duplicate by Mike Pierce, max_zorn, Xander Henderson, John Ma, Jyrki Lahtonen abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Jul 31 at 4:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Mike Pierce, max_zorn, Xander Henderson, John Ma, Jyrki Lahtonen abstract-algebra
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Jul 31 at 4:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
1
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47
add a comment |Â
1
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
1
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47
1
1
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
1
1
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47
add a comment |Â
1 Answer
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Let $G=textrmAut_KFcong S_3$. Then the fixed field of $G$, $K'$ say,
contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$
is a Galois extension with degree $|G|=6$ and Galois group $G$.
From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1,2)in S_3$. Then $|L:K|=3$.
Let $ain L$, $anotin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $langle(1,2)rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $G=textrmAut_KFcong S_3$. Then the fixed field of $G$, $K'$ say,
contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$
is a Galois extension with degree $|G|=6$ and Galois group $G$.
From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1,2)in S_3$. Then $|L:K|=3$.
Let $ain L$, $anotin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $langle(1,2)rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
add a comment |Â
up vote
1
down vote
accepted
Let $G=textrmAut_KFcong S_3$. Then the fixed field of $G$, $K'$ say,
contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$
is a Galois extension with degree $|G|=6$ and Galois group $G$.
From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1,2)in S_3$. Then $|L:K|=3$.
Let $ain L$, $anotin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $langle(1,2)rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $G=textrmAut_KFcong S_3$. Then the fixed field of $G$, $K'$ say,
contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$
is a Galois extension with degree $|G|=6$ and Galois group $G$.
From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1,2)in S_3$. Then $|L:K|=3$.
Let $ain L$, $anotin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $langle(1,2)rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
Let $G=textrmAut_KFcong S_3$. Then the fixed field of $G$, $K'$ say,
contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$
is a Galois extension with degree $|G|=6$ and Galois group $G$.
From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1,2)in S_3$. Then $|L:K|=3$.
Let $ain L$, $anotin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $langle(1,2)rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
edited Jul 30 at 22:33
Mike Pierce
10.9k93573
10.9k93573
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
answered Jul 29 at 8:10
Lord Shark the Unknown
84.5k950111
84.5k950111
add a comment |Â
add a comment |Â
1
In a separable extension $F:K$, if the degree $[F:K]$ equals the order of the the group $Aut_KF$, then the extension is Galois. This means that "not necessarily a Galois extension" means "not necessarily a separable extension".
– Crostul
Jul 29 at 7:07
@Crostul Yes, but in this example $F/K$ will be Galois.
– Lord Shark the Unknown
Jul 29 at 7:09
@LordSharktheUnknown I don't see why that must be
– Mike Pierce
Jul 29 at 7:46
1
@MikePierce It's because if $G$ is a finite group of automorphisms of a field $F$, then $F/F^G$ is a Galois extension of degree $|G|$.
– Lord Shark the Unknown
Jul 29 at 7:47