A cyclic subgroup of order 3 of the Galois group $Gal(Bbb{R}(x)/Bbb{R})$
$begingroup$
Let's consider the field $mathbb{R}(x)$ formed by the quotients of $mathbb{R}[x]$. We know that $A=begin{pmatrix}
frac{-1}{2} &frac{-sqrt{3}}{2} \
frac{sqrt{3}}{2} & frac{-1}{2}
end{pmatrix}$ -the 120 degree rotation matrix- determines an element $$varphi_A in Gal(mathbb{R}(x)/mathbb{R})$$ such that $$varphi_A(x)=frac{ax+b}{cx+d}.$$
Now I need to prove that the cyclic subgroup generated by the equivalence class of A
$$bar{A} in M_{2x2}/Ker(varphi)=Gal(mathbb{R}(x)/mathbb{R})$$
has order 3, also calculate the fixed field
$$F=mathbb{R}(x)^{<bar{A}>},$$ show that
$$mathbb{R}/F$$ is a Galois extention and find $u$ such that $F=mathbb{R}(u)$. Finally, if $x in mathbb{R}(x)$ I need to calculate $min(x,F) in F[y]$ and find its roots
abstract-algebra galois-theory finite-fields
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
$endgroup$
add a comment |
$begingroup$
Let's consider the field $mathbb{R}(x)$ formed by the quotients of $mathbb{R}[x]$. We know that $A=begin{pmatrix}
frac{-1}{2} &frac{-sqrt{3}}{2} \
frac{sqrt{3}}{2} & frac{-1}{2}
end{pmatrix}$ -the 120 degree rotation matrix- determines an element $$varphi_A in Gal(mathbb{R}(x)/mathbb{R})$$ such that $$varphi_A(x)=frac{ax+b}{cx+d}.$$
Now I need to prove that the cyclic subgroup generated by the equivalence class of A
$$bar{A} in M_{2x2}/Ker(varphi)=Gal(mathbb{R}(x)/mathbb{R})$$
has order 3, also calculate the fixed field
$$F=mathbb{R}(x)^{<bar{A}>},$$ show that
$$mathbb{R}/F$$ is a Galois extention and find $u$ such that $F=mathbb{R}(u)$. Finally, if $x in mathbb{R}(x)$ I need to calculate $min(x,F) in F[y]$ and find its roots
abstract-algebra galois-theory finite-fields
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
$endgroup$
1
$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31
add a comment |
$begingroup$
Let's consider the field $mathbb{R}(x)$ formed by the quotients of $mathbb{R}[x]$. We know that $A=begin{pmatrix}
frac{-1}{2} &frac{-sqrt{3}}{2} \
frac{sqrt{3}}{2} & frac{-1}{2}
end{pmatrix}$ -the 120 degree rotation matrix- determines an element $$varphi_A in Gal(mathbb{R}(x)/mathbb{R})$$ such that $$varphi_A(x)=frac{ax+b}{cx+d}.$$
Now I need to prove that the cyclic subgroup generated by the equivalence class of A
$$bar{A} in M_{2x2}/Ker(varphi)=Gal(mathbb{R}(x)/mathbb{R})$$
has order 3, also calculate the fixed field
$$F=mathbb{R}(x)^{<bar{A}>},$$ show that
$$mathbb{R}/F$$ is a Galois extention and find $u$ such that $F=mathbb{R}(u)$. Finally, if $x in mathbb{R}(x)$ I need to calculate $min(x,F) in F[y]$ and find its roots
abstract-algebra galois-theory finite-fields
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
$endgroup$
Let's consider the field $mathbb{R}(x)$ formed by the quotients of $mathbb{R}[x]$. We know that $A=begin{pmatrix}
frac{-1}{2} &frac{-sqrt{3}}{2} \
frac{sqrt{3}}{2} & frac{-1}{2}
end{pmatrix}$ -the 120 degree rotation matrix- determines an element $$varphi_A in Gal(mathbb{R}(x)/mathbb{R})$$ such that $$varphi_A(x)=frac{ax+b}{cx+d}.$$
Now I need to prove that the cyclic subgroup generated by the equivalence class of A
$$bar{A} in M_{2x2}/Ker(varphi)=Gal(mathbb{R}(x)/mathbb{R})$$
has order 3, also calculate the fixed field
$$F=mathbb{R}(x)^{<bar{A}>},$$ show that
$$mathbb{R}/F$$ is a Galois extention and find $u$ such that $F=mathbb{R}(u)$. Finally, if $x in mathbb{R}(x)$ I need to calculate $min(x,F) in F[y]$ and find its roots
abstract-algebra galois-theory finite-fields
abstract-algebra galois-theory finite-fields
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
edited Dec 13 '18 at 4:40
Jyrki Lahtonen
110k13172389
110k13172389
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
asked Dec 12 '18 at 23:17
Tina MartínezTina Martínez
12
12
1
$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31
add a comment |
1
$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31
1
1
$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31
$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31
add a comment |
1 Answer
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$begingroup$
Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
$endgroup$
add a comment |
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$begingroup$
Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
$endgroup$
add a comment |
$begingroup$
Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
$endgroup$
add a comment |
$begingroup$
Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
$endgroup$
Hint: you have your group of three elements. It often happens (but it’s not guaranteed!) that the fixed field is generated by the Trace or the Norm of your special generating element, $x$ in this case. Try both.
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
answered Dec 13 '18 at 3:54
LubinLubin
45.5k44688
45.5k44688
add a comment |
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$begingroup$
Where are you stuck ? For any $u(x) in mathbb{R}(x)$ and non-constant $x mapsto u(x)$ is a field morphism $mathbb{R}(x) to mathbb{R}(x)$. That morphism is surjective iff $u(x)$ has only one pole and zero.
$endgroup$
– reuns
Dec 12 '18 at 23:31