Mixing
Field isomorphic to a transcendental extension of itself
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
Let $k$ be any field and $K:=k(X_1,X_2,dots)$ be the field of rational functions over $k$ in countably many variables. Now $K$ has the interesting property that it is isomorphic to a transcendental extension of itself namely $Kcong K(X)$.
Are there any other examples of this phenomenon or is the following true?
When $K$ is a field that is isomorphic to a transcendental extension of itself, then there is some field $k$ s.t. $Kcong k(X_1,X_2,dots)$.
abstract-algebra field-theory
asked Jul 18 at 21:03
Achilles
819416
add a comment |Â
up vote
5
down vote
favorite
Let $k$ be any field and $K:=k(X_1,X_2,dots)$ be the field of rational functions over $k$ in countably many variables. Now $K$ has the interesting property that it is isomorphic to a transcendental extension of itself namely $Kcong K(X)$.
Are there any other examples of this phenomenon or is the following true?
When $K$ is a field that is isomorphic to a transcendental extension of itself, then there is some field $k$ s.t. $Kcong k(X_1,X_2,dots)$.
abstract-algebra field-theory
asked Jul 18 at 21:03
Achilles
819416
1
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $k$ be any field and $K:=k(X_1,X_2,dots)$ be the field of rational functions over $k$ in countably many variables. Now $K$ has the interesting property that it is isomorphic to a transcendental extension of itself namely $Kcong K(X)$.
Are there any other examples of this phenomenon or is the following true?
When $K$ is a field that is isomorphic to a transcendental extension of itself, then there is some field $k$ s.t. $Kcong k(X_1,X_2,dots)$.
abstract-algebra field-theory
asked Jul 18 at 21:03
Achilles
819416
Let $k$ be any field and $K:=k(X_1,X_2,dots)$ be the field of rational functions over $k$ in countably many variables. Now $K$ has the interesting property that it is isomorphic to a transcendental extension of itself namely $Kcong K(X)$.
Are there any other examples of this phenomenon or is the following true?
When $K$ is a field that is isomorphic to a transcendental extension of itself, then there is some field $k$ s.t. $Kcong k(X_1,X_2,dots)$.
abstract-algebra field-theory
asked Jul 18 at 21:03
Achilles
819416
edited Jul 18 at 21:14
asked Jul 18 at 21:03
Achilles
819416
asked Jul 18 at 21:03
Achilles
819416
asked Jul 18 at 21:03
Achilles
819416
819416
1
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06
add a comment |Â
1
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06
1
1
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
As written, no.
$mathbb C$ is isomorphic to $overlinemathbb C(X)$ which is a transcendental extension, but $mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).
answered Jul 18 at 21:34
Henning Makholm
226k16291519
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
As written, no.
$mathbb C$ is isomorphic to $overlinemathbb C(X)$ which is a transcendental extension, but $mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).
answered Jul 18 at 21:34
Henning Makholm
226k16291519
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
add a comment |Â
up vote
6
down vote
accepted
As written, no.
$mathbb C$ is isomorphic to $overlinemathbb C(X)$ which is a transcendental extension, but $mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).
answered Jul 18 at 21:34
Henning Makholm
226k16291519
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
As written, no.
$mathbb C$ is isomorphic to $overlinemathbb C(X)$ which is a transcendental extension, but $mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).
answered Jul 18 at 21:34
Henning Makholm
226k16291519
As written, no.
$mathbb C$ is isomorphic to $overlinemathbb C(X)$ which is a transcendental extension, but $mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).
answered Jul 18 at 21:34
Henning Makholm
226k16291519
edited Jul 19 at 1:03
answered Jul 18 at 21:34
Henning Makholm
226k16291519
answered Jul 18 at 21:34
Henning Makholm
226k16291519
answered Jul 18 at 21:34
Henning Makholm
226k16291519
226k16291519
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
add a comment |Â
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
Why is $mathbbC$ isomorphic to $overlinemathbbC(X)$?
– Unit
Jul 18 at 22:05
2
2
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
@Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic.
– Achilles
Jul 18 at 23:24
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
I see that by considering the theory consisting of the diagram of $K(X_1,X_2,dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$?
– Achilles
Jul 18 at 23:29
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
@Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim.
– Henning Makholm
Jul 19 at 1:04
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856003%2ffield-isomorphic-to-a-transcendental-extension-of-itself%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Follow-up question which is what I thought your question was at first: math.stackexchange.com/questions/2856060/…
– Eric Wofsey
Jul 18 at 22:06