# If a polynomial has a real and a complex roots, its Galois group is non-abelian.

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)?

• $\Bbb Q$ has characteristic $0$, and thus any irreducible polynomial is automatically separable. So they don't need to mention it. The same is true for finite fields. Irreducible yet non-separable polynomials can only exist over infinite fields with positive characteristic. Jun 18 '19 at 13:40

My question is, how can it be proved that the automorphisms that permute the roots in such fashion are really automorphisms?

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.

The short answer: $$F\mid \mathbb{Q}$$ is a normal extension and thus $$\sigma(F)=F$$. That means the restriction $$\sigma|_F$$ defines in fact an automorphism of $$F$$.

Here is a sketch of how to prove this. Let $$S$$ be the set of roots of $$f$$. Then $$F=\mathbb{Q}(S)$$. Let $$\rho: F\to \mathbb{C}$$ be any homomorphism. Since $$F$$ is a field, $$\rho$$ is injective. Moreover $$\rho(\alpha)\in S$$ for each $$\alpha \in S$$, and thus $$\rho$$ induces a bijection on the finite set $$S$$. Hence $$\rho(F)=\rho(\mathbb{Q}(S))=\mathbb{Q}(\rho(S))=\mathbb{Q}(S)=F$$.

• As I understand you prove that any homomorphism defined by a permutation of roots is an automorphism. I designed my question poorly, the real question was about the existence of such homomorphisms. Maybe you can check update 2 in my question. Jun 19 '19 at 7:02
• @DeuzharNickens This answer only claims that complex conjugaction (= the automorphism you denoted by $\sigma$) is an automorphism. And that is kinda obvious, no? Jun 20 '19 at 21:47
• The question in Update 2 is only about the existence of $\tau$. That is done in all the textbooks on Galois theory. Because $f$ is irreducible, there exists an isomorphism $\tau:\Bbb{Q}(r)\to\Bbb{Q}(t)$ such that $\tau(r)=t$. This can then be extended to an isomorphism between the splitting fields (the argument is the same as the proof of uniqueness of the splitting field). but both splitting fields can be taken to be $F$, and the claim follows. Jun 20 '19 at 21:51
• @DeuzharNickens If you want to ping me, don't add space between the at-sign and my name. I wasn't notified of your comment, and only saw it because I checked out this tag again. Recall the proof of uniqueness-up-to-isomorphism of a splitting field of a polynomial. At the first step, when adjoining the first zero, you get an isomorphism $\tau:\Bbb{Q}(r)\to \Bbb{Q}(s)$. Then you can adjoin the other roots in a codependent order, and get to extend the isomorphism at all the steps. If you adjoin the roots of $f(x)$ in $L$ to $\Bbb{Q}$ like that, but A) begin with $r$, B) begin with $t$. Jun 23 '19 at 5:06
• (cont'd) Then you 1) end up with $L$ as the splitting in both orders, 2) extend $\tau$ to an automorphism of $L$. Mind you, there are other ways of reaching the same conclusion, depending on exactly which consequences of Galois correspondence have been covered in your course material already. This is just one of the earlier ways, not the prettiest. Jun 23 '19 at 5:08