Is there any invariant that is preserved when you change the underlying field of a vector space?
1
$begingroup$
If you change the field you change the vector space.
The vector space is defined by the triad: field + set + operations, if you change one of these (in your case field), you change the space, so you can not have an endormorphism.
I am in this hypothetical vector space over A complex field which there is a passage or 'transition' from this complex field into a real field
Complex Field $rightarrow$ Real Field
I'm not interested to change the vector space but seems that is not possible if you change field because you change vector space if you change field.
Is there any invariant that is preserved when you change the underlying field of a vector space ?
What should be preserved, what operation or structure emerges if we pass from the vector space over a complex field to a vector space over real field or viceversa?
vector-spaces field-theory
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
$endgroup$
|
show 3 more comments
1
$begingroup$
If you change the field you change the vector space.
The vector space is defined by the triad: field + set + operations, if you change one of these (in your case field), you change the space, so you can not have an endormorphism.
I am in this hypothetical vector space over A complex field which there is a passage or 'transition' from this complex field into a real field
Complex Field $rightarrow$ Real Field
I'm not interested to change the vector space but seems that is not possible if you change field because you change vector space if you change field.
Is there any invariant that is preserved when you change the underlying field of a vector space ?
What should be preserved, what operation or structure emerges if we pass from the vector space over a complex field to a vector space over real field or viceversa?
vector-spaces field-theory
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
$endgroup$
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30
|
show 3 more comments
1
1
1
0
$begingroup$
If you change the field you change the vector space.
The vector space is defined by the triad: field + set + operations, if you change one of these (in your case field), you change the space, so you can not have an endormorphism.
I am in this hypothetical vector space over A complex field which there is a passage or 'transition' from this complex field into a real field
Complex Field $rightarrow$ Real Field
I'm not interested to change the vector space but seems that is not possible if you change field because you change vector space if you change field.
Is there any invariant that is preserved when you change the underlying field of a vector space ?
What should be preserved, what operation or structure emerges if we pass from the vector space over a complex field to a vector space over real field or viceversa?
vector-spaces field-theory
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
$endgroup$
If you change the field you change the vector space.
The vector space is defined by the triad: field + set + operations, if you change one of these (in your case field), you change the space, so you can not have an endormorphism.
I am in this hypothetical vector space over A complex field which there is a passage or 'transition' from this complex field into a real field
Complex Field $rightarrow$ Real Field
I'm not interested to change the vector space but seems that is not possible if you change field because you change vector space if you change field.
Is there any invariant that is preserved when you change the underlying field of a vector space ?
What should be preserved, what operation or structure emerges if we pass from the vector space over a complex field to a vector space over real field or viceversa?
vector-spaces field-theory
vector-spaces field-theory
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
edited Dec 19 '18 at 13:49
Emilio Novati
52.2k43474
52.2k43474
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
asked Dec 19 '18 at 13:14
Peter LongPeter Long
194
194
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30
|
show 3 more comments
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30
|
show 3 more comments
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
});
}
});
draft saved
draft discarded
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%2f3046376%2fis-there-any-invariant-that-is-preserved-when-you-change-the-underlying-field-of%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3046376%2fis-there-any-invariant-that-is-preserved-when-you-change-the-underlying-field-of%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
$begingroup$
You cannot "change field" in such generality. What you can do is constructing a complex vector space if you are given a real one; this is called complexification. The opposite operation can be done in greater generality; if you are given fields $F_1subset F_2$ then an $F_2$ vector space can be regarded as a $F_1$ vector space. (For example, any complex vector space is also a real vector space). That's all you can do.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 13:38
$begingroup$
So when I "change field" is it really the very definition of a field that must be preserved? I read that for complexification we need to set a relation between field extension and its subfield or about morphism of rings, but..this is an generic Ring homomorphism or is it something more specific (inner automorphism, endomorphism..)? For realization or decomplexification how removes the possibility of complex multiplication of scalars ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:00
$begingroup$
And if I want to change from a real field to rational field we need to talk about rationalization ? Inverse process from rational field to real field is a sort of realization but not like a decomplexification ?
$endgroup$
– Peter Long
Dec 20 '18 at 10:03
$begingroup$
I ask about "how removes the possibility of complex multiplication of scalars" because I read that When the underlying ring is commutative, for example, the real or complex number field, these two multiplications are the same, and are simply called scalar multiplication. $rightarrow$ It gives me the impression that here is removed the scalar multiplication itself before, not only that "removes the possibility" of complex multiplication of scalars then
$endgroup$
– Peter Long
Dec 20 '18 at 10:11
$begingroup$
I have a hard time following what you wrote, it is very complicated. These things should be easy; if you have a complex vector space, it is obviously also a real one; if you have a real one, it is also obviously a rational one, and this is true for all fields which include a subfield. That's it. Complexification is a different operation, that is tailored to complex numbers.
$endgroup$
– Giuseppe Negro
Dec 20 '18 at 11:30