Mixing
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
asked Jul 22 at 17:45
Dean Young
1,403719
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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
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 |Â
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
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
asked Jul 22 at 17:45
Dean Young
1,403719
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 |Â
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 |Â
2 Answers
2
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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 $k$.
answered Jul 22 at 18:25
Eric Wofsey
162k12189300
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up vote
-1
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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.
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 |Â
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 $k$.
answered Jul 22 at 18:25
Eric Wofsey
162k12189300
add a comment |Â
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 $k$.
answered Jul 22 at 18:25
Eric Wofsey
162k12189300
add a comment |Â
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 $k$.
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 $k$.
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
answered Jul 22 at 18:25
Eric Wofsey
162k12189300
162k12189300
add a comment |Â
add a comment |Â
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.
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 |Â
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.
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 |Â
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.
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.
answered Jul 22 at 18:28
AlgebraicsAnonymous
69111
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 |Â
@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 |Â
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1
So $K/k$ is not necessarily module-finite. I see.
– AlgebraicsAnonymous
Jul 22 at 19:38