Let $f\in\mathbb{Q}[X]$ be an irreducible polynomial. Show, that its Galois group is non-abelian if $f$ has roots in $\mathbb{R}$ and $\mathbb{C}\backslash\mathbb{R}.$
First of all, the separability is not assumed. Is Galois group defined in a same way for this case, i.e. the group of automorphisms of the splitting field of $f$? Supposing it is, define $$\alpha_1 = a +br,~\alpha_2 = c+di; ~a,b,c,d\in\mathbb{Q}, r\in\mathbb{R\backslash\mathbb{Q}}$$ the real and complex roots. Now $d\neq0$ by assumption and $b\neq 0$ because f is irreducible, so $$\alpha_3 = a – br,~\alpha_4 = c-di$$ are also roots, because coefficients of f are rational. Now we have two pairs of conjugate roots, and an argument similiar to one from the answer to the question here can be presumably used:
Non-abelian Galois group
"Let the two nonreal roots be $r,s$, let a real root be $t$. Write $\sigma$ for complex conjugation: then $\sigma(r)=s$, $\sigma(s)=r$, $\sigma(t)=t$. Now Galois groups are transitive, so there's an automorphism $\tau$ with $\tau(r)=t$. Then $\sigma(\tau(r))=\sigma(t)=t$, while $\tau(\sigma(r))=\tau(s)\ne t$."
My question is, how can it be proved that the automorphisms that permute the roots in such fashion are really automorphisms?
Update: as was pointed out in the comments, irreducible polynomial over a field with characteristic $0$ is separable, hence the splitting field $F$ of $f$ is a Galois extension and $$Gal(F/\mathbb{Q}) = [F:\mathbb{Q}]\geq 4,$$ because $$[F:\mathbb{Q}]\geq[\mathbb{Q(\alpha_1,\alpha_2)}:\mathbb{Q}] = [\mathbb{Q(\alpha_1,\alpha_2)}:\mathbb{Q(\alpha_1)}]\cdot[\mathbb{Q(\alpha_1)}:\mathbb{Q}]\geq2\cdot 2 = 4.$$ So there are at least three automorphisms apart from identity.
Update 2: It seems like I formulated my question vaguely, so I'll try it again. Suppose $deg(f) = n, |Gal(F/\mathbb{Q})| = m$. The only method I know so far to find all permutations of the roots that define $\mathbb{Q}$-automorphism on $F$ is to eliminate $k = n! – m$ permutations that are not homomorphic. Is there a way to check a specific permutation for being continuable to $\mathbb{Q}$-automorphism? In the problem described above, neither $n$ nor $m$ can be found. Can it be proved, that $\sigma,\tau$ from the cited answer, described only by their action on $2$ and $1$ roots correspondingly, are indeed homomorphisms (and by argument from @rts automorphisms)?
Best Answer
It is part of the main result of Galois theory that the Galois group of $f$ acts transitively on the set of roots. Hence there exists a $\tau\in G$ that maps the given complex root $r$ to the given real root $t$. On the other hand, complex conjugation $\sigma$ is an automorphism of $\Bbb C$ (and hence also of $\overline{\Bbb Q}$ and of the splitting filed of $f$) and leaves $t$ fixed while mapping $r$ to a different root $s$. In general, you cannot prescribe more than one value of the permutation of roots (or: The action of $G$ on the thes of roots is not necessarily $2$-transitive). But for the arfument at hand, this is not needed. Observing that $\sigma(\tau(t))=\sigma(r)=s$ and $\tau(\sigma(t))=\tau(t)=r$ is enough.