Determine the fixed field of a field of rational functions and its Galois group.
$begingroup$
Let $k$ be a field and $K=k(x)$ be the field of rational functions of $k(x)$. Let $sigma $ and $tau$ automorphisms of $K$ defined by $sigma left( frac{f(x)}{g(x)} right) = frac{f(frac{1}{x})}{g(frac{1}{x})}$ and $tau left( frac{f(x)}{g(x)} right)= frac{f(1-x)}{g(1-x)}$. Determine the fixed field $F=K^S$, where $S={sigma, tau}$ and $Gal(K/F)$. Find an $hin F$ such that $F=k(h)$.
The fixed field is defined as $K^S = {xin K ;|; alpha(x)=x ; ; forallalpha in S}$, so, in this case I have that an element $frac{f(x)}{g(x)}$ belongs to the fixed field iff $frac{f(x)}{g(x)} = frac{f(1-x)}{g(1-x)}=frac{f(frac{1}{x})}{g(frac{1}{x})}$, and then I literally don't have any clue of what to do. Any hint would be very appreciated, thanks!
abstract-algebra field-theory galois-theory extension-field galois-extensions
asked Dec 14 '18 at 4:59
user392559user392559
37918
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $K=k(x)$ be the field of rational functions of $k(x)$. Let $sigma $ and $tau$ automorphisms of $K$ defined by $sigma left( frac{f(x)}{g(x)} right) = frac{f(frac{1}{x})}{g(frac{1}{x})}$ and $tau left( frac{f(x)}{g(x)} right)= frac{f(1-x)}{g(1-x)}$. Determine the fixed field $F=K^S$, where $S={sigma, tau}$ and $Gal(K/F)$. Find an $hin F$ such that $F=k(h)$.
The fixed field is defined as $K^S = {xin K ;|; alpha(x)=x ; ; forallalpha in S}$, so, in this case I have that an element $frac{f(x)}{g(x)}$ belongs to the fixed field iff $frac{f(x)}{g(x)} = frac{f(1-x)}{g(1-x)}=frac{f(frac{1}{x})}{g(frac{1}{x})}$, and then I literally don't have any clue of what to do. Any hint would be very appreciated, thanks!
abstract-algebra field-theory galois-theory extension-field galois-extensions
asked Dec 14 '18 at 4:59
user392559user392559
37918
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field and $K=k(x)$ be the field of rational functions of $k(x)$. Let $sigma $ and $tau$ automorphisms of $K$ defined by $sigma left( frac{f(x)}{g(x)} right) = frac{f(frac{1}{x})}{g(frac{1}{x})}$ and $tau left( frac{f(x)}{g(x)} right)= frac{f(1-x)}{g(1-x)}$. Determine the fixed field $F=K^S$, where $S={sigma, tau}$ and $Gal(K/F)$. Find an $hin F$ such that $F=k(h)$.
The fixed field is defined as $K^S = {xin K ;|; alpha(x)=x ; ; forallalpha in S}$, so, in this case I have that an element $frac{f(x)}{g(x)}$ belongs to the fixed field iff $frac{f(x)}{g(x)} = frac{f(1-x)}{g(1-x)}=frac{f(frac{1}{x})}{g(frac{1}{x})}$, and then I literally don't have any clue of what to do. Any hint would be very appreciated, thanks!
abstract-algebra field-theory galois-theory extension-field galois-extensions
asked Dec 14 '18 at 4:59
user392559user392559
37918
$endgroup$
Let $k$ be a field and $K=k(x)$ be the field of rational functions of $k(x)$. Let $sigma $ and $tau$ automorphisms of $K$ defined by $sigma left( frac{f(x)}{g(x)} right) = frac{f(frac{1}{x})}{g(frac{1}{x})}$ and $tau left( frac{f(x)}{g(x)} right)= frac{f(1-x)}{g(1-x)}$. Determine the fixed field $F=K^S$, where $S={sigma, tau}$ and $Gal(K/F)$. Find an $hin F$ such that $F=k(h)$.
The fixed field is defined as $K^S = {xin K ;|; alpha(x)=x ; ; forallalpha in S}$, so, in this case I have that an element $frac{f(x)}{g(x)}$ belongs to the fixed field iff $frac{f(x)}{g(x)} = frac{f(1-x)}{g(1-x)}=frac{f(frac{1}{x})}{g(frac{1}{x})}$, and then I literally don't have any clue of what to do. Any hint would be very appreciated, thanks!
abstract-algebra field-theory galois-theory extension-field galois-extensions
abstract-algebra field-theory galois-theory extension-field galois-extensions
asked Dec 14 '18 at 4:59
user392559user392559
37918
asked Dec 14 '18 at 4:59
user392559user392559
37918
asked Dec 14 '18 at 4:59
user392559user392559
37918
asked Dec 14 '18 at 4:59
user392559user392559
37918
asked Dec 14 '18 at 4:59
user392559user392559
37918
37918
add a comment |
add a comment |
1 Answer
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$begingroup$
$sigma$ and $tau$ generate a group of order six. The images of $x$
under its six elements are
$$x,quadfrac1x,quad 1-x,quadfrac1{1-x},quadfrac{x-1}x,quadfrac x{x-1}.$$
Any symmetric function of these is in the fixed field $F$.
The sum of their squares is
begin{align}
x^2+(1-x)^2+frac{1+(x-1)^2}{x^2}+frac{1+x^2}{(1-x)^2}
&=frac{H(x)}{x^2(1-x)^2}
end{align}
for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $hin F$
and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$,
so $F=k(h)$.
Warning: this might not work in characteristic two.
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$sigma$ and $tau$ generate a group of order six. The images of $x$
under its six elements are
$$x,quadfrac1x,quad 1-x,quadfrac1{1-x},quadfrac{x-1}x,quadfrac x{x-1}.$$
Any symmetric function of these is in the fixed field $F$.
The sum of their squares is
begin{align}
x^2+(1-x)^2+frac{1+(x-1)^2}{x^2}+frac{1+x^2}{(1-x)^2}
&=frac{H(x)}{x^2(1-x)^2}
end{align}
for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $hin F$
and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$,
so $F=k(h)$.
Warning: this might not work in characteristic two.
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
$begingroup$
$sigma$ and $tau$ generate a group of order six. The images of $x$
under its six elements are
$$x,quadfrac1x,quad 1-x,quadfrac1{1-x},quadfrac{x-1}x,quadfrac x{x-1}.$$
Any symmetric function of these is in the fixed field $F$.
The sum of their squares is
begin{align}
x^2+(1-x)^2+frac{1+(x-1)^2}{x^2}+frac{1+x^2}{(1-x)^2}
&=frac{H(x)}{x^2(1-x)^2}
end{align}
for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $hin F$
and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$,
so $F=k(h)$.
Warning: this might not work in characteristic two.
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
add a comment |
$begingroup$
$sigma$ and $tau$ generate a group of order six. The images of $x$
under its six elements are
$$x,quadfrac1x,quad 1-x,quadfrac1{1-x},quadfrac{x-1}x,quadfrac x{x-1}.$$
Any symmetric function of these is in the fixed field $F$.
The sum of their squares is
begin{align}
x^2+(1-x)^2+frac{1+(x-1)^2}{x^2}+frac{1+x^2}{(1-x)^2}
&=frac{H(x)}{x^2(1-x)^2}
end{align}
for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $hin F$
and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$,
so $F=k(h)$.
Warning: this might not work in characteristic two.
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$sigma$ and $tau$ generate a group of order six. The images of $x$
under its six elements are
$$x,quadfrac1x,quad 1-x,quadfrac1{1-x},quadfrac{x-1}x,quadfrac x{x-1}.$$
Any symmetric function of these is in the fixed field $F$.
The sum of their squares is
begin{align}
x^2+(1-x)^2+frac{1+(x-1)^2}{x^2}+frac{1+x^2}{(1-x)^2}
&=frac{H(x)}{x^2(1-x)^2}
end{align}
for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $hin F$
and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$,
so $F=k(h)$.
Warning: this might not work in characteristic two.
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Dec 14 '18 at 5:13
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
add a comment |
add a comment |
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