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

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?

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

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).

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, $$x^2+1=0$$.

• 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? Sep 17, 2020 at 8:26
• 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. Sep 17, 2020 at 8:47
• 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 $\beta$ Sep 17, 2020 at 8:55
• Right you are! I have edited my answer. Thank you. Sep 17, 2020 at 9:53

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 $$x^2 + 1 = 0$$.

• 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 $x^2+1=0$ Sep 17, 2020 at 7:52
• 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.
– user208649
Sep 17, 2020 at 7:57
• 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.
– user208649
Sep 17, 2020 at 7:59
• You are right to be honest, this proof is not very well written I must say Sep 17, 2020 at 8:09