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An easy trick to compute the Galois group of a product of high degree polynomials
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If $f(X)=(X^4+3)(X^9-1) in mathbbQ[X]$ is there a trick to find its Galois group?
For example, if $f$ has only factos of degree $3$ and $2$ then its easy calculated because it can only be a direct product of $S_3, A_3, S_2$ and $id$ using the discriminant Theorem.
Now with high orders it became impossible to use discrimant.
Is there a easy way to solve it without enumerating all $f's$ roots and combining them with the automorphism?
group-theory galois-theory finite-fields extension-field irreducible-polynomials
edited 32 mins ago
pisco
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
add a comment |Â
up vote
3
down vote
favorite
If $f(X)=(X^4+3)(X^9-1) in mathbbQ[X]$ is there a trick to find its Galois group?
For example, if $f$ has only factos of degree $3$ and $2$ then its easy calculated because it can only be a direct product of $S_3, A_3, S_2$ and $id$ using the discriminant Theorem.
Now with high orders it became impossible to use discrimant.
Is there a easy way to solve it without enumerating all $f's$ roots and combining them with the automorphism?
group-theory galois-theory finite-fields extension-field irreducible-polynomials
edited 32 mins ago
pisco
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
1
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $f(X)=(X^4+3)(X^9-1) in mathbbQ[X]$ is there a trick to find its Galois group?
For example, if $f$ has only factos of degree $3$ and $2$ then its easy calculated because it can only be a direct product of $S_3, A_3, S_2$ and $id$ using the discriminant Theorem.
Now with high orders it became impossible to use discrimant.
Is there a easy way to solve it without enumerating all $f's$ roots and combining them with the automorphism?
group-theory galois-theory finite-fields extension-field irreducible-polynomials
edited 32 mins ago
pisco
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
If $f(X)=(X^4+3)(X^9-1) in mathbbQ[X]$ is there a trick to find its Galois group?
For example, if $f$ has only factos of degree $3$ and $2$ then its easy calculated because it can only be a direct product of $S_3, A_3, S_2$ and $id$ using the discriminant Theorem.
Now with high orders it became impossible to use discrimant.
Is there a easy way to solve it without enumerating all $f's$ roots and combining them with the automorphism?
group-theory galois-theory finite-fields extension-field irreducible-polynomials
group-theory galois-theory finite-fields extension-field irreducible-polynomials
edited 32 mins ago
pisco
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
edited 32 mins ago
pisco
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
edited 32 mins ago
pisco
10.6k21336
edited 32 mins ago
pisco
10.6k21336
edited 32 mins ago
pisco
10.6k21336
10.6k21336
asked 4 hours ago
Eduardo Silva
62739
asked 4 hours ago
Eduardo Silva
62739
asked 4 hours ago
Eduardo Silva
62739
62739
1
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago
add a comment |Â
1
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago
1
1
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago
add a comment |Â
2 Answers
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up vote
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The splitting field of $x^4+3$ is $mathbbQ(zeta_4, sqrt[4]-3)$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=mathbbQ(zeta_36, sqrt[4]-3)=mathbbQ(zeta_36, sqrtzeta_3+1/2)$, evidently it has degree $24$ over $mathbbQ$.
Let $alpha = sqrtzeta_3+1/2$, then $sigmain G=textGal(L/mathbbQ)$ is determined by $(a,b) in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$:
$$sigma:zeta_36 mapsto zeta_36^a qquad alpha mapsto i^b alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$zeta_3 mapsto zeta_3^a qquad alpha^2 = zeta_3 + frac12 mapsto (-1)^b (zeta_3 + frac12)$$
therefore $$zeta_3^a + frac12 = (-1)^b (zeta_3 + frac12)$$
this implies $$tag1 aequiv 1 pmod3 iff bequiv 0 pmod2 qquad aequiv 2 pmod3 iff bequiv 1 pmod2$$
Therefore elements in $G$ can be regarded as those $(a,b)in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$ that satisfies $(1)$, the group operation is given by
$$(a_1,b_1)(a_2,b_2) = (a_1a_2,a_1b_2 + b_1)$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 times D_4$.
answered 49 mins ago
pisco
10.6k21336
add a comment |Â
up vote
1
down vote
Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
answered 4 hours ago
Kapil
22112
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The splitting field of $x^4+3$ is $mathbbQ(zeta_4, sqrt[4]-3)$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=mathbbQ(zeta_36, sqrt[4]-3)=mathbbQ(zeta_36, sqrtzeta_3+1/2)$, evidently it has degree $24$ over $mathbbQ$.
Let $alpha = sqrtzeta_3+1/2$, then $sigmain G=textGal(L/mathbbQ)$ is determined by $(a,b) in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$:
$$sigma:zeta_36 mapsto zeta_36^a qquad alpha mapsto i^b alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$zeta_3 mapsto zeta_3^a qquad alpha^2 = zeta_3 + frac12 mapsto (-1)^b (zeta_3 + frac12)$$
therefore $$zeta_3^a + frac12 = (-1)^b (zeta_3 + frac12)$$
this implies $$tag1 aequiv 1 pmod3 iff bequiv 0 pmod2 qquad aequiv 2 pmod3 iff bequiv 1 pmod2$$
Therefore elements in $G$ can be regarded as those $(a,b)in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$ that satisfies $(1)$, the group operation is given by
$$(a_1,b_1)(a_2,b_2) = (a_1a_2,a_1b_2 + b_1)$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 times D_4$.
answered 49 mins ago
pisco
10.6k21336
add a comment |Â
up vote
2
down vote
The splitting field of $x^4+3$ is $mathbbQ(zeta_4, sqrt[4]-3)$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=mathbbQ(zeta_36, sqrt[4]-3)=mathbbQ(zeta_36, sqrtzeta_3+1/2)$, evidently it has degree $24$ over $mathbbQ$.
Let $alpha = sqrtzeta_3+1/2$, then $sigmain G=textGal(L/mathbbQ)$ is determined by $(a,b) in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$:
$$sigma:zeta_36 mapsto zeta_36^a qquad alpha mapsto i^b alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$zeta_3 mapsto zeta_3^a qquad alpha^2 = zeta_3 + frac12 mapsto (-1)^b (zeta_3 + frac12)$$
therefore $$zeta_3^a + frac12 = (-1)^b (zeta_3 + frac12)$$
this implies $$tag1 aequiv 1 pmod3 iff bequiv 0 pmod2 qquad aequiv 2 pmod3 iff bequiv 1 pmod2$$
Therefore elements in $G$ can be regarded as those $(a,b)in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$ that satisfies $(1)$, the group operation is given by
$$(a_1,b_1)(a_2,b_2) = (a_1a_2,a_1b_2 + b_1)$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 times D_4$.
answered 49 mins ago
pisco
10.6k21336
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The splitting field of $x^4+3$ is $mathbbQ(zeta_4, sqrt[4]-3)$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=mathbbQ(zeta_36, sqrt[4]-3)=mathbbQ(zeta_36, sqrtzeta_3+1/2)$, evidently it has degree $24$ over $mathbbQ$.
Let $alpha = sqrtzeta_3+1/2$, then $sigmain G=textGal(L/mathbbQ)$ is determined by $(a,b) in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$:
$$sigma:zeta_36 mapsto zeta_36^a qquad alpha mapsto i^b alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$zeta_3 mapsto zeta_3^a qquad alpha^2 = zeta_3 + frac12 mapsto (-1)^b (zeta_3 + frac12)$$
therefore $$zeta_3^a + frac12 = (-1)^b (zeta_3 + frac12)$$
this implies $$tag1 aequiv 1 pmod3 iff bequiv 0 pmod2 qquad aequiv 2 pmod3 iff bequiv 1 pmod2$$
Therefore elements in $G$ can be regarded as those $(a,b)in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$ that satisfies $(1)$, the group operation is given by
$$(a_1,b_1)(a_2,b_2) = (a_1a_2,a_1b_2 + b_1)$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 times D_4$.
answered 49 mins ago
pisco
10.6k21336
The splitting field of $x^4+3$ is $mathbbQ(zeta_4, sqrt[4]-3)$. Hence the splitting field of $(x^4+3)(x^9-1)$ is $L=mathbbQ(zeta_36, sqrt[4]-3)=mathbbQ(zeta_36, sqrtzeta_3+1/2)$, evidently it has degree $24$ over $mathbbQ$.
Let $alpha = sqrtzeta_3+1/2$, then $sigmain G=textGal(L/mathbbQ)$ is determined by $(a,b) in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$:
$$sigma:zeta_36 mapsto zeta_36^a qquad alpha mapsto i^b alpha$$
Note that $(a,b)$ has to satisfy certain condition:
$$zeta_3 mapsto zeta_3^a qquad alpha^2 = zeta_3 + frac12 mapsto (-1)^b (zeta_3 + frac12)$$
therefore $$zeta_3^a + frac12 = (-1)^b (zeta_3 + frac12)$$
this implies $$tag1 aequiv 1 pmod3 iff bequiv 0 pmod2 qquad aequiv 2 pmod3 iff bequiv 1 pmod2$$
Therefore elements in $G$ can be regarded as those $(a,b)in (mathbbZ/36mathbbZ)^times times (mathbbZ/4mathbbZ)$ that satisfies $(1)$, the group operation is given by
$$(a_1,b_1)(a_2,b_2) = (a_1a_2,a_1b_2 + b_1)$$
Sage says $G$ as an abstract group, is isomorphic to $C_3 times D_4$.
answered 49 mins ago
pisco
10.6k21336
edited 37 mins ago
answered 49 mins ago
pisco
10.6k21336
answered 49 mins ago
pisco
10.6k21336
answered 49 mins ago
pisco
10.6k21336
10.6k21336
add a comment |Â
add a comment |Â
up vote
1
down vote
Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
answered 4 hours ago
Kapil
22112
add a comment |Â
up vote
1
down vote
Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
answered 4 hours ago
Kapil
22112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
answered 4 hours ago
Kapil
22112
Algorithms to compute Galois groups are described in the answers to a question posed on Math Overflow.
In the specific example which you gave, it may be possible to calculate the Galois group by other simpler means.
answered 4 hours ago
Kapil
22112
answered 4 hours ago
Kapil
22112
answered 4 hours ago
Kapil
22112
answered 4 hours ago
Kapil
22112
22112
add a comment |Â
add a comment |Â
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1
The splitting field is $mathbbQ(zeta_36,sqrtzeta_3+1/2)$ a quadratic extension of $mathbbQ(zeta_36)$, its Galois group should be $Z/(9)^times times textAff_1(Z/(4))$ where $textAff_1(Z/(4))$ acts on the $m$ of $i^m sqrt[4]-3$ and $Z/(9)^times$ acts on the $l$ of $zeta_9^l$
– reuns
2 hours ago