Mixing
Clarification on size of Galois group
$begingroup$
I was under the impression that if $K/k$ is Galois then $G$ the Galois group of this extension had the size of the degree of the extension. However, I don't think this is correct anymore. Under what circumstances is this true? I think this is true for number fields? Because if $K/mathbb{Q}$ is a Galois number field then we can express $K$ as $K = mathbb{Q}[alpha]$ which means we each automorphism in $G$ corresponds to where we send $alpha$ to among its $n$ conjugates where $n = [K:mathbb{Q}].$ In particular, if $K/k$ is Galois and $K = k[alpha]$ I guess this would be true but are there other notable examples for which this is true?
Do all finite Galois extensions have this property? I suspect this is true as we can write such an extension as the base field adjoined with finitely many elements. But I am not sure about the other cases...for example infinite Galois extensions? I have not encountered such examples yet.
galois-theory
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
$endgroup$
add a comment |
$begingroup$
I was under the impression that if $K/k$ is Galois then $G$ the Galois group of this extension had the size of the degree of the extension. However, I don't think this is correct anymore. Under what circumstances is this true? I think this is true for number fields? Because if $K/mathbb{Q}$ is a Galois number field then we can express $K$ as $K = mathbb{Q}[alpha]$ which means we each automorphism in $G$ corresponds to where we send $alpha$ to among its $n$ conjugates where $n = [K:mathbb{Q}].$ In particular, if $K/k$ is Galois and $K = k[alpha]$ I guess this would be true but are there other notable examples for which this is true?
Do all finite Galois extensions have this property? I suspect this is true as we can write such an extension as the base field adjoined with finitely many elements. But I am not sure about the other cases...for example infinite Galois extensions? I have not encountered such examples yet.
galois-theory
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
$endgroup$
1
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
1
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
1
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15
add a comment |
$begingroup$
I was under the impression that if $K/k$ is Galois then $G$ the Galois group of this extension had the size of the degree of the extension. However, I don't think this is correct anymore. Under what circumstances is this true? I think this is true for number fields? Because if $K/mathbb{Q}$ is a Galois number field then we can express $K$ as $K = mathbb{Q}[alpha]$ which means we each automorphism in $G$ corresponds to where we send $alpha$ to among its $n$ conjugates where $n = [K:mathbb{Q}].$ In particular, if $K/k$ is Galois and $K = k[alpha]$ I guess this would be true but are there other notable examples for which this is true?
Do all finite Galois extensions have this property? I suspect this is true as we can write such an extension as the base field adjoined with finitely many elements. But I am not sure about the other cases...for example infinite Galois extensions? I have not encountered such examples yet.
galois-theory
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
$endgroup$
I was under the impression that if $K/k$ is Galois then $G$ the Galois group of this extension had the size of the degree of the extension. However, I don't think this is correct anymore. Under what circumstances is this true? I think this is true for number fields? Because if $K/mathbb{Q}$ is a Galois number field then we can express $K$ as $K = mathbb{Q}[alpha]$ which means we each automorphism in $G$ corresponds to where we send $alpha$ to among its $n$ conjugates where $n = [K:mathbb{Q}].$ In particular, if $K/k$ is Galois and $K = k[alpha]$ I guess this would be true but are there other notable examples for which this is true?
Do all finite Galois extensions have this property? I suspect this is true as we can write such an extension as the base field adjoined with finitely many elements. But I am not sure about the other cases...for example infinite Galois extensions? I have not encountered such examples yet.
galois-theory
galois-theory
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
asked Jan 21 at 22:18
伽罗瓦伽罗瓦
1,205615
1,205615
1
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
1
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
1
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15
add a comment |
1
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
1
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
1
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15
1
1
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
1
1
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
1
1
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15
add a comment |
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1
$begingroup$
All finite Galois extensions are separable and simple so they have this property by more or less the same argument you used for number fields. Infinite Galois extensions don't have this property. The group may have a higher cardinality than the extension degree. Linking to that thread mostly because while trying to answer it I dug up links to other related threads. I'm afraid it's not really too well organized.
$endgroup$
– Jyrki Lahtonen
Jan 21 at 22:32
1
$begingroup$
Also you should try starting with this definition : $K/k$ Galois iff $k = K^G$ (the subfield fixed by $G$) for some finite subgroup $G le Aut(K)$. Normal and separable means you can always find $sigma in Aut(K/k)$ moving $alpha in K,alpha not in k$ so $K/k$ is Galois. From $k = K^G$ you know all the minimal polynomials.
$endgroup$
– reuns
Jan 21 at 23:22
1
$begingroup$
As I understand things, infinite Galois groups are always uncountable. The extensions, not so.
$endgroup$
– Lubin
Jan 22 at 1:15