Degree of extension of fixed field by infinite set of automorphisms.












0














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







share|cite|improve this question




















  • 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
















0














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







share|cite|improve this question




















  • 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














0












0








0


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 21:45









user26857

39.3k123983




39.3k123983










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














  • 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










2 Answers
2






active

oldest

votes


















5














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.






share|cite|improve this answer























  • 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



















2














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.






share|cite|improve this answer





















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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.






    share|cite|improve this answer























    • 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
















    5














    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.






    share|cite|improve this answer























    • 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














    5












    5








    5






    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 18:32









    Kenny Lau

    19.8k2159




    19.8k2159










    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


















    • 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











    2














    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.






    share|cite|improve this answer


























      2














      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.






      share|cite|improve this answer
























        2












        2








        2






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 23:00









        Lubin

        43.8k44585




        43.8k44585






























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