Mixing
Question about the fundamental theorem of field theory
$begingroup$
In this proof, I understand the logic that E contains a copy of F and so we are considering E as an extension of F . But, why should we? Why is E considered as an extension of F even though F is not actually a subfield of E ? .
field-theory
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
$endgroup$
|
show 1 more comment
$begingroup$
In this proof, I understand the logic that E contains a copy of F and so we are considering E as an extension of F . But, why should we? Why is E considered as an extension of F even though F is not actually a subfield of E ? .
field-theory
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
$endgroup$
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54
|
show 1 more comment
$begingroup$
In this proof, I understand the logic that E contains a copy of F and so we are considering E as an extension of F . But, why should we? Why is E considered as an extension of F even though F is not actually a subfield of E ? .
field-theory
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
$endgroup$
In this proof, I understand the logic that E contains a copy of F and so we are considering E as an extension of F . But, why should we? Why is E considered as an extension of F even though F is not actually a subfield of E ? .
field-theory
field-theory
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
edited Feb 1 at 8:38
José Carlos Santos
173k23133242
173k23133242
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
asked Feb 1 at 8:26
GaneshbabuGaneshbabu
82
82
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54
|
show 1 more comment
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
$textit{Just submitting my comments from above as answer to resolve this question.}$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism.
Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[x] / (x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[x]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
$endgroup$
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
add a comment |
$begingroup$
If f is a polynomial of degree d, then an element of $F[x]/(f)$, looks like $a_0+a_1X+...a_{d-1}X^{d-1}$, ie polynomials in F[x] of degree at most d-1. In particular, it contains all polynomials which are constant, ie it contains F.
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
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%2f3095979%2fquestion-about-the-fundamental-theorem-of-field-theory%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$textit{Just submitting my comments from above as answer to resolve this question.}$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism.
Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[x] / (x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[x]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
$endgroup$
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
add a comment |
$begingroup$
$textit{Just submitting my comments from above as answer to resolve this question.}$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism.
Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[x] / (x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[x]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
$endgroup$
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
add a comment |
$begingroup$
$textit{Just submitting my comments from above as answer to resolve this question.}$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism.
Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[x] / (x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[x]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
$endgroup$
$textit{Just submitting my comments from above as answer to resolve this question.}$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism.
Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[x] / (x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[x]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
answered Feb 2 at 12:58
Adam HigginsAdam Higgins
615113
615113
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
add a comment |
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
1
1
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
Thank you . Just to clarify, I've accepted this answer and also upvoted it . But, it's not getting displayed since I don't have 15 points yet :) .
$endgroup$
– Ganeshbabu
Feb 2 at 13:08
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
$begingroup$
@Ganeshbabu oh sure! No worries :)
$endgroup$
– Adam Higgins
Feb 2 at 13:09
add a comment |
$begingroup$
If f is a polynomial of degree d, then an element of $F[x]/(f)$, looks like $a_0+a_1X+...a_{d-1}X^{d-1}$, ie polynomials in F[x] of degree at most d-1. In particular, it contains all polynomials which are constant, ie it contains F.
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
If f is a polynomial of degree d, then an element of $F[x]/(f)$, looks like $a_0+a_1X+...a_{d-1}X^{d-1}$, ie polynomials in F[x] of degree at most d-1. In particular, it contains all polynomials which are constant, ie it contains F.
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
If f is a polynomial of degree d, then an element of $F[x]/(f)$, looks like $a_0+a_1X+...a_{d-1}X^{d-1}$, ie polynomials in F[x] of degree at most d-1. In particular, it contains all polynomials which are constant, ie it contains F.
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
$endgroup$
If f is a polynomial of degree d, then an element of $F[x]/(f)$, looks like $a_0+a_1X+...a_{d-1}X^{d-1}$, ie polynomials in F[x] of degree at most d-1. In particular, it contains all polynomials which are constant, ie it contains F.
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
answered Feb 1 at 9:08
AlexandrosAlexandros
1,0151413
1,0151413
add a comment |
add a comment |
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%2f3095979%2fquestion-about-the-fundamental-theorem-of-field-theory%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
$begingroup$
Because in abstract algebra, we often (read almost always) classify things only up to isomorphism. There are many reasons for this, but perhaps the most simple is that we only really care about understanding the structure rather than the specific realisation of this structure. You might have read the statement that there is a unique finite field of each prime power order, but this is a statement only up to isomorphism. Thus we say that one field is an extension of another, whenever there is an injection from one into the other.
$endgroup$
– Adam Higgins
Feb 1 at 10:05
$begingroup$
You might have seen before that the Algebraic construction of the complex numbers is given by $mathbb{R}[X]/(x^{2} + 1)$. So technically speaking in this construction, every element of $mathbb{C}$ is a representative for an equivalence class in $mathbb{R}[X]$. Thus you could say that set theoretically, in this construction, $mathbb{C}$ does not literally contain $mathbb{R}$ as a subfield, but there is an obvious injection from $mathbb{R}$ into $mathbb{C}$. Thus $mathbb{R}$ is a subfield of $mathbb{C}$ in the only way it really matters.
$endgroup$
– Adam Higgins
Feb 1 at 10:09
$begingroup$
Ah, I got it, thank you !
$endgroup$
– Ganeshbabu
Feb 1 at 23:25
$begingroup$
you might wanna accept an answer, or at least upvote the answers you found useful. If only to take the question out of the unresolved pile. If you don’t think the question is resolved and still have queries feel free to ask and I might try and write out a longer answer?
$endgroup$
– Adam Higgins
Feb 2 at 10:28
$begingroup$
No, i think your answer is great and enough . And I tried to upvote your answer, but, apparently i cannot . It says "You cannot vote for this comment" .
$endgroup$
– Ganeshbabu
Feb 2 at 12:54