# Galois group of a quintic with 3 real roots. How to conclude that there's one cycle of order 5?

I understand perfectly the argument making use of Cauchy's theorem, which I'll lay down for clarity's sake: take $$p(x)$$ of degree 5 irreducible over $$\mathbb{Q}$$. Let $$K$$ be the root field of $$p(x)$$ over $$\mathbb{Q}$$ and $$G$$ its galois group.

Take $$r_1$$ a root of $$p(x)$$, then $$[\mathbb{Q}(r_1):\mathbb{Q}]=5$$ and

$$[K:\mathbb{Q}]=[K:\mathbb{Q}(r_1)][\mathbb{Q}(r_1):\mathbb{Q}]\Longrightarrow 5\mid [K:\mathbb{Q}]$$

Cauchy's theorem gives that $$G$$ has an element of order 5. Call $$\sigma$$ a 5-cycle permutation. Now, if $$p(x)$$ has 2 complex roots, $$G$$ has a transposition $$\tau$$. $$G$$ contains $$\sigma\tau\sigma^{-1}$$, $$\sigma^{2}\tau\sigma^{-2}$$, ..., $$\sigma^{4}\tau\sigma^{-4}$$, which are all possible transpositions and they generate $$S_5$$, hence $$G=S_5$$ and $$p(x)$$ is unsolvable by radicals because $$S_5$$ is an unsolvable group.

$$\blacksquare$$

This is clear and Cauchy's theorem is very elementary, but it puzzles me to imagine a 5-cycle that would always be a valid automorphism when there are 2 complex roots.

As an example of what I'm saying: take $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$, then $$\phi : \sqrt{2}\mapsto\sqrt{3}$$ is not a valid automorphism

$$2 = \phi(2)=\phi(\sqrt{2}\sqrt{2})=\phi(\sqrt{2})\phi(\sqrt{2})=\sqrt{3}\sqrt{3}=3$$

So the question is: how could I ensure that in such situations ($$n$$ is a prime and there's a pair of complex roots) the automorphisms are valid and couldn't end up in a case like above?

• Are you equally puzzled by the fact that $X^3-2$ has $S_3$ as Galois Group? It does (!) and we can see an automorphism carrying a real number to a non-real one. – ancientmathematician Apr 8 at 16:03
• Re: the edited version. You won't ever mix $\sqrt2$ and $\sqrt3$ together. This is because they are not zeros of the same irreducible polynomial. With an irreducible quintic $f(x)$ with two zeros, $\alpha_1$ and $\alpha_2$, the fields $\Bbb{Q}(\alpha_1)$ and $\Bbb{Q}(\alpha_2)$ are trivially both isomorphic to $\Bbb{Q}[x]/\langle f(x)\rangle$ by an automorphism $\sigma$ such that $\sigma(\alpha_1)=\alpha_2$. Furthermore, $\sigma$ can be extended to an automorphism of $\Bbb{Q}(\alpha_1,\alpha_2,\ldots,\alpha_5)$ such that we still have $\sigma(\alpha_1)=\alpha_2$. – Jyrki Lahtonen Apr 9 at 5:30
• Proving that can be done by mimicking the proof of uniqueness of the splitting field (or by other means). Anyway, a corollary is that the automorphism group acts transitively on the set of zeros of an irreducible polynomial. – Jyrki Lahtonen Apr 9 at 5:34
• What may be holding you back is lack of specific examples, similar to $X^3-2$ that @ancientmathematician discussed, but quintic. Remember that when the Galois group is forced to be $S_5$, the splitting field is a degree $120$ extension of the rationals. That's a bit high for us to describe in a compact form, so we prefer this roundabout way of arguing about the automorphisms. – Jyrki Lahtonen Apr 9 at 5:38
• Mind you, there are easy to describe degree 120 Galois extensions with Galois group $S_5$, but the ones I have in mind don't have $\Bbb{Q}$ as the base field, and therefore the complex conjugation drops out of sight. – Jyrki Lahtonen Apr 9 at 5:39

• Presumably we know that the quintic is irreducible over $$\Bbb{Q}$$. For otherwise the order of its Galois group will not be divisible by five, and it cannot contain a 5-cycle. Anyway, we can think of the elements of the Galois group $$G$$ of a polynomial as permutations of the roots. In the case of an irreducible quintic this means that we can think of $$G$$ as a subgroup of $$S_5$$. The question in the title can then be answered by recalling that irreducibility is equivalent to $$G$$ acting transitively on the set of five roots. By the orbit-stabilizer theorem this implies that $$|G|=5\cdot|H|$$, where $$H$$ is the stabilizer of one of the roots in $$G$$. The conclusion is that $$5\mid |G|$$. By Cauchy's theorem from elementary group theory we can then conclude that $$G$$ has an element of order five. But, the only elements of order five in $$S_5$$ are the $$5$$-cycles.
• The question in the body may (I hope?) be answered by recalling that there is no reason to expect the complex conjugation to be in the center of $$G$$. The real roots are the fixed points of the complex conjugation. But if another permutation from $$G$$ does not commute with the complex conjugation, there is no need to expect it to take a real root to a real root. If the complex conjugation corresponds to the 2-cycle $$\phi=(12)$$ (under some numbering of the roots), it keeps the real roots (numbered $$3,4,5$$) fixed. Yet, we may easily have the permutation $$\sigma=(234)$$ in $$G$$. Because $$\sigma\phi\neq\phi\sigma$$ there is no need for $$\sigma(4)$$ to be a fixed point of $$\phi$$ even though $$4$$ is.
Do observe that if $$G$$ is known to be abelian, then the complex conjugation $$\phi$$ will be in the center. If $$\sigma$$ is another element of $$G$$ and $$r$$ is a real root, then $$\phi(r)=r$$, and therefore also $$\phi(\sigma(r))=(\phi\sigma)(r)=(\sigma\phi)(r)=\sigma(\phi(r))=\sigma(r)$$ implying that $$\sigma(r)$$ is real also. Note the role played by commutativity.