Why does $\deg p(x)=2$ in this older question about finite extension of $\Bbb R$

Question

I was reading this older question and at some point the OP says "Surely $\deg (p(x))=2$ since otherwise it would be reducible."

Why is that true?

Answer 1

It really depends on what you already know (or, at least, what you're logically allowed to assume). For instance, if you know that $\Bbb C$ is the algebraic closure of $\Bbb R$, then it's because an irreducible polynomial of degree $n$ gives rise to an algebraic extension of degree at least $n$, and any algebraic extension of $\Bbb R$ is either trivial (i.e. just $\Bbb R$ itself again) or $\Bbb C$ (which has degree $2$).

If you don't know this, then what you're looking for is a proof of the fundamental theorem of algebra. There are several proofs just in that wikipedia article, and there are many more around. It's one of those theorems mathematicians like to come up with new proofs for.