Suppose that $f(x)$ has a nonreal root. Show that 2 divides $|\operatorname{Gal}(f(x))|$

Question

If f(x) has n roots and $\Sigma$ is the splitting field of $f(x)$, then $|\operatorname{Gal}(f(x))| = [\Sigma:Q]=[\Sigma:Q(\alpha_{n-1},\alpha_{n-2}, \ldots,\alpha_{1})]\cdots[Q(\alpha_{1}):Q]$. If, for example, $\alpha_{1} = i$, then $[Q(\alpha_{1}):Q]=2$, correct? (Can this statement be made WLOG?) So then $|\operatorname{Gal}(f(x))|$ is divisible by $2$? That's my thought process; I would appreciate feedback on if this thinking is correct or if I made an error somewhere. Thanks in advance!

Answer 1

This part " If, for example, $\alpha_1=i$, then $[\mathbb Q (\alpha_1):\mathbb Q] = 2$", while correct is not proving the statement.

Yes, if $\alpha_1=i$ then $[\mathbb Q (\alpha_1):\mathbb Q]=2$, but what if $\alpha_1 \neq i$? It is possible $\alpha_1$ to be non-real and $[\mathbb Q (\alpha_1):\mathbb Q] = \text{odd}$.

Hint: If $f$ has one non-real root, and real coefficient, then $\sigma(z)=\bar{z} \in \operatorname{Gal}(f)$ and $\sigma \neq \operatorname{Id}$.