Mixing
Normal Subgroup of Galois Group
$begingroup$
Let $L/K$ be a Galois extension, and let $Rsubseteq L$ be a subring such that $tau(R)=R$ for every $tauintext{Gal}(L/K)$.
Let $alphain R$. How would I show that $H={tauintext{Gal}(L/K):tau(alpha)=alpha}$ a normal subgroup of $text{Gal}(L/K)$?
I've tried using the Fundamental Theorem of Galois Theory, by which it would be equivalent to show that $L^H/K$ is a Galois extension. I would think that $L^H=K(alpha)$, but haven't been able to show this, or that assuming this the extension is Galois.
group-theory field-theory galois-theory galois-extensions
asked Jan 27 at 11:26
DaveDave
597
$endgroup$
add a comment |
$begingroup$
Let $L/K$ be a Galois extension, and let $Rsubseteq L$ be a subring such that $tau(R)=R$ for every $tauintext{Gal}(L/K)$.
Let $alphain R$. How would I show that $H={tauintext{Gal}(L/K):tau(alpha)=alpha}$ a normal subgroup of $text{Gal}(L/K)$?
I've tried using the Fundamental Theorem of Galois Theory, by which it would be equivalent to show that $L^H/K$ is a Galois extension. I would think that $L^H=K(alpha)$, but haven't been able to show this, or that assuming this the extension is Galois.
group-theory field-theory galois-theory galois-extensions
asked Jan 27 at 11:26
DaveDave
597
$endgroup$
$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45
add a comment |
$begingroup$
Let $L/K$ be a Galois extension, and let $Rsubseteq L$ be a subring such that $tau(R)=R$ for every $tauintext{Gal}(L/K)$.
Let $alphain R$. How would I show that $H={tauintext{Gal}(L/K):tau(alpha)=alpha}$ a normal subgroup of $text{Gal}(L/K)$?
I've tried using the Fundamental Theorem of Galois Theory, by which it would be equivalent to show that $L^H/K$ is a Galois extension. I would think that $L^H=K(alpha)$, but haven't been able to show this, or that assuming this the extension is Galois.
group-theory field-theory galois-theory galois-extensions
asked Jan 27 at 11:26
DaveDave
597
$endgroup$
Let $L/K$ be a Galois extension, and let $Rsubseteq L$ be a subring such that $tau(R)=R$ for every $tauintext{Gal}(L/K)$.
Let $alphain R$. How would I show that $H={tauintext{Gal}(L/K):tau(alpha)=alpha}$ a normal subgroup of $text{Gal}(L/K)$?
I've tried using the Fundamental Theorem of Galois Theory, by which it would be equivalent to show that $L^H/K$ is a Galois extension. I would think that $L^H=K(alpha)$, but haven't been able to show this, or that assuming this the extension is Galois.
group-theory field-theory galois-theory galois-extensions
group-theory field-theory galois-theory galois-extensions
asked Jan 27 at 11:26
DaveDave
597
asked Jan 27 at 11:26
DaveDave
597
edited Jan 27 at 11:53
Dave
asked Jan 27 at 11:26
DaveDave
597
asked Jan 27 at 11:26
DaveDave
597
asked Jan 27 at 11:26
DaveDave
597
597
$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45
add a comment |
$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45
$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Unless I misunderstood something the claim is false.
Take the field extension $mathbb{Q} subset mathbb{Q}(sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = mathbb{Q}(sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $alpha = sqrt[4]{2}$. Then it's not hard to show that $L^H = mathbb{Q}(sqrt[4]{2})$. But this is obviously not a Galois extension of $mathbb{Q}$, so H can't be a normal group of $text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(alpha)$. To prove this first note that $K(alpha) subseteq L^H$, as obviously $K(alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(alpha) = L^{H'}$ for some subgroup $H'$ of $text{Gal}(L/K)$. Now as $H'$ fixes $alpha$ we have that $H' le H$. From this $L^{H} subseteq L^{H'} = K(alpha)$. Hence $L^H = K(alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $alpha$ are in $R$, they might not be in $K(alpha)$, which has to be the case if it were a Galois extension of $K$.
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
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add a comment |
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$begingroup$
Unless I misunderstood something the claim is false.
Take the field extension $mathbb{Q} subset mathbb{Q}(sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = mathbb{Q}(sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $alpha = sqrt[4]{2}$. Then it's not hard to show that $L^H = mathbb{Q}(sqrt[4]{2})$. But this is obviously not a Galois extension of $mathbb{Q}$, so H can't be a normal group of $text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(alpha)$. To prove this first note that $K(alpha) subseteq L^H$, as obviously $K(alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(alpha) = L^{H'}$ for some subgroup $H'$ of $text{Gal}(L/K)$. Now as $H'$ fixes $alpha$ we have that $H' le H$. From this $L^{H} subseteq L^{H'} = K(alpha)$. Hence $L^H = K(alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $alpha$ are in $R$, they might not be in $K(alpha)$, which has to be the case if it were a Galois extension of $K$.
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
$endgroup$
add a comment |
$begingroup$
Unless I misunderstood something the claim is false.
Take the field extension $mathbb{Q} subset mathbb{Q}(sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = mathbb{Q}(sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $alpha = sqrt[4]{2}$. Then it's not hard to show that $L^H = mathbb{Q}(sqrt[4]{2})$. But this is obviously not a Galois extension of $mathbb{Q}$, so H can't be a normal group of $text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(alpha)$. To prove this first note that $K(alpha) subseteq L^H$, as obviously $K(alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(alpha) = L^{H'}$ for some subgroup $H'$ of $text{Gal}(L/K)$. Now as $H'$ fixes $alpha$ we have that $H' le H$. From this $L^{H} subseteq L^{H'} = K(alpha)$. Hence $L^H = K(alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $alpha$ are in $R$, they might not be in $K(alpha)$, which has to be the case if it were a Galois extension of $K$.
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
$endgroup$
add a comment |
$begingroup$
Unless I misunderstood something the claim is false.
Take the field extension $mathbb{Q} subset mathbb{Q}(sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = mathbb{Q}(sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $alpha = sqrt[4]{2}$. Then it's not hard to show that $L^H = mathbb{Q}(sqrt[4]{2})$. But this is obviously not a Galois extension of $mathbb{Q}$, so H can't be a normal group of $text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(alpha)$. To prove this first note that $K(alpha) subseteq L^H$, as obviously $K(alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(alpha) = L^{H'}$ for some subgroup $H'$ of $text{Gal}(L/K)$. Now as $H'$ fixes $alpha$ we have that $H' le H$. From this $L^{H} subseteq L^{H'} = K(alpha)$. Hence $L^H = K(alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $alpha$ are in $R$, they might not be in $K(alpha)$, which has to be the case if it were a Galois extension of $K$.
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
$endgroup$
Unless I misunderstood something the claim is false.
Take the field extension $mathbb{Q} subset mathbb{Q}(sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = mathbb{Q}(sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $alpha = sqrt[4]{2}$. Then it's not hard to show that $L^H = mathbb{Q}(sqrt[4]{2})$. But this is obviously not a Galois extension of $mathbb{Q}$, so H can't be a normal group of $text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(alpha)$. To prove this first note that $K(alpha) subseteq L^H$, as obviously $K(alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(alpha) = L^{H'}$ for some subgroup $H'$ of $text{Gal}(L/K)$. Now as $H'$ fixes $alpha$ we have that $H' le H$. From this $L^{H} subseteq L^{H'} = K(alpha)$. Hence $L^H = K(alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $alpha$ are in $R$, they might not be in $K(alpha)$, which has to be the case if it were a Galois extension of $K$.
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
edited Jan 27 at 12:55
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
answered Jan 27 at 12:44
Stefan4024Stefan4024
30.6k63579
30.6k63579
add a comment |
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$begingroup$
What do you think?
$endgroup$
– ÍgjøgnumMeg
Jan 27 at 11:37
$begingroup$
Apologies for the phrasing, I've been trying to follow a proof which assumes that $H$ is normal and haven't been able to show this either way myself yet. Based on the provenance I think it is true, I'll rephrase the question
$endgroup$
– Dave
Jan 27 at 11:45