Degree of extension of fixed field by infinite set of automorphisms.
If $G$ is a finite group of automorphism $E rightarrow E$, then Dedekind-Artin theorem tells us that $[E:E^G]=; mid G mid$ where $E^G$ is the subfield of $E$ fixed by the automorphisms of $G$. Is this still true if $G$ is infinite or are they simple counter-examples?
abstract-algebra field-theory galois-theory extension-field automorphism-group
edited Jan 2 at 21:45
user26857
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
add a comment |
If $G$ is a finite group of automorphism $E rightarrow E$, then Dedekind-Artin theorem tells us that $[E:E^G]=; mid G mid$ where $E^G$ is the subfield of $E$ fixed by the automorphisms of $G$. Is this still true if $G$ is infinite or are they simple counter-examples?
abstract-algebra field-theory galois-theory extension-field automorphism-group
edited Jan 2 at 21:45
user26857
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
1
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
1
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago
add a comment |
If $G$ is a finite group of automorphism $E rightarrow E$, then Dedekind-Artin theorem tells us that $[E:E^G]=; mid G mid$ where $E^G$ is the subfield of $E$ fixed by the automorphisms of $G$. Is this still true if $G$ is infinite or are they simple counter-examples?
abstract-algebra field-theory galois-theory extension-field automorphism-group
edited Jan 2 at 21:45
user26857
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
If $G$ is a finite group of automorphism $E rightarrow E$, then Dedekind-Artin theorem tells us that $[E:E^G]=; mid G mid$ where $E^G$ is the subfield of $E$ fixed by the automorphisms of $G$. Is this still true if $G$ is infinite or are they simple counter-examples?
abstract-algebra field-theory galois-theory extension-field automorphism-group
abstract-algebra field-theory galois-theory extension-field automorphism-group
edited Jan 2 at 21:45
user26857
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
edited Jan 2 at 21:45
user26857
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
edited Jan 2 at 21:45
user26857
39.3k123983
edited Jan 2 at 21:45
user26857
39.3k123983
edited Jan 2 at 21:45
user26857
39.3k123983
39.3k123983
asked Jan 2 at 16:13
roi_saumon
42028
asked Jan 2 at 16:13
roi_saumon
42028
asked Jan 2 at 16:13
roi_saumon
42028
42028
1
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
1
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago
add a comment |
1
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
1
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago
1
1
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
1
1
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
No. If $E$ is the algebraic closure of the field $Bbb{F}_p$, then we can think of $E$ as the nested union
$$K_0subset K_1subset K_2subset cdots,$$
where the field $K_ell$ is th unique (up to isomorphism) field of $p^{ell!}$ elements.
Then $E=bigcup_{i=0}^infty K_i$ is a countably infinite set, and hence $[E:Bbb{F}_p]$ is countably infinite.
But the group of automorphisms $Gal(E/Bbb{F}_p)$ is uncountable.
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
|
show 1 more comment
An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $Esupset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $tmapsto t+n$, where $ninBbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.
Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $tmapsto t+n$, there being infinitely many different such substitutions.
And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions ${frac1{t-alpha}}$ are all $k$-linearly independent.
Thus you have a situation where the original group $GcongBbb Z$ is countable, and even its completion $hat{Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.
answered Jan 3 at 23:00
Lubin
43.8k44585
add a comment |
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2 Answers
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2 Answers
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No. If $E$ is the algebraic closure of the field $Bbb{F}_p$, then we can think of $E$ as the nested union
$$K_0subset K_1subset K_2subset cdots,$$
where the field $K_ell$ is th unique (up to isomorphism) field of $p^{ell!}$ elements.
Then $E=bigcup_{i=0}^infty K_i$ is a countably infinite set, and hence $[E:Bbb{F}_p]$ is countably infinite.
But the group of automorphisms $Gal(E/Bbb{F}_p)$ is uncountable.
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
|
show 1 more comment
No. If $E$ is the algebraic closure of the field $Bbb{F}_p$, then we can think of $E$ as the nested union
$$K_0subset K_1subset K_2subset cdots,$$
where the field $K_ell$ is th unique (up to isomorphism) field of $p^{ell!}$ elements.
Then $E=bigcup_{i=0}^infty K_i$ is a countably infinite set, and hence $[E:Bbb{F}_p]$ is countably infinite.
But the group of automorphisms $Gal(E/Bbb{F}_p)$ is uncountable.
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
|
show 1 more comment
No. If $E$ is the algebraic closure of the field $Bbb{F}_p$, then we can think of $E$ as the nested union
$$K_0subset K_1subset K_2subset cdots,$$
where the field $K_ell$ is th unique (up to isomorphism) field of $p^{ell!}$ elements.
Then $E=bigcup_{i=0}^infty K_i$ is a countably infinite set, and hence $[E:Bbb{F}_p]$ is countably infinite.
But the group of automorphisms $Gal(E/Bbb{F}_p)$ is uncountable.
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
No. If $E$ is the algebraic closure of the field $Bbb{F}_p$, then we can think of $E$ as the nested union
$$K_0subset K_1subset K_2subset cdots,$$
where the field $K_ell$ is th unique (up to isomorphism) field of $p^{ell!}$ elements.
Then $E=bigcup_{i=0}^infty K_i$ is a countably infinite set, and hence $[E:Bbb{F}_p]$ is countably infinite.
But the group of automorphisms $Gal(E/Bbb{F}_p)$ is uncountable.
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
edited Jan 3 at 18:32
Kenny Lau
19.8k2159
19.8k2159
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
answered Jan 2 at 20:48
Jyrki Lahtonen
108k12166367
108k12166367
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
|
show 1 more comment
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
The same holds for $E=$ the algebraic closure of $Bbb{Q}$ in $Bbb{C}$.
– Jyrki Lahtonen
Jan 2 at 20:49
1
1
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
@roi_saumon $Bbb{F}_{27}$ has no subfield isomorphic to $Bbb{F}_9$. Remember that $Bbb{F}_{p^n}$ is a subfield of $Bbb{F}_{p^m}$ only when $nmid m$. That is why I use the factorial $ell!$ as the exponent in the cardinality of $K_ell$.
– Jyrki Lahtonen
Jan 2 at 23:33
1
1
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
See Wikipedia for a description of the Galois group of $E/Bbb{F}_p$. Over $Bbb{Q}$ we have a very complicated group.
– Jyrki Lahtonen
Jan 2 at 23:42
1
1
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
Also here for a nice but brief account of why $hat{Bbb{Z}}$ really parametrizes the elements of the Galois group
– Jyrki Lahtonen
Jan 2 at 23:46
1
1
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
This thread describes $hat{Bbb{Z}}$ in terms of $p$-adic rings of integers. I have written an answer explaining why the union $F$ of $Bbb{F}_{p^{2^ell}}$ has uncountably many automorphisms (in that case the automorphism group is the ring of 2-adic integers), but may be I shouldn't link to it. A) self-advertising is bad, B) I skipped why automorphisms of $F$ can be extended to the algebraic closure.
– Jyrki Lahtonen
Jan 3 at 8:18
|
show 1 more comment
An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $Esupset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $tmapsto t+n$, where $ninBbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.
Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $tmapsto t+n$, there being infinitely many different such substitutions.
And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions ${frac1{t-alpha}}$ are all $k$-linearly independent.
Thus you have a situation where the original group $GcongBbb Z$ is countable, and even its completion $hat{Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.
answered Jan 3 at 23:00
Lubin
43.8k44585
add a comment |
An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $Esupset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $tmapsto t+n$, where $ninBbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.
Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $tmapsto t+n$, there being infinitely many different such substitutions.
And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions ${frac1{t-alpha}}$ are all $k$-linearly independent.
Thus you have a situation where the original group $GcongBbb Z$ is countable, and even its completion $hat{Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.
answered Jan 3 at 23:00
Lubin
43.8k44585
add a comment |
An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $Esupset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $tmapsto t+n$, where $ninBbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.
Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $tmapsto t+n$, there being infinitely many different such substitutions.
And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions ${frac1{t-alpha}}$ are all $k$-linearly independent.
Thus you have a situation where the original group $GcongBbb Z$ is countable, and even its completion $hat{Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.
answered Jan 3 at 23:00
Lubin
43.8k44585
An example never to be ignored is the case when $k$ is a field of characteristic zero, and $E=k(t)$, $t$ being an indeterminate, so that the extension $Esupset k$ is simple transcendental. Now consider the group $G$ of all automorphisms of $E$ of form $tmapsto t+n$, where $ninBbb Z$. That is, if $f(t)$ is a $k$-rational expression in $t$ the image of $f$ is to be $f(t+n)$. You see that this is a good automorphism of $E$, and certainly the group it generates is countable infinite.
Now, the fixed field of $G$ is $k$ itself, ’cause only a constant rational function is unchanged by the substitution $tmapsto t+n$, there being infinitely many different such substitutions.
And what is the dimension of $k(t)=E$ as a $k$-vector space? It is at least the cardinality of $k$ (!), since you see that the functions ${frac1{t-alpha}}$ are all $k$-linearly independent.
Thus you have a situation where the original group $GcongBbb Z$ is countable, and even its completion $hat{Bbb Z}$ is merely of continuum cardinality, whereas $k$ itself might have any cardinality whatever that’s at least countable.
answered Jan 3 at 23:00
Lubin
43.8k44585
answered Jan 3 at 23:00
Lubin
43.8k44585
answered Jan 3 at 23:00
Lubin
43.8k44585
answered Jan 3 at 23:00
Lubin
43.8k44585
43.8k44585
add a comment |
add a comment |
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1
To make sure you have heard of this. The Galois correspondence for infinite extensions takes a slightly different form. We need to give the Galois group the structure of a topological group. That way the correspondence survives in the form that the intermediate field are the fixed fields of closed subgroups. The fixed fields of a subgroup and its closure are the same. You need to consult a textbook for those pieces of theory.
– Jyrki Lahtonen
Jan 3 at 18:37
Oh, this is already Galois theory level 16.
– roi_saumon
Jan 3 at 18:54
1
Of course, @JyrkiLahtonen, you’re assuming that the total extension is algebraic, which OP did not do.
– Lubin
Jan 3 at 23:02
@Lubin Correct. That was an oversight.
– Jyrki Lahtonen
2 days ago