Finite field extension of $\mathbb{R}$, why must it admit an element such that $x^2+1=0$ Proof understanding.

Question

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?

Answer 1

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, $x^2+1=0$.

Answer 2

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).

Answer 3

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 $x^2 + 1 = 0$.