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Does there exist $f(x) \in \mathbb Q[x]$, $\deg f(x) = 3$, such that for $E$ being a splitting field of $f(x)$ over $\mathbb Q$, we have $\text{Gal}(E/\mathbb Q) = \mathbb Z_3$?

Motivation for the question: I am asked to prove that if the Galois group of a splitting field of a cubic over $\mathbb Q$ is cyclic of order $3$ then all the roots of this cubic are real. So I'm wondering what is an example of such a cubic to begin with. Thanks in advance.

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  • $\begingroup$ That's a bit hard to answer since difficulty is relative. But I will say that it involves some trickery that I'd see myself taking quite some time to figure out from scratch, especially if it were my first time around. If you want to see how far you can get on your own, I will say it involves looking at the discriminant and its square root, which is defined as $\displaystyle D = \prod_{i \neq j} (x_i - x_j)$, where the $x_k$'s are the roots of $f$. $\endgroup$
    – Kaj Hansen
    Aug 16 '17 at 19:58
  • $\begingroup$ @KajHansen sorry for deleting my comment. I thought you weren't going to reply. $\endgroup$
    – Cauchy
    Aug 16 '17 at 20:00
  • $\begingroup$ Ha, no worries @Cauchy $\endgroup$
    – Kaj Hansen
    Aug 16 '17 at 20:14
  • $\begingroup$ Start by saying the splitting field of $f(x) = \prod_{i=1}^3 (x-\alpha_i) \in \mathbb{Q}[x]$ is $K=\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3) =\mathbb{Q}(\alpha_i,\sqrt{\Delta})$ where $\Delta = \prod_{i \ne j} (\alpha_i-\alpha_j) \in \mathbb{Q}$. Thus there are two possible cases : $\sqrt{\Delta} \in \mathbb{Q}$ so $[K:\mathbb{Q}] = 3$ and $\text{Gal}(K/\mathbb{Q}) = C_3$, or $\sqrt{\Delta} \not \in \mathbb{Q}$ so $[K:\mathbb{Q}] = 6$ and $\text{Gal}(K/\mathbb{Q}) = S_3$ $\endgroup$
    – reuns
    Aug 17 '17 at 3:22
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Let $p\equiv1\pmod3$ be a prime. Then $\Bbb Q(\zeta_p)/\Bbb Q$ is a Galois extension, cyclic of degree $p-1$ (where $\zeta_p= \exp(2\pi i/p)$). This group has a quotient $C_3$, so has a cyclic cubic subextension.

Simple example: $p=7$. Let $\alpha=\zeta+\zeta^{-1}$. As $$\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$$ then $$\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0.$$ This gives $$\alpha^3+\alpha^2-2\alpha-1=0$$ so take $f(x)=x^3+x^2-2x-1$.

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  • $\begingroup$ $\text{Gal}\left(\mathbb Q(\zeta_p)/\mathbb Q\right)$ has order $p-1$, and $3 \mid (p-1)$.. from this, how do we get that it has a quotient $C_3$? $\endgroup$
    – Cauchy
    Aug 16 '17 at 19:59
  • $\begingroup$ @Cauchy It's a cyclic group of order $3k$. It's a pity we are after quotients, not subgroups, since your theorem guarantees the existence of a subgroup of order $3$.... $\endgroup$ Aug 16 '17 at 20:02
  • $\begingroup$ Nice answer, +1. Although the OP seems to be asking from a more elementary perspective, it might be worth noting that the key words here are ''cyclotomic extension'', since (as I'm sure you are aware) every abelian Galois extension of $\mathbb{Q}$ is contained in some sufficiently large cyclotomic extension by Kronecker-Weber. $\endgroup$ Aug 16 '17 at 20:02
  • $\begingroup$ Ouch.. I should've read your answer more closely.. "..is a Galois extension, cyclic of degree $p-1$.." $\endgroup$
    – Cauchy
    Aug 16 '17 at 20:05
  • $\begingroup$ Thank you! As always, great answer. $\endgroup$
    – Cauchy
    Aug 16 '17 at 20:12
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Underneath "Claim" in my post here, I show that the Galois group of any cubic polynomial in $\mathbb{Q}[x]$ is determined by its discriminant. However, we can attack this with a more naive approach:

First, we know that the Galois group of a polynomial $f \in \mathbb{Q}[x]$ is a subgroup of the symmetric group $S_{\deg(f)}$, and when $f$ is irreducible, this subgroup is transitive$^\dagger$. Therefore, the Galois group of any irreducible cubic polynomial must be either $S_3$ or $A_3 \cong \mathbb{Z}_3$. If the cubic polynomial in question has complex roots, complex conjugation will be a nontrivial automorphism of the splitting field. Since complex conjugation is an element of the Galois group of order $2$, we'll have $\text{Gal}(f) \cong S_3$ per Lagrange's theorem.

So for irreducible cubic polynomials, we know $\text{Complex roots} \implies S_3$ Galois group. But what of cubics with all real roots? Will we necessarily have $\text{Gal}(f) \cong \mathbb{Z}_3$? Is $\text{Gal}(f) \cong S_3$ still possible? At this point, I think the discriminant argument is the easiest way to proceed. Some hints are below:

It is always the case that $D \in \mathbb{Q}$, and anything in $\mathbb{Q}$ is fixed by elements of $\text{Gal}(f)$. Next, consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt{D}) \subset \mathbb{Q}(\sqrt{D}, \alpha)$, where $\alpha$ is one of the roots of $f$. Note that the latter field is the splitting field of $f$ (why?).


$\dagger$ For a proof of this fact, see Theorem 2.9(b) here.

$\ddagger$ Alternatively, note that complex conjugation, in general, is an odd permutation and thus not an element of an alternating group. From this, we can conclude $\text{Gal}(f) \cong S_3$.

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  • $\begingroup$ Nice answer, Kaj. Two small nitpicks: (1), although it's true that the Galois group of a polynomial $f \in \mathbb{Q}[X]$ must be a subgroup of $S_{\deg(f)}$, I think what you really want to say here is that the Galois group of an irreducible polynomial $f \in \mathbb{Q}[X]$ must be a transitive subgroup of $S_{\deg(f)}$. (Hence, the Galois group of any irreducible cubic polynomial must be either $S_{3}$ or $A_{3}$). (2) Complex conjugation will always be a valid automorphism of the splitting field whether the poly has complex roots or not; if you do have complex roots, it's nontrivial $\endgroup$ Aug 16 '17 at 20:05
  • $\begingroup$ Thanks a lot Hansen! I think I'll postpone the question you posed until I study discriminants and other stuff. I'm still a newbie in Galois theory. $\endgroup$
    – Cauchy
    Aug 16 '17 at 20:11
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    $\begingroup$ Rather than invoking Cauchy's theorem, complex conjugation on a degree 3 polynomial over $\mathbb{Q}$ with a complex root is necessarily a transposition, which would not be an element of $A_3$. $\endgroup$
    – Jacob Bond
    Aug 16 '17 at 20:21
  • $\begingroup$ You're right @JacobBond! Thanks for the suggestion; I added a note to that effect. $\endgroup$
    – Kaj Hansen
    Aug 16 '17 at 20:36

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