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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 ? .

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  • $\begingroup$ 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
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$\textit{Just submitting my comments from above as answer to resolve this question.}$


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.

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    $\begingroup$ 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
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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.

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