Prove that every finite extension field of $\Bbb R$ is either $\Bbb R$ itself or isomorphic to $\Bbb C$.
I tried in this way.Let $E$ be a a finite extension of $\Bbb R$. Then $E$ is an algebraic extension of $\Bbb R$. Let $\alpha \in E-R$ .then $\exists p(x)\in \Bbb R[x]$ such that $p(\alpha)=0$. Surely $\deg(p(x)=2$ since otherwise it would be reducible.
Now $\Bbb R(\alpha)\cong \Bbb R[x]/<p(x)>$. Now $[\Bbb R(\alpha):\Bbb R]=2$ .Also $[\Bbb C:\Bbb R]=2\implies \Bbb R(\alpha)\cong \Bbb C$.
Since $\Bbb C$ is algebraically closed so is $\Bbb R(\alpha)$ and so it has no proper algebraic extensions.
Can I conclude from here that $E=\Bbb C$?
Please suggest steps if needed.I would be very grateful