Countable ordered subfield of any Ordered Field
$begingroup$
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
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
$endgroup$
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
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
$endgroup$
$begingroup$
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.
$endgroup$
– 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 |
$begingroup$
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
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
$endgroup$
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
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
edited Dec 22 '18 at 7:14
Satwata Hans
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
asked Dec 22 '18 at 7:04
Satwata HansSatwata Hans
52
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049182%2fcountable-ordered-subfield-of-any-ordered-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049182%2fcountable-ordered-subfield-of-any-ordered-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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