Example of when $E^H neq E^textAut(E/E^H)$?
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Let $E/F$ be a field extension of $F$, and let $H leq textAut(E/F)$ be a subgroup of the automorphisms of $E$ fixing $F$. If $H$ is finite, then it is well-known that $H = textAut(E/E^H)$ and so $E^H = E^textAut(E/E^H)$, where $E^H$ is the fixed field of $H$.
Therefore, to find an example where equality does not hold, we necessarily require $H$ to be infinite. So I tried looking at $E = mathbbR(x)$ and $F = mathbbR$, which gives
beginalign
textAut(mathbbR(x)/mathbbR) cong GL_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix,
endalign
with $lambda in mathbbR^times$. An automorphism can be constructed by considering $g = beginpmatrixa & b \ c & dendpmatrix in GL_2(mathbbR)$ and defining a map $phi: GL_2(mathbbR) to textAut(mathbbR(x)/mathbbR) $ by $g mapsto phi(g)$ where, for $f(x) in mathbbR(x)$, we have
beginalign
phi(g)(f) = fleft(fracax + cbx + dright).
endalign
The only problem is that I don't know how to find the fixed field. Say if I took $H = SO_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix < textAut(mathbbR(x)/mathbbR)$, then, to obtain the fixed field $mathbbR(x)^H$, I would require all rational functions $f$ satisfying
beginalign
fleft(fracxcostheta + sintheta-xsintheta + costhetaright) = f(x).
endalign
How should I proceed from here, or, is there an easier example I can examine?
abstract-algebra field-theory galois-theory examples-counterexamples automorphism-group
asked Sep 1 at 0:36
user1239334
716
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up vote
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Let $E/F$ be a field extension of $F$, and let $H leq textAut(E/F)$ be a subgroup of the automorphisms of $E$ fixing $F$. If $H$ is finite, then it is well-known that $H = textAut(E/E^H)$ and so $E^H = E^textAut(E/E^H)$, where $E^H$ is the fixed field of $H$.
Therefore, to find an example where equality does not hold, we necessarily require $H$ to be infinite. So I tried looking at $E = mathbbR(x)$ and $F = mathbbR$, which gives
beginalign
textAut(mathbbR(x)/mathbbR) cong GL_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix,
endalign
with $lambda in mathbbR^times$. An automorphism can be constructed by considering $g = beginpmatrixa & b \ c & dendpmatrix in GL_2(mathbbR)$ and defining a map $phi: GL_2(mathbbR) to textAut(mathbbR(x)/mathbbR) $ by $g mapsto phi(g)$ where, for $f(x) in mathbbR(x)$, we have
beginalign
phi(g)(f) = fleft(fracax + cbx + dright).
endalign
The only problem is that I don't know how to find the fixed field. Say if I took $H = SO_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix < textAut(mathbbR(x)/mathbbR)$, then, to obtain the fixed field $mathbbR(x)^H$, I would require all rational functions $f$ satisfying
beginalign
fleft(fracxcostheta + sintheta-xsintheta + costhetaright) = f(x).
endalign
How should I proceed from here, or, is there an easier example I can examine?
abstract-algebra field-theory galois-theory examples-counterexamples automorphism-group
asked Sep 1 at 0:36
user1239334
716
2
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $E/F$ be a field extension of $F$, and let $H leq textAut(E/F)$ be a subgroup of the automorphisms of $E$ fixing $F$. If $H$ is finite, then it is well-known that $H = textAut(E/E^H)$ and so $E^H = E^textAut(E/E^H)$, where $E^H$ is the fixed field of $H$.
Therefore, to find an example where equality does not hold, we necessarily require $H$ to be infinite. So I tried looking at $E = mathbbR(x)$ and $F = mathbbR$, which gives
beginalign
textAut(mathbbR(x)/mathbbR) cong GL_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix,
endalign
with $lambda in mathbbR^times$. An automorphism can be constructed by considering $g = beginpmatrixa & b \ c & dendpmatrix in GL_2(mathbbR)$ and defining a map $phi: GL_2(mathbbR) to textAut(mathbbR(x)/mathbbR) $ by $g mapsto phi(g)$ where, for $f(x) in mathbbR(x)$, we have
beginalign
phi(g)(f) = fleft(fracax + cbx + dright).
endalign
The only problem is that I don't know how to find the fixed field. Say if I took $H = SO_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix < textAut(mathbbR(x)/mathbbR)$, then, to obtain the fixed field $mathbbR(x)^H$, I would require all rational functions $f$ satisfying
beginalign
fleft(fracxcostheta + sintheta-xsintheta + costhetaright) = f(x).
endalign
How should I proceed from here, or, is there an easier example I can examine?
abstract-algebra field-theory galois-theory examples-counterexamples automorphism-group
asked Sep 1 at 0:36
user1239334
716
Let $E/F$ be a field extension of $F$, and let $H leq textAut(E/F)$ be a subgroup of the automorphisms of $E$ fixing $F$. If $H$ is finite, then it is well-known that $H = textAut(E/E^H)$ and so $E^H = E^textAut(E/E^H)$, where $E^H$ is the fixed field of $H$.
Therefore, to find an example where equality does not hold, we necessarily require $H$ to be infinite. So I tried looking at $E = mathbbR(x)$ and $F = mathbbR$, which gives
beginalign
textAut(mathbbR(x)/mathbbR) cong GL_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix,
endalign
with $lambda in mathbbR^times$. An automorphism can be constructed by considering $g = beginpmatrixa & b \ c & dendpmatrix in GL_2(mathbbR)$ and defining a map $phi: GL_2(mathbbR) to textAut(mathbbR(x)/mathbbR) $ by $g mapsto phi(g)$ where, for $f(x) in mathbbR(x)$, we have
beginalign
phi(g)(f) = fleft(fracax + cbx + dright).
endalign
The only problem is that I don't know how to find the fixed field. Say if I took $H = SO_2(mathbbR)bigg/beginpmatrixlambda & 0 \ 0 & lambdaendpmatrix < textAut(mathbbR(x)/mathbbR)$, then, to obtain the fixed field $mathbbR(x)^H$, I would require all rational functions $f$ satisfying
beginalign
fleft(fracxcostheta + sintheta-xsintheta + costhetaright) = f(x).
endalign
How should I proceed from here, or, is there an easier example I can examine?
abstract-algebra field-theory galois-theory examples-counterexamples automorphism-group
abstract-algebra field-theory galois-theory examples-counterexamples automorphism-group
asked Sep 1 at 0:36
user1239334
716
asked Sep 1 at 0:36
user1239334
716
edited Sep 1 at 8:01
asked Sep 1 at 0:36
user1239334
716
asked Sep 1 at 0:36
user1239334
716
asked Sep 1 at 0:36
user1239334
716
716
2
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38
add a comment |Â
2
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38
2
2
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38
add a comment |Â
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2
You cannot have $E^Hneq E^Aut(E/E^H)$. Note that $Hleq Aut(E/E^H)$ always holds, and it implies $E^Aut(E/E^H)leq E^H$. On the other hand, $E^Hleq E^Aut(E/E^H)$ is straight-forward: if $xin E^H$ and $gin Aut(E/E^H)$ then obviously $g(x)=x$, hence $xin E^Aut(E/E^H)$. Maybe you want to ask for an example that $Hneq Aut(E/E^H)$? Or that $E^Aut(E/L)neq L$ for an intermediate field $L$?
– SMM
Sep 2 at 16:42
This was the answer I was looking for. I could show myself that $E^H subset E^textAut(E/E^H)$ in the way you said, but I missed $H leq textAut(E/E^H)$.
– user1239334
Sep 2 at 21:38