Mixing
Equation with Galois group twisted $S_{3}$
$begingroup$
I note that a Galois group is not just a Galois group. Let $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, denote the roots of a quartic equation. Then $x^4-5x^2+6$ has Galois group $Z_{2}^2$, where the permutations of the roots are given by ($r$'s and fixed roots omitted) $()$, $(12)$, $(34)$, and $(12)(34)$, while $x^4+1$ also has Galois group $Z_{2}^2$, but its permutations are $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$.
These situations are different. The first polynomial can be factored, but the second one is irreducible over the rationals. So actually it might be better to refer to a Galois group as a group acting on a set or a permutation group instead of just a group.
In that case, I wonder if there is a quintic equation whose Galois group/set is $()$, $(123)$, $(132)$, $(12)(45)$, $(23)(45)$, $(13)(45)$. This group is $S_{3}$, but its $2$-cycles are paired up with $(45)$; I have seen this called "twisted $S_{3}$" on some sites. If so, it can't be like the usual $S_{3}$ equation such as $x^3-2$. Since this group permutes $5$ elements, it must be the group of a quintic, not a cubic, but it looks like it has cubic roots. Further, the usual cubic equation such as $x^3-2$ has a non-square discriminant (in this case, $108$), but twisted $S_{3}$ is a subgroup of $A_{5}$, which means that the quintic has to have a square discriminant. So can there be such an equation?
finite-groups galois-theory symmetric-groups quintic-equations
edited Jan 21 at 4:27
the_fox
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
$endgroup$
add a comment |
$begingroup$
I note that a Galois group is not just a Galois group. Let $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, denote the roots of a quartic equation. Then $x^4-5x^2+6$ has Galois group $Z_{2}^2$, where the permutations of the roots are given by ($r$'s and fixed roots omitted) $()$, $(12)$, $(34)$, and $(12)(34)$, while $x^4+1$ also has Galois group $Z_{2}^2$, but its permutations are $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$.
These situations are different. The first polynomial can be factored, but the second one is irreducible over the rationals. So actually it might be better to refer to a Galois group as a group acting on a set or a permutation group instead of just a group.
In that case, I wonder if there is a quintic equation whose Galois group/set is $()$, $(123)$, $(132)$, $(12)(45)$, $(23)(45)$, $(13)(45)$. This group is $S_{3}$, but its $2$-cycles are paired up with $(45)$; I have seen this called "twisted $S_{3}$" on some sites. If so, it can't be like the usual $S_{3}$ equation such as $x^3-2$. Since this group permutes $5$ elements, it must be the group of a quintic, not a cubic, but it looks like it has cubic roots. Further, the usual cubic equation such as $x^3-2$ has a non-square discriminant (in this case, $108$), but twisted $S_{3}$ is a subgroup of $A_{5}$, which means that the quintic has to have a square discriminant. So can there be such an equation?
finite-groups galois-theory symmetric-groups quintic-equations
edited Jan 21 at 4:27
the_fox
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
$endgroup$
add a comment |
$begingroup$
I note that a Galois group is not just a Galois group. Let $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, denote the roots of a quartic equation. Then $x^4-5x^2+6$ has Galois group $Z_{2}^2$, where the permutations of the roots are given by ($r$'s and fixed roots omitted) $()$, $(12)$, $(34)$, and $(12)(34)$, while $x^4+1$ also has Galois group $Z_{2}^2$, but its permutations are $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$.
These situations are different. The first polynomial can be factored, but the second one is irreducible over the rationals. So actually it might be better to refer to a Galois group as a group acting on a set or a permutation group instead of just a group.
In that case, I wonder if there is a quintic equation whose Galois group/set is $()$, $(123)$, $(132)$, $(12)(45)$, $(23)(45)$, $(13)(45)$. This group is $S_{3}$, but its $2$-cycles are paired up with $(45)$; I have seen this called "twisted $S_{3}$" on some sites. If so, it can't be like the usual $S_{3}$ equation such as $x^3-2$. Since this group permutes $5$ elements, it must be the group of a quintic, not a cubic, but it looks like it has cubic roots. Further, the usual cubic equation such as $x^3-2$ has a non-square discriminant (in this case, $108$), but twisted $S_{3}$ is a subgroup of $A_{5}$, which means that the quintic has to have a square discriminant. So can there be such an equation?
finite-groups galois-theory symmetric-groups quintic-equations
edited Jan 21 at 4:27
the_fox
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
$endgroup$
I note that a Galois group is not just a Galois group. Let $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$, denote the roots of a quartic equation. Then $x^4-5x^2+6$ has Galois group $Z_{2}^2$, where the permutations of the roots are given by ($r$'s and fixed roots omitted) $()$, $(12)$, $(34)$, and $(12)(34)$, while $x^4+1$ also has Galois group $Z_{2}^2$, but its permutations are $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$.
These situations are different. The first polynomial can be factored, but the second one is irreducible over the rationals. So actually it might be better to refer to a Galois group as a group acting on a set or a permutation group instead of just a group.
In that case, I wonder if there is a quintic equation whose Galois group/set is $()$, $(123)$, $(132)$, $(12)(45)$, $(23)(45)$, $(13)(45)$. This group is $S_{3}$, but its $2$-cycles are paired up with $(45)$; I have seen this called "twisted $S_{3}$" on some sites. If so, it can't be like the usual $S_{3}$ equation such as $x^3-2$. Since this group permutes $5$ elements, it must be the group of a quintic, not a cubic, but it looks like it has cubic roots. Further, the usual cubic equation such as $x^3-2$ has a non-square discriminant (in this case, $108$), but twisted $S_{3}$ is a subgroup of $A_{5}$, which means that the quintic has to have a square discriminant. So can there be such an equation?
finite-groups galois-theory symmetric-groups quintic-equations
finite-groups galois-theory symmetric-groups quintic-equations
edited Jan 21 at 4:27
the_fox
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
edited Jan 21 at 4:27
the_fox
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
edited Jan 21 at 4:27
the_fox
2,90021537
edited Jan 21 at 4:27
the_fox
2,90021537
edited Jan 21 at 4:27
the_fox
2,90021537
2,90021537
asked Jan 21 at 1:17
jimvb13jimvb13
1652
asked Jan 21 at 1:17
jimvb13jimvb13
1652
asked Jan 21 at 1:17
jimvb13jimvb13
1652
1652
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=mathbb Q(r_1,dots,r_n)$ of descomposition of $P$
is a Galois extension of $mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(mathbb Q^P/ mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying ${r_1,dots r_n}$ with ${1,dots n}$.
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$.
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $mathbb Q^P= mathbb Q(sqrt 2, i)=mathbb Q^{(x^2 -2)(x^2+1)}$.
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/mathbb Q$.
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents
$Gal(mathbb Q^Q/mathbb Q)$ canonically as the group of units $(mathbb Z/ 8mathbb Z)^times$ which is more than a group of permutations.
answered Jan 21 at 15:42
eduardeduard
47427
$endgroup$
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=mathbb Q(r_1,dots,r_n)$ of descomposition of $P$
is a Galois extension of $mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(mathbb Q^P/ mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying ${r_1,dots r_n}$ with ${1,dots n}$.
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$.
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $mathbb Q^P= mathbb Q(sqrt 2, i)=mathbb Q^{(x^2 -2)(x^2+1)}$.
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/mathbb Q$.
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents
$Gal(mathbb Q^Q/mathbb Q)$ canonically as the group of units $(mathbb Z/ 8mathbb Z)^times$ which is more than a group of permutations.
answered Jan 21 at 15:42
eduardeduard
47427
$endgroup$
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
add a comment |
$begingroup$
An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=mathbb Q(r_1,dots,r_n)$ of descomposition of $P$
is a Galois extension of $mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(mathbb Q^P/ mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying ${r_1,dots r_n}$ with ${1,dots n}$.
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$.
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $mathbb Q^P= mathbb Q(sqrt 2, i)=mathbb Q^{(x^2 -2)(x^2+1)}$.
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/mathbb Q$.
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents
$Gal(mathbb Q^Q/mathbb Q)$ canonically as the group of units $(mathbb Z/ 8mathbb Z)^times$ which is more than a group of permutations.
answered Jan 21 at 15:42
eduardeduard
47427
$endgroup$
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
add a comment |
$begingroup$
An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=mathbb Q(r_1,dots,r_n)$ of descomposition of $P$
is a Galois extension of $mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(mathbb Q^P/ mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying ${r_1,dots r_n}$ with ${1,dots n}$.
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$.
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $mathbb Q^P= mathbb Q(sqrt 2, i)=mathbb Q^{(x^2 -2)(x^2+1)}$.
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/mathbb Q$.
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents
$Gal(mathbb Q^Q/mathbb Q)$ canonically as the group of units $(mathbb Z/ 8mathbb Z)^times$ which is more than a group of permutations.
answered Jan 21 at 15:42
eduardeduard
47427
$endgroup$
An example
$(x^3 - 2)(x^2+3)$.
Edit: With respect to the claim that a Galois group is not just a Galois group: let $P$ be a polynomial with rational coefficients and let $r_1,dots, r_n$ the roots of $P$. For simplicity assume that $P$ has simple roots. The field $Q^P:=mathbb Q(r_1,dots,r_n)$ of descomposition of $P$
is a Galois extension of $mathbb Q$.
To the polynomial $P$ one usually attaches the group $G(P):=Gal(mathbb Q^P/ mathbb Q)$ and a natural representation
as a group of permutation, that is an embedding
$$
G(P) hookrightarrow S_n
$$
into the symmetric group of $n$ elements; just by identifying ${r_1,dots r_n}$ with ${1,dots n}$.
This embedding depends strongly on $P$ but $G(P)$ weakly on $P$.
For example let $P=x^4 + 1$, and let $r$ be a root of $P$. Then $mathbb Q^P= Q(r)$. As the OP mentioned the group attached to $P$ is represented as a transitive permutation group by () , (12)(34), (13)(24), and (14)(23).
Since $r$ is a $8$th roof of unity it is direct by Euler's identity that $mathbb Q^P= mathbb Q(sqrt 2, i)=mathbb Q^{(x^2 -2)(x^2+1)}$.
The permutation group attached to $Q=(x^2 -2)(x^2+1)$ is $(), (1,2),(3,4),(1,2)(3,4)$, a non-transitive permutation group. As abstract groups they are isomorphic.
I would say that a Galois group is a group of field automorphisms but a particular realization of it as a permutation group is not automatically inherent of $G$ but a consequence of the procedure used to obtain $K/mathbb Q$.
Number theorists have been interested in more exotic representations. For example the theory of cyclotomic fields represents
$Gal(mathbb Q^Q/mathbb Q)$ canonically as the group of units $(mathbb Z/ 8mathbb Z)^times$ which is more than a group of permutations.
answered Jan 21 at 15:42
eduardeduard
47427
edited Jan 21 at 22:33
answered Jan 21 at 15:42
eduardeduard
47427
answered Jan 21 at 15:42
eduardeduard
47427
answered Jan 21 at 15:42
eduardeduard
47427
47427
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
add a comment |
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
1
1
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
$begingroup$
It is remarkable that if $P$ is irreducible the group of permutations attached to it is transitive. The twisted $S_3$ is not transitive.
$endgroup$
– eduard
Jan 21 at 15:44
1
1
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
$begingroup$
Also it is the normal basis theorem that shows : $K$ is a $n=|G|$ dimensional $mathbb{Q}$-vector space, choosing a basis then the action of $G$ on $K cong mathbb{Q}^n$ becomes a representation $G to GL_n(mathbb{Q})$, that a change of basis makes this representation the left regular representation
$endgroup$
– reuns
Jan 21 at 23:14
1
1
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
$begingroup$
Interesting. Let R be your example. Then to extend $mathbb Q$ to include the roots of R, solve $x^3-2=0$ first. One first has to include the cube roots of unity, so extend $mathbb Q$ by adding $sqrt{-3}$, but that automatically brings the roots of $x^2+3$ into the field, so one needs only extend to a degree 6 extension of $mathbb Q$. I had found a similar example earlier. I thought that no quartic equation could have the Galois group consisting of () and (12)(34), but then I found $x^4-4x^3+3x^2+2x-1 = (x^2-x-1)(x^2-3x+1)$. Solving the first factor automatically solves the second.
$endgroup$
– jimvb13
Jan 22 at 2:53
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