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The proposition is as follows and I read the proof but I am not so certain regarding a particular point the proof has made:

Any finite extension of $\mathbb{R}$ is at most degree $2$.

Proof: Suppose the field extension $\mathbb{F}$ is non-trivial and thus there must exist $\alpha\in\mathbb{F}\setminus\mathbb{R}.$ Since the extension is finite then $\alpha$ must be $\mathbb{F}$-algebraic. In particular its minimal polynomial must be quadratic since it is not in $\mathbb{R}.$ Hence there must exist an element $x\in\mathbb{F}$ such that $x^2+1=0.$ [The rest of the proof is quite understandable.]

My question is why is it guaranteed that such $x$ must exist? I get that the minimal polynomial must be in the form of $m_\alpha(x)=(x-p)(x-\overline{p})$ for some $p\in\mathbb{C}\setminus\mathbb{R}$ but does that do much?

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    $\begingroup$ Are you confused about why it is quadratic or why you can find an element satisfying $x^2+1=0$? $\endgroup$
    – user208649
    Sep 17, 2020 at 7:27
  • $\begingroup$ @TokenToucan Hello, I am more confused about the latter; why can we always find $x$ that $x^2+1=0$ $\endgroup$ Sep 17, 2020 at 7:30

3 Answers 3

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Since $\mathbb{C}$ is algebraically closed all finite extension of $\mathbb{R}$ embedded in $\mathbb{C}$ but since the degree of
$\mathbb{C}$ over $\mathbb{R}$ is two and all non trivial extension of $\mathbb{R}$ have degree more than two all non trivial finite extension of $\mathbb{R}$ have degree 2 and by equality of degree all finite extension is equal to $\mathbb{C}$ hence have x such that $x^2+1=0$

Another proof : If the minimal polynomial is $(x-\alpha )(x +\bar{\alpha} )$ then $\alpha - \bar{\alpha}=2\operatorname{Im}(\alpha)i $ and since $2\operatorname{Im}(\alpha )\in \mathbb{R}$ we know that $\frac{2\operatorname{Im}(\alpha)i }{2\operatorname{Im}(\alpha)} = i$ is in $\mathbb F$ (by closure).

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The minimal polynomial of $\alpha$ is of the form $x^2+\beta x+\gamma$. Since it is irreducible over $\Bbb R$, $\beta^2-4\gamma<0$. You know then that $\alpha^2+\beta\alpha+\gamma=0$. In other words,$$\left(\alpha-\frac\beta2\right)^2+\gamma-\frac\beta4=0.$$So, take$$x=\frac{\alpha-\frac\beta2}{\sqrt{\gamma-\frac{\beta^2}4}}$$and then$$x^2=\frac{\left(\alpha-\frac\beta2\right)^2}{\gamma-\frac{\beta^2}4}=-1.$$In other words, $x^2+1=0$.

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  • $\begingroup$ Hi Jose, thank you for your answer! I am just thinking would your solution potentially assume that $\mathbb{F}\subset\mathbb{C}$? This is because looking at $x=\frac{\alpha-\beta/2}{\sqrt{\gamma-\beta/4}}$, it could be the case that the denominator is actually an imaginary number. Henceforth, for $x\in\mathbb{F}$ wouldn't we need $\mathbb{F}\subset\mathbb{C}$ as an assumption at the start? $\endgroup$ Sep 17, 2020 at 8:26
  • $\begingroup$ My answer has nothing to do with $\Bbb C$ and, since $\beta^2-4\gamma<0$, it is clear that $\gamma-\frac{\beta^2}4>0$. So, no, $\sqrt{\gamma-\frac{\beta^2}4}$ cannot be a complex non-real number. $\endgroup$ Sep 17, 2020 at 8:47
  • $\begingroup$ Ohh right, your answer is exactly what I am looking for! By the way, there were some typos in your answer, in particular you missed the square in $\beta$ $\endgroup$ Sep 17, 2020 at 8:55
  • $\begingroup$ Right you are! I have edited my answer. Thank you. $\endgroup$ Sep 17, 2020 at 9:53
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If you know $\mathbb C$ is algebraically closed, then we may assume $\mathbb F$ is embedded in $\mathbb C$, and in this way view $\alpha$ as being an element of $\mathbb C$.

That means you can write $\alpha = a+bi$ with $a,b$ real. The only time $\alpha$ is not in $\mathbb R$ is when $b\neq 0$ and so $i = \frac{\alpha - a}{b}$ will satisfy $x^2 + 1 = 0$.

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    $\begingroup$ Thank you for your great answer. However though, if I could take your first paragraph as an assumption then would I be right in saying that the proposition becomes trivial? I guess what I am trying to say is if we can assume $\mathbb{F}$ is embedded in $\mathbb{C}$, since $\mathbb{C}$ is a degree $2$ extension of $\mathbb{R}$ then automatically the proposition at the very beginning is proved. Hence there will no longer need to find $x$ such that $x^2+1=0$ $\endgroup$ Sep 17, 2020 at 7:52
  • $\begingroup$ Yes, I suppose it might. However, you can prove that $\mathbb C$ is algebraically closed without using your proposition, so the argument would not be circular, at least. $\endgroup$
    – user208649
    Sep 17, 2020 at 7:57
  • $\begingroup$ Actually, in the proof you wrote, I'm not sure how you'd know that $\alpha$ is quadratic over $\mathbb R$ without already knowing that $\mathbb C$ is the algebraic closure of $\mathbb R$ or something very nearly equivalent. $\endgroup$
    – user208649
    Sep 17, 2020 at 7:59
  • $\begingroup$ You are right to be honest, this proof is not very well written I must say $\endgroup$ Sep 17, 2020 at 8:09

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