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.


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?

  • 2
    $\begingroup$ 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. $\endgroup$ – ancientmathematician Apr 8 at 16:03
  • 1
    $\begingroup$ 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 $\sigma(\alpha_1)=\alpha_2$. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:30
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:34
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:38
  • 1
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:39

Now obsolete. The question was edited. That's fine, waiting for the issues to be isolated :-)

I'm not sure this exactly what you find puzzling, but it seems to me that it may fit. So, two observations:

  • 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.

The complex conjugation is just another element of the Galois group. It has no special status. Most notably, it is not always an element of the center of Galois group.

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.

  • $\begingroup$ That's fine. I will delete this when it becomes clear what exactly an answer should address. $\endgroup$ – Jyrki Lahtonen Apr 9 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy