Countable ordered subfield of any Ordered Field
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When we talk about any ordered field($mathbb{k}$), we can always generate the rational elements of the field($mathbb{Q_{k}}$) by first generating the natural elements($mathbb{N_{k}}$) and the integer elements($mathbb{Z_{k}}$) entirely from the multiplicative identity($mathbb{1_{k}}$). We all know that $mathbb{Q_{k}}$ is a countable subfield $mathbb{k}$.
Now I want to ask this question that is it possible to find another countable subfield of $mathbb{k}$ other than that of $mathbb{Q_{k}}$? If that is possible to find, will it be order isomorphic to $mathbb{Q_{k}}$?
Will the field $mathbb{k}$ being Archimedean or non-Archimedean alter the results to the previous question?
P.S. I was trying to find this countable subfield in an ordered field both in particular examples such as that of $mathbb{R}$(Archimedean) and Rational functions defined on an integral domain(non-Archimedean). In both the cases I was unable to find so. Although I strongly believe that it is not possible to find another countable field in an Archimedean field, but I am not sure for the non-Archimedean case.
Thanks in advance!
field-theory ordered-fields
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add a comment |
$begingroup$
When we talk about any ordered field($mathbb{k}$), we can always generate the rational elements of the field($mathbb{Q_{k}}$) by first generating the natural elements($mathbb{N_{k}}$) and the integer elements($mathbb{Z_{k}}$) entirely from the multiplicative identity($mathbb{1_{k}}$). We all know that $mathbb{Q_{k}}$ is a countable subfield $mathbb{k}$.
Now I want to ask this question that is it possible to find another countable subfield of $mathbb{k}$ other than that of $mathbb{Q_{k}}$? If that is possible to find, will it be order isomorphic to $mathbb{Q_{k}}$?
Will the field $mathbb{k}$ being Archimedean or non-Archimedean alter the results to the previous question?
P.S. I was trying to find this countable subfield in an ordered field both in particular examples such as that of $mathbb{R}$(Archimedean) and Rational functions defined on an integral domain(non-Archimedean). In both the cases I was unable to find so. Although I strongly believe that it is not possible to find another countable field in an Archimedean field, but I am not sure for the non-Archimedean case.
Thanks in advance!
field-theory ordered-fields
$endgroup$
$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
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– Torsten Schoeneberg
Dec 22 '18 at 7:18
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Yes, I got it. Thanks!
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– Satwata Hans
Dec 22 '18 at 7:20
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@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
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– Satwata Hans
Dec 22 '18 at 8:07
1
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14
add a comment |
$begingroup$
When we talk about any ordered field($mathbb{k}$), we can always generate the rational elements of the field($mathbb{Q_{k}}$) by first generating the natural elements($mathbb{N_{k}}$) and the integer elements($mathbb{Z_{k}}$) entirely from the multiplicative identity($mathbb{1_{k}}$). We all know that $mathbb{Q_{k}}$ is a countable subfield $mathbb{k}$.
Now I want to ask this question that is it possible to find another countable subfield of $mathbb{k}$ other than that of $mathbb{Q_{k}}$? If that is possible to find, will it be order isomorphic to $mathbb{Q_{k}}$?
Will the field $mathbb{k}$ being Archimedean or non-Archimedean alter the results to the previous question?
P.S. I was trying to find this countable subfield in an ordered field both in particular examples such as that of $mathbb{R}$(Archimedean) and Rational functions defined on an integral domain(non-Archimedean). In both the cases I was unable to find so. Although I strongly believe that it is not possible to find another countable field in an Archimedean field, but I am not sure for the non-Archimedean case.
Thanks in advance!
field-theory ordered-fields
$endgroup$
When we talk about any ordered field($mathbb{k}$), we can always generate the rational elements of the field($mathbb{Q_{k}}$) by first generating the natural elements($mathbb{N_{k}}$) and the integer elements($mathbb{Z_{k}}$) entirely from the multiplicative identity($mathbb{1_{k}}$). We all know that $mathbb{Q_{k}}$ is a countable subfield $mathbb{k}$.
Now I want to ask this question that is it possible to find another countable subfield of $mathbb{k}$ other than that of $mathbb{Q_{k}}$? If that is possible to find, will it be order isomorphic to $mathbb{Q_{k}}$?
Will the field $mathbb{k}$ being Archimedean or non-Archimedean alter the results to the previous question?
P.S. I was trying to find this countable subfield in an ordered field both in particular examples such as that of $mathbb{R}$(Archimedean) and Rational functions defined on an integral domain(non-Archimedean). In both the cases I was unable to find so. Although I strongly believe that it is not possible to find another countable field in an Archimedean field, but I am not sure for the non-Archimedean case.
Thanks in advance!
field-theory ordered-fields
field-theory ordered-fields
edited Dec 22 '18 at 7:14
Satwata Hans
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
52
$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 7:18
$begingroup$
Yes, I got it. Thanks!
$endgroup$
– Satwata Hans
Dec 22 '18 at 7:20
$begingroup$
@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
$endgroup$
– Satwata Hans
Dec 22 '18 at 8:07
1
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14
add a comment |
$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 7:18
$begingroup$
Yes, I got it. Thanks!
$endgroup$
– Satwata Hans
Dec 22 '18 at 7:20
$begingroup$
@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
$endgroup$
– Satwata Hans
Dec 22 '18 at 8:07
1
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14
$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 7:18
$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 7:18
$begingroup$
Yes, I got it. Thanks!
$endgroup$
– Satwata Hans
Dec 22 '18 at 7:20
$begingroup$
Yes, I got it. Thanks!
$endgroup$
– Satwata Hans
Dec 22 '18 at 7:20
$begingroup$
@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
$endgroup$
– Satwata Hans
Dec 22 '18 at 8:07
$begingroup$
@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
$endgroup$
– Satwata Hans
Dec 22 '18 at 8:07
1
1
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14
add a comment |
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$begingroup$
All number fields are countable, and many of them are contained in $Bbb R$, hence have an archimedean order. The first non-trivial examples are the quadratic number fields $Bbb Q (sqrt d)$ for square-free positive integers $d$. -- As for non-archimedean orders: for any countable field $F$, the rational function field $ F (x)$ is also countable.
$endgroup$
– Torsten Schoeneberg
Dec 22 '18 at 7:18
$begingroup$
Yes, I got it. Thanks!
$endgroup$
– Satwata Hans
Dec 22 '18 at 7:20
$begingroup$
@TorstenSchoeneberg, I want to ask another question. Like we can uniquely characterise $mathbb(R)$ as a complete ordered field, is there a possible unique charactisation of $mathbb(Q)$?
$endgroup$
– Satwata Hans
Dec 22 '18 at 8:07
1
$begingroup$
It's the minimal ordered field.
$endgroup$
– Slade
Dec 22 '18 at 8:14