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Prove that the Galois group of an irreducible cubic polynomial over $mathbbQ$ is isomorphic to $S_3$ or $mathbbZ_3$
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I know that in order to show that the Galois group of this polynomial is isomorphic to $mathbbZ_3$ we must show that, if $alpha$ is a root of the polynomial, then $[mathbbQ(alpha):mathbbQ]=|G(mathbbQ(alpha):mathbbQ)|=3$ , but how can I show that and also how can I show that it can also be $S_3$? Really don't know where to start, but I am thinking that if $alpha in mathbbC$ then $G(mathbbQ(alpha)/mathbbQ)$ must have $id,sigma,sigma^2$ where $sigma$ is complex conjugation. But what about, $S_3$?
galois-theory
asked Jul 16 at 23:40
math4everyone
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I know that in order to show that the Galois group of this polynomial is isomorphic to $mathbbZ_3$ we must show that, if $alpha$ is a root of the polynomial, then $[mathbbQ(alpha):mathbbQ]=|G(mathbbQ(alpha):mathbbQ)|=3$ , but how can I show that and also how can I show that it can also be $S_3$? Really don't know where to start, but I am thinking that if $alpha in mathbbC$ then $G(mathbbQ(alpha)/mathbbQ)$ must have $id,sigma,sigma^2$ where $sigma$ is complex conjugation. But what about, $S_3$?
galois-theory
asked Jul 16 at 23:40
math4everyone
38918
1
This is a good place to start your self-studies.
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Jul 16 at 23:58
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up vote
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favorite
I know that in order to show that the Galois group of this polynomial is isomorphic to $mathbbZ_3$ we must show that, if $alpha$ is a root of the polynomial, then $[mathbbQ(alpha):mathbbQ]=|G(mathbbQ(alpha):mathbbQ)|=3$ , but how can I show that and also how can I show that it can also be $S_3$? Really don't know where to start, but I am thinking that if $alpha in mathbbC$ then $G(mathbbQ(alpha)/mathbbQ)$ must have $id,sigma,sigma^2$ where $sigma$ is complex conjugation. But what about, $S_3$?
galois-theory
asked Jul 16 at 23:40
math4everyone
38918
I know that in order to show that the Galois group of this polynomial is isomorphic to $mathbbZ_3$ we must show that, if $alpha$ is a root of the polynomial, then $[mathbbQ(alpha):mathbbQ]=|G(mathbbQ(alpha):mathbbQ)|=3$ , but how can I show that and also how can I show that it can also be $S_3$? Really don't know where to start, but I am thinking that if $alpha in mathbbC$ then $G(mathbbQ(alpha)/mathbbQ)$ must have $id,sigma,sigma^2$ where $sigma$ is complex conjugation. But what about, $S_3$?
galois-theory
asked Jul 16 at 23:40
math4everyone
38918
asked Jul 16 at 23:40
math4everyone
38918
asked Jul 16 at 23:40
math4everyone
38918
asked Jul 16 at 23:40
math4everyone
38918
38918
1
This is a good place to start your self-studies.
– Bumblebee
Jul 16 at 23:58
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1
This is a good place to start your self-studies.
– Bumblebee
Jul 16 at 23:58
1
1
This is a good place to start your self-studies.
– Bumblebee
Jul 16 at 23:58
This is a good place to start your self-studies.
– Bumblebee
Jul 16 at 23:58
add a comment |Â
2 Answers
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Here is the key fact:
The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.
Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.
Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.
It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.
answered Jul 17 at 1:41
lhf
156k9160367
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Let $finmathbbQ[X]$ be an irreducible cubic polynomial with roots $alpha_1,alpha_2,alpha_3$ in $mathbbC$. We set $F = mathbbQ(omega)$, where $omega = e^2pi i/3$, and let $E = F(alpha_1,alpha_2, alpha_3)$ be the splitting field over $F$ of $f$ in $mathbbC$. We regard the Galois group $G = operatornameGal(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3mid |G|$, so the possible orders of $G$ are $3$, and $6$.
For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n le 3$, and so $|A_3|=3!/2=3$ which implies $A_3cong mathbbZ/(3)$. Hence $Gcong A_3$ if $|G|=3$.
For order $6$ we have the two groups $mathbbZ/(6)$ or $S_3$ to consider. Since $mathbbZ/(6)nsubseteq S_3$, we have $Gcong S_3$ if $|G|=6$.
Note that $f$ is irreducible in $mathbbQ[X]$ if and only if $G=operatornameGal(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $mathbbQ$. If $operatornameDisc f=square$ then $Gcong A_3$, and if $operatornameDisc fneqsquare$ then $Gcong S_3$.
Hence an irreducible cubic has Galois group $Gcong A_3$ or $Gcong S_3$.
answered Jul 18 at 0:52
Daniel Buck
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is the key fact:
The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.
Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.
Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.
It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.
answered Jul 17 at 1:41
lhf
156k9160367
add a comment |Â
up vote
1
down vote
Here is the key fact:
The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.
Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.
Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.
It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.
answered Jul 17 at 1:41
lhf
156k9160367
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is the key fact:
The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.
Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.
Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.
It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.
answered Jul 17 at 1:41
lhf
156k9160367
Here is the key fact:
The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.
Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.
Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.
It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.
answered Jul 17 at 1:41
lhf
156k9160367
edited Jul 17 at 1:50
answered Jul 17 at 1:41
lhf
156k9160367
answered Jul 17 at 1:41
lhf
156k9160367
answered Jul 17 at 1:41
lhf
156k9160367
156k9160367
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $finmathbbQ[X]$ be an irreducible cubic polynomial with roots $alpha_1,alpha_2,alpha_3$ in $mathbbC$. We set $F = mathbbQ(omega)$, where $omega = e^2pi i/3$, and let $E = F(alpha_1,alpha_2, alpha_3)$ be the splitting field over $F$ of $f$ in $mathbbC$. We regard the Galois group $G = operatornameGal(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3mid |G|$, so the possible orders of $G$ are $3$, and $6$.
For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n le 3$, and so $|A_3|=3!/2=3$ which implies $A_3cong mathbbZ/(3)$. Hence $Gcong A_3$ if $|G|=3$.
For order $6$ we have the two groups $mathbbZ/(6)$ or $S_3$ to consider. Since $mathbbZ/(6)nsubseteq S_3$, we have $Gcong S_3$ if $|G|=6$.
Note that $f$ is irreducible in $mathbbQ[X]$ if and only if $G=operatornameGal(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $mathbbQ$. If $operatornameDisc f=square$ then $Gcong A_3$, and if $operatornameDisc fneqsquare$ then $Gcong S_3$.
Hence an irreducible cubic has Galois group $Gcong A_3$ or $Gcong S_3$.
answered Jul 18 at 0:52
Daniel Buck
2,3141624
add a comment |Â
up vote
0
down vote
Let $finmathbbQ[X]$ be an irreducible cubic polynomial with roots $alpha_1,alpha_2,alpha_3$ in $mathbbC$. We set $F = mathbbQ(omega)$, where $omega = e^2pi i/3$, and let $E = F(alpha_1,alpha_2, alpha_3)$ be the splitting field over $F$ of $f$ in $mathbbC$. We regard the Galois group $G = operatornameGal(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3mid |G|$, so the possible orders of $G$ are $3$, and $6$.
For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n le 3$, and so $|A_3|=3!/2=3$ which implies $A_3cong mathbbZ/(3)$. Hence $Gcong A_3$ if $|G|=3$.
For order $6$ we have the two groups $mathbbZ/(6)$ or $S_3$ to consider. Since $mathbbZ/(6)nsubseteq S_3$, we have $Gcong S_3$ if $|G|=6$.
Note that $f$ is irreducible in $mathbbQ[X]$ if and only if $G=operatornameGal(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $mathbbQ$. If $operatornameDisc f=square$ then $Gcong A_3$, and if $operatornameDisc fneqsquare$ then $Gcong S_3$.
Hence an irreducible cubic has Galois group $Gcong A_3$ or $Gcong S_3$.
answered Jul 18 at 0:52
Daniel Buck
2,3141624
add a comment |Â
up vote
0
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up vote
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down vote
Let $finmathbbQ[X]$ be an irreducible cubic polynomial with roots $alpha_1,alpha_2,alpha_3$ in $mathbbC$. We set $F = mathbbQ(omega)$, where $omega = e^2pi i/3$, and let $E = F(alpha_1,alpha_2, alpha_3)$ be the splitting field over $F$ of $f$ in $mathbbC$. We regard the Galois group $G = operatornameGal(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3mid |G|$, so the possible orders of $G$ are $3$, and $6$.
For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n le 3$, and so $|A_3|=3!/2=3$ which implies $A_3cong mathbbZ/(3)$. Hence $Gcong A_3$ if $|G|=3$.
For order $6$ we have the two groups $mathbbZ/(6)$ or $S_3$ to consider. Since $mathbbZ/(6)nsubseteq S_3$, we have $Gcong S_3$ if $|G|=6$.
Note that $f$ is irreducible in $mathbbQ[X]$ if and only if $G=operatornameGal(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $mathbbQ$. If $operatornameDisc f=square$ then $Gcong A_3$, and if $operatornameDisc fneqsquare$ then $Gcong S_3$.
Hence an irreducible cubic has Galois group $Gcong A_3$ or $Gcong S_3$.
answered Jul 18 at 0:52
Daniel Buck
2,3141624
Let $finmathbbQ[X]$ be an irreducible cubic polynomial with roots $alpha_1,alpha_2,alpha_3$ in $mathbbC$. We set $F = mathbbQ(omega)$, where $omega = e^2pi i/3$, and let $E = F(alpha_1,alpha_2, alpha_3)$ be the splitting field over $F$ of $f$ in $mathbbC$. We regard the Galois group $G = operatornameGal(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3mid |G|$, so the possible orders of $G$ are $3$, and $6$.
For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n le 3$, and so $|A_3|=3!/2=3$ which implies $A_3cong mathbbZ/(3)$. Hence $Gcong A_3$ if $|G|=3$.
For order $6$ we have the two groups $mathbbZ/(6)$ or $S_3$ to consider. Since $mathbbZ/(6)nsubseteq S_3$, we have $Gcong S_3$ if $|G|=6$.
Note that $f$ is irreducible in $mathbbQ[X]$ if and only if $G=operatornameGal(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $mathbbQ$. If $operatornameDisc f=square$ then $Gcong A_3$, and if $operatornameDisc fneqsquare$ then $Gcong S_3$.
Hence an irreducible cubic has Galois group $Gcong A_3$ or $Gcong S_3$.
answered Jul 18 at 0:52
Daniel Buck
2,3141624
answered Jul 18 at 0:52
Daniel Buck
2,3141624
answered Jul 18 at 0:52
Daniel Buck
2,3141624
answered Jul 18 at 0:52
Daniel Buck
2,3141624
2,3141624
add a comment |Â
add a comment |Â
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1
This is a good place to start your self-studies.
– Bumblebee
Jul 16 at 23:58