Filtered Colimit of Subalgebras

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Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $mathcalS$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $textHom_k text-linear (K, K) cong textcolimit_F in mathcalS textHom_F (K, K)$ as $k$-algebras under composition?

field-theory limits-colimits

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edited Jul 22 at 19:11

asked Jul 22 at 17:45

Dean Young

1,403719

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  1
  

  
  
  
  
  
  

  
  So $K/k$ is not necessarily module-finite. I see.
  – AlgebraicsAnonymous
  Jul 22 at 19:38
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $mathcalS$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $textHom_k text-linear (K, K) cong textcolimit_F in mathcalS textHom_F (K, K)$ as $k$-algebras under composition?

field-theory limits-colimits

share|cite|improve this question

edited Jul 22 at 19:11

asked Jul 22 at 17:45

Dean Young

1,403719

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So $K/k$ is not necessarily module-finite. I see.
  – AlgebraicsAnonymous
  Jul 22 at 19:38
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $mathcalS$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $textHom_k text-linear (K, K) cong textcolimit_F in mathcalS textHom_F (K, K)$ as $k$-algebras under composition?

field-theory limits-colimits

share|cite|improve this question

edited Jul 22 at 19:11

asked Jul 22 at 17:45

Dean Young

1,403719

Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $mathcalS$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $textHom_k text-linear (K, K) cong textcolimit_F in mathcalS textHom_F (K, K)$ as $k$-algebras under composition?

field-theory limits-colimits

share|cite|improve this question

edited Jul 22 at 19:11

asked Jul 22 at 17:45

Dean Young

1,403719

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 19:11

edited Jul 22 at 19:11

edited Jul 22 at 19:11

asked Jul 22 at 17:45

Dean Young

1,403719

asked Jul 22 at 17:45

Dean Young

1,403719

asked Jul 22 at 17:45

Dean Young

1,403719

1,403719

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So $K/k$ is not necessarily module-finite. I see.
  – AlgebraicsAnonymous
  Jul 22 at 19:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So $K/k$ is not necessarily module-finite. I see.
  – AlgebraicsAnonymous
  Jul 22 at 19:38
  

  
  
  
  

1

1

So $K/k$ is not necessarily module-finite. I see.
– AlgebraicsAnonymous
Jul 22 at 19:38

So $K/k$ is not necessarily module-finite. I see.
– AlgebraicsAnonymous
Jul 22 at 19:38

add a comment | 

2 Answers
2

active

oldest

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2
down vote



accepted

No. For instance, if $k=mathbbQ$ and $K=mathbbQ(x)$, then $mathcalS$ is countable and $operatornameHom_F(K,K)$ is countable for any $Fin S$, so the colimit is countable. However, $operatornameHom_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

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answered Jul 22 at 18:25

Eric Wofsey

162k12189300

add a comment | 

up vote
-1
down vote

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k le S le T le K$ the existence of a forgetful functor
$$
mathcal C_T to mathcal C_S,
$$
where $mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
operatornameAut_T(K) to operatornameAut_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EricWofsey Please consider my notation.
  – AlgebraicsAnonymous
  Jul 22 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
  – Dean Young
  Jul 22 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
  – AlgebraicsAnonymous
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

No. For instance, if $k=mathbbQ$ and $K=mathbbQ(x)$, then $mathcalS$ is countable and $operatornameHom_F(K,K)$ is countable for any $Fin S$, so the colimit is countable. However, $operatornameHom_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

share|cite|improve this answer

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

add a comment | 

up vote
2
down vote



accepted

No. For instance, if $k=mathbbQ$ and $K=mathbbQ(x)$, then $mathcalS$ is countable and $operatornameHom_F(K,K)$ is countable for any $Fin S$, so the colimit is countable. However, $operatornameHom_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

share|cite|improve this answer

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

No. For instance, if $k=mathbbQ$ and $K=mathbbQ(x)$, then $mathcalS$ is countable and $operatornameHom_F(K,K)$ is countable for any $Fin S$, so the colimit is countable. However, $operatornameHom_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

share|cite|improve this answer

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

No. For instance, if $k=mathbbQ$ and $K=mathbbQ(x)$, then $mathcalS$ is countable and $operatornameHom_F(K,K)$ is countable for any $Fin S$, so the colimit is countable. However, $operatornameHom_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

share|cite|improve this answer

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

answered Jul 22 at 18:25

Eric Wofsey

162k12189300

162k12189300

add a comment | 

add a comment | 

up vote
-1
down vote

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k le S le T le K$ the existence of a forgetful functor
$$
mathcal C_T to mathcal C_S,
$$
where $mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
operatornameAut_T(K) to operatornameAut_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EricWofsey Please consider my notation.
  – AlgebraicsAnonymous
  Jul 22 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
  – Dean Young
  Jul 22 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
  – AlgebraicsAnonymous
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

up vote
-1
down vote

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k le S le T le K$ the existence of a forgetful functor
$$
mathcal C_T to mathcal C_S,
$$
where $mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
operatornameAut_T(K) to operatornameAut_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EricWofsey Please consider my notation.
  – AlgebraicsAnonymous
  Jul 22 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
  – Dean Young
  Jul 22 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
  – AlgebraicsAnonymous
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

up vote
-1
down vote

up vote
-1
down vote

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k le S le T le K$ the existence of a forgetful functor
$$
mathcal C_T to mathcal C_S,
$$
where $mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
operatornameAut_T(K) to operatornameAut_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k le S le T le K$ the existence of a forgetful functor
$$
mathcal C_T to mathcal C_S,
$$
where $mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
operatornameAut_T(K) to operatornameAut_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

answered Jul 22 at 18:28

AlgebraicsAnonymous

69111

69111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EricWofsey Please consider my notation.
  – AlgebraicsAnonymous
  Jul 22 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
  – Dean Young
  Jul 22 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
  – AlgebraicsAnonymous
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EricWofsey Please consider my notation.
  – AlgebraicsAnonymous
  Jul 22 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
  – Dean Young
  Jul 22 at 19:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
  – AlgebraicsAnonymous
  Jul 22 at 19:58
  

  
  
  
  

@EricWofsey Please consider my notation.
– AlgebraicsAnonymous
Jul 22 at 19:10

@EricWofsey Please consider my notation.
– AlgebraicsAnonymous
Jul 22 at 19:10

Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
– Dean Young
Jul 22 at 19:56

Sorry for the typo. I meant for $S$ to be the finite index intermediate fields of $K/k$.
– Dean Young
Jul 22 at 19:56

That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
– AlgebraicsAnonymous
Jul 22 at 19:58

That's what I guessed, but I didn't assume that the extension $K/k$ may be infinite.
– AlgebraicsAnonymous
Jul 22 at 19:58

add a comment | 

 

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