Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$

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Let $E$ be an extension field of $F$ and suppose that $K$ and $L$ are two intermediate fields. If there exists an element $sigma in G(E/F)$ such that $sigma(K) = L$ then K and L are said to be conjugate fields. Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$.

This direction ($Rightarrow$):

Consider $sigma G(E/K) sigma^-1$ where $sigma in G(E/F)$ and $sigma(K) = L$ and consider some element $alpha in L$. I believe that if I can show that $sigma G(E/K) sigma^-1$ fixes $alpha$, then we are done.

So consider any $gamma in G(E/K)$. As $sigma (K) = L$ then $K = sigma^-1(L)$.

Thus, $sigma^-1(alpha)=beta$ for some $betain K$.

And then, $gamma(sigma^-1(alpha))=gamma(beta)=beta$ as $gamma$ fixes all $beta in K$

Finally, $sigma(gamma(sigma^-1(alpha)))=sigma(beta)=alpha$. So, $sigma G(E/K) sigma^-1$ consists of all automorphisms of E that fix $alpha in L$ and so it is $G(E/L)$

This direction ($Leftarrow$):

Well, if $G(E/K)$ and $G(E/L)$ are conjugate then there exists some $sigma in G(E/F)$ such that $sigma G(E/K) sigma^-1 = G(E/L)$. So consider some $alpha in K$. If $alpha$ is also in $F$ (as $K$ is an extension field of $F$ because it is an intermediate field between $E$ and $F$) then $alpha$ is fixed by $sigma G(E/K) sigma^-1$ and as $alpha in F$ then $alpha$ is also in $L$.

If $alpha in Kbackslash F$ then I don't know how to proceed.

galois-theory

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asked Jul 18 at 16:48

math4everyone

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Let $E$ be an extension field of $F$ and suppose that $K$ and $L$ are two intermediate fields. If there exists an element $sigma in G(E/F)$ such that $sigma(K) = L$ then K and L are said to be conjugate fields. Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$.

This direction ($Rightarrow$):

Consider $sigma G(E/K) sigma^-1$ where $sigma in G(E/F)$ and $sigma(K) = L$ and consider some element $alpha in L$. I believe that if I can show that $sigma G(E/K) sigma^-1$ fixes $alpha$, then we are done.

So consider any $gamma in G(E/K)$. As $sigma (K) = L$ then $K = sigma^-1(L)$.

Thus, $sigma^-1(alpha)=beta$ for some $betain K$.

And then, $gamma(sigma^-1(alpha))=gamma(beta)=beta$ as $gamma$ fixes all $beta in K$

Finally, $sigma(gamma(sigma^-1(alpha)))=sigma(beta)=alpha$. So, $sigma G(E/K) sigma^-1$ consists of all automorphisms of E that fix $alpha in L$ and so it is $G(E/L)$

This direction ($Leftarrow$):

Well, if $G(E/K)$ and $G(E/L)$ are conjugate then there exists some $sigma in G(E/F)$ such that $sigma G(E/K) sigma^-1 = G(E/L)$. So consider some $alpha in K$. If $alpha$ is also in $F$ (as $K$ is an extension field of $F$ because it is an intermediate field between $E$ and $F$) then $alpha$ is fixed by $sigma G(E/K) sigma^-1$ and as $alpha in F$ then $alpha$ is also in $L$.

If $alpha in Kbackslash F$ then I don't know how to proceed.

galois-theory

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asked Jul 18 at 16:48

math4everyone

38918

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Let $E$ be an extension field of $F$ and suppose that $K$ and $L$ are two intermediate fields. If there exists an element $sigma in G(E/F)$ such that $sigma(K) = L$ then K and L are said to be conjugate fields. Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$.

This direction ($Rightarrow$):

Consider $sigma G(E/K) sigma^-1$ where $sigma in G(E/F)$ and $sigma(K) = L$ and consider some element $alpha in L$. I believe that if I can show that $sigma G(E/K) sigma^-1$ fixes $alpha$, then we are done.

So consider any $gamma in G(E/K)$. As $sigma (K) = L$ then $K = sigma^-1(L)$.

Thus, $sigma^-1(alpha)=beta$ for some $betain K$.

And then, $gamma(sigma^-1(alpha))=gamma(beta)=beta$ as $gamma$ fixes all $beta in K$

Finally, $sigma(gamma(sigma^-1(alpha)))=sigma(beta)=alpha$. So, $sigma G(E/K) sigma^-1$ consists of all automorphisms of E that fix $alpha in L$ and so it is $G(E/L)$

This direction ($Leftarrow$):

Well, if $G(E/K)$ and $G(E/L)$ are conjugate then there exists some $sigma in G(E/F)$ such that $sigma G(E/K) sigma^-1 = G(E/L)$. So consider some $alpha in K$. If $alpha$ is also in $F$ (as $K$ is an extension field of $F$ because it is an intermediate field between $E$ and $F$) then $alpha$ is fixed by $sigma G(E/K) sigma^-1$ and as $alpha in F$ then $alpha$ is also in $L$.

If $alpha in Kbackslash F$ then I don't know how to proceed.

galois-theory

share|cite|improve this question

asked Jul 18 at 16:48

math4everyone

38918

Let $E$ be an extension field of $F$ and suppose that $K$ and $L$ are two intermediate fields. If there exists an element $sigma in G(E/F)$ such that $sigma(K) = L$ then K and L are said to be conjugate fields. Prove that $K$ and $L$ are conjugate iff $G(E/K)$ and $G(E/L)$ are conjugate subgroups of $G(E/F)$.

This direction ($Rightarrow$):

Consider $sigma G(E/K) sigma^-1$ where $sigma in G(E/F)$ and $sigma(K) = L$ and consider some element $alpha in L$. I believe that if I can show that $sigma G(E/K) sigma^-1$ fixes $alpha$, then we are done.

So consider any $gamma in G(E/K)$. As $sigma (K) = L$ then $K = sigma^-1(L)$.

Thus, $sigma^-1(alpha)=beta$ for some $betain K$.

And then, $gamma(sigma^-1(alpha))=gamma(beta)=beta$ as $gamma$ fixes all $beta in K$

Finally, $sigma(gamma(sigma^-1(alpha)))=sigma(beta)=alpha$. So, $sigma G(E/K) sigma^-1$ consists of all automorphisms of E that fix $alpha in L$ and so it is $G(E/L)$

This direction ($Leftarrow$):

Well, if $G(E/K)$ and $G(E/L)$ are conjugate then there exists some $sigma in G(E/F)$ such that $sigma G(E/K) sigma^-1 = G(E/L)$. So consider some $alpha in K$. If $alpha$ is also in $F$ (as $K$ is an extension field of $F$ because it is an intermediate field between $E$ and $F$) then $alpha$ is fixed by $sigma G(E/K) sigma^-1$ and as $alpha in F$ then $alpha$ is also in $L$.

If $alpha in Kbackslash F$ then I don't know how to proceed.

galois-theory

share|cite|improve this question

asked Jul 18 at 16:48

math4everyone

38918

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 16:48

math4everyone

38918

asked Jul 18 at 16:48

math4everyone

38918

asked Jul 18 at 16:48

math4everyone

38918

38918

add a comment | 

add a comment | 

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I believe that the you need the assumption that $F subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $mathbbQ subseteq E = mathbbQ(sqrt[3]2)$. The Galois group is the trivial group of a single element. Hence $G(E/mathbbQ) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $sigma in G(E/mathbbQ)$ s.t. $sigma(E) = mathbbQ$, as then $sigma$ must be the identity and we know $E not = mathbbQ$

If we assume the extension is Galois the claim holds. In particular if $sigma G(E/K)sigma^-1 = G(E/L)$, then we have that $sigma(K) = L$.

First we'll prove $sigma(K) subseteq L$. Consider $sigma(alpha)$ for some $alpha in K$. Also let $gamma$ be an arbitrary element of $G(E/L)$. Then we have that $exists beta in G(E/K)$ s.t. $gamma = sigma beta sigma^-1$. Then we have:

$$gamma(sigma(alpha)) = sigma beta sigma^-1(sigma(alpha)) = sigma(alpha)$$

So we have that $sigma(alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).

Now similarly you can prove that $sigma^-1(L) subseteq K$ to finish off the proof.

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answered Jul 19 at 22:50

Stefan4024

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I believe that the you need the assumption that $F subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $mathbbQ subseteq E = mathbbQ(sqrt[3]2)$. The Galois group is the trivial group of a single element. Hence $G(E/mathbbQ) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $sigma in G(E/mathbbQ)$ s.t. $sigma(E) = mathbbQ$, as then $sigma$ must be the identity and we know $E not = mathbbQ$

If we assume the extension is Galois the claim holds. In particular if $sigma G(E/K)sigma^-1 = G(E/L)$, then we have that $sigma(K) = L$.

First we'll prove $sigma(K) subseteq L$. Consider $sigma(alpha)$ for some $alpha in K$. Also let $gamma$ be an arbitrary element of $G(E/L)$. Then we have that $exists beta in G(E/K)$ s.t. $gamma = sigma beta sigma^-1$. Then we have:

$$gamma(sigma(alpha)) = sigma beta sigma^-1(sigma(alpha)) = sigma(alpha)$$

So we have that $sigma(alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).

Now similarly you can prove that $sigma^-1(L) subseteq K$ to finish off the proof.

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answered Jul 19 at 22:50

Stefan4024

28.3k53174

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up vote
0
down vote

I believe that the you need the assumption that $F subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $mathbbQ subseteq E = mathbbQ(sqrt[3]2)$. The Galois group is the trivial group of a single element. Hence $G(E/mathbbQ) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $sigma in G(E/mathbbQ)$ s.t. $sigma(E) = mathbbQ$, as then $sigma$ must be the identity and we know $E not = mathbbQ$

If we assume the extension is Galois the claim holds. In particular if $sigma G(E/K)sigma^-1 = G(E/L)$, then we have that $sigma(K) = L$.

First we'll prove $sigma(K) subseteq L$. Consider $sigma(alpha)$ for some $alpha in K$. Also let $gamma$ be an arbitrary element of $G(E/L)$. Then we have that $exists beta in G(E/K)$ s.t. $gamma = sigma beta sigma^-1$. Then we have:

$$gamma(sigma(alpha)) = sigma beta sigma^-1(sigma(alpha)) = sigma(alpha)$$

So we have that $sigma(alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).

Now similarly you can prove that $sigma^-1(L) subseteq K$ to finish off the proof.

share|cite|improve this answer

answered Jul 19 at 22:50

Stefan4024

28.3k53174

add a comment | 

up vote
0
down vote

up vote
0
down vote

I believe that the you need the assumption that $F subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $mathbbQ subseteq E = mathbbQ(sqrt[3]2)$. The Galois group is the trivial group of a single element. Hence $G(E/mathbbQ) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $sigma in G(E/mathbbQ)$ s.t. $sigma(E) = mathbbQ$, as then $sigma$ must be the identity and we know $E not = mathbbQ$

If we assume the extension is Galois the claim holds. In particular if $sigma G(E/K)sigma^-1 = G(E/L)$, then we have that $sigma(K) = L$.

First we'll prove $sigma(K) subseteq L$. Consider $sigma(alpha)$ for some $alpha in K$. Also let $gamma$ be an arbitrary element of $G(E/L)$. Then we have that $exists beta in G(E/K)$ s.t. $gamma = sigma beta sigma^-1$. Then we have:

$$gamma(sigma(alpha)) = sigma beta sigma^-1(sigma(alpha)) = sigma(alpha)$$

So we have that $sigma(alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).

Now similarly you can prove that $sigma^-1(L) subseteq K$ to finish off the proof.

share|cite|improve this answer

answered Jul 19 at 22:50

Stefan4024

28.3k53174

I believe that the you need the assumption that $F subseteq E$ is a Galois extension. Indeed consider the non-Galois extension $mathbbQ subseteq E = mathbbQ(sqrt[3]2)$. The Galois group is the trivial group of a single element. Hence $G(E/mathbbQ) = G(E/E)$ and in particular they are conjugates. However it is not true that there exists $sigma in G(E/mathbbQ)$ s.t. $sigma(E) = mathbbQ$, as then $sigma$ must be the identity and we know $E not = mathbbQ$

If we assume the extension is Galois the claim holds. In particular if $sigma G(E/K)sigma^-1 = G(E/L)$, then we have that $sigma(K) = L$.

First we'll prove $sigma(K) subseteq L$. Consider $sigma(alpha)$ for some $alpha in K$. Also let $gamma$ be an arbitrary element of $G(E/L)$. Then we have that $exists beta in G(E/K)$ s.t. $gamma = sigma beta sigma^-1$. Then we have:

$$gamma(sigma(alpha)) = sigma beta sigma^-1(sigma(alpha)) = sigma(alpha)$$

So we have that $sigma(alpha)$ is in the fixed field of $G(E/L)$, which is $L$ itself. (Here's where we use the fact that the extension is Galois).

Now similarly you can prove that $sigma^-1(L) subseteq K$ to finish off the proof.

share|cite|improve this answer

answered Jul 19 at 22:50

Stefan4024

28.3k53174

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 22:50

Stefan4024

28.3k53174

answered Jul 19 at 22:50

Stefan4024

28.3k53174

answered Jul 19 at 22:50

Stefan4024

28.3k53174

28.3k53174

add a comment | 

add a comment | 

 

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