Let $f\in\mathbb{Q}[x]$ be irreducible of odd degree and let $\alpha\in\mathbb{C}$ be a root of $f$ (i.e. $f(\alpha)=0$). Prove that $i\notin\mathbb{Q}(\alpha)$.
I now know that one can prove this via considering the order of the extension, but I at first sight came up with the following proof:
Since $f$ is of odd degree, there exists $\beta\in\mathbb{R}$ such that $f(\beta)=0$. Since $\alpha$ and $\beta$ have the same irreducible polynomial, namely $f$, we have $\mathbb{Q}(\alpha)\simeq\mathbb{Q}(\beta)$ via an isomorphism, say $\varphi$, which fixes $\mathbb{Q}$. Suppose $i\in\mathbb{Q}(\alpha)$, then $$ -1=\varphi(-1)=\varphi(i^2)=\left(\varphi(i)\right)^2. $$ But as $\mathbb{Q}(\beta)\subset \mathbb{R}$ inherits the order of $\mathbb{R}$, we have a contradiction, since $x^2≥0>-1\forall x\in\mathbb{R}$.
Is this proof alright? Is this a useful idea which might be generalized? The proof seemed rather strange to me, as we implicitly order a subfield of $\mathbb{C}$. Furthermore, the properties of $\mathbb{R}$ as an ordered field don't turn up that often in the basic theory of field extensions.