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I want to prove the following:

Let $f\in \mathbb{Q}[x]$ be an irreducible cubic polynomial, whose Galois group is cyclic. Prove that all of the roots of $f$ are real.

I know that the Galois group $G$ is a subgroup of $S_{3}$, and I tried to go other the different possibilities: ${e},\mathbb{Z}_2, A_3$, but couldn't find a way to use this. I also tried thinking about the discriminant, since I know that the discriminant is a square iff $G\leq A_3 $ but once again this did not help me.

Tian Vlašić
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Ariel Yael
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    Think about the action of complex conjugation – Mathmo123 Jul 02 '23 at 19:43
  • @Mathmo123 yes I thought about this as well, knowing that if I have a complex root the other root is it's conjugate (and of course a cubic polynomial always has at least one real root), so if I don't have a complex root the conjugate is simply the identity. Still I couldn't finish this up to a complete proof – Ariel Yael Jul 02 '23 at 19:44
  • What happens if complex conjugation acts non-trivially? Can $G$ still be cyclic of order $3$? – Mathmo123 Jul 02 '23 at 19:46
  • @Mathmo123 Why does $G$ have to be be of order 3? Also, I don't see why it couldn't be of order 3 if conjugation acts non-trivially – Ariel Yael Jul 02 '23 at 19:52
  • $G$ has to be of order $3$ because $f$ is irreducible, so it's Galois group has to be of order at least $3$ (it has to be a transitive subgroup of $S_3$). What is the order of complex conjugation? – Mathmo123 Jul 02 '23 at 19:53
  • @Mathmo123 so the order of complex conjugation is 2, but in $\mathbb{Z}_3$ there are no elements of order 2 (only 1 or 3), finishing the proof(?), But I still don't understand why $f$ being irreducible implies $G$ having order 3 – Ariel Yael Jul 02 '23 at 19:56
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    Well, for one thing, if $f$ is irreducible, $\mathrm{Gal}(f)$ acts transitively on the roots of $f$. Or because $|\mathrm{Gal}(f)| = [\mathbb Q_f : \mathbb Q]$, where $\mathbb Q_f$ is the splitting field of $f$, which has degree divisible by the degree of $f$. In general, a degree $3$ irreducible polynomial can only have Galois group $S_3$ or $C_3\cong A_3$. – Mathmo123 Jul 02 '23 at 19:57
  • Does this answer your question? Does a cubic polynomial with 3 real roots have Galois group C3? – Galois group Jul 02 '23 at 21:55
  • In this question it is said that:"If an irreducible cubic polynomial with coefficients in ℚ has Galois group 3 then, since no order 2 symmetry lies in the Galois group no complex conjugation acts on the roots, it's roots must all be real." – Galois group Jul 02 '23 at 22:31
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    The polynomial has galois group of order $3$ and hence splitting field is of degree $3$. And since the polynomial is irreducible adjoining any root gives a degree $3$ extension. It follows that splitting field is obtained by adjoining any root. Since one root is real all of them are real. – Paramanand Singh Jul 04 '23 at 05:29

4 Answers4

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Just wanted to post an organized answer based on my conversation with @Mathmo123 in the comments - all credit goes to him.

Denote the roots of $f$ as $\alpha_1,\alpha_2,\alpha_3$ if $f$ is irreducible, it is the minimal polynomial of it's roots. This means that $[\mathbb{Q}(\alpha_1):\mathbb{Q}]=\deg f=3$ meaning $|G|=[\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3):\mathbb{Q}]=[\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3):\mathbb{Q}(\alpha_1)]\cdot[\mathbb{Q}(\alpha_1):\mathbb{Q}]$ is dibisible by 3.

Note that if there are repeating roots, it means that we can't have a complex root, since every polynomial of off degree has a real root, and if there is a complex root it's conjugate is also a root, and these are all the roots (since $\deg f=3$) and they are all different. So we can now assume the other case: there are no repeating roots.

Since each element in $G$ is a permutation of $f$'s roots, $G\leq S_3$, meaning $|G|=3$ or $|G|=6$, but for $G$ since $S_3$ isn't cyclic, $|G|\neq 6$ therefore $|G|=3$ and $G\simeq \mathbb{Z}_3$. Each element in $G$ therefore is either of order 1 (the neutral element) or of order 3.

Assume towards a contradiction $f$ has a complex root. As we said before, we know that a polynomial of an odd degree must have a real root, meaning f has one real root, a complex root, and the conjugate of that complex root as it's final root.

This means the conjugation automorphism is non-trivial, and is of order 2, but we said that $G\simeq \mathbb{Z}_3$ has no elements of order 2 - which is a contradiction. Meaning all roots of $f$ must be real.

Ariel Yael
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Complex conjugation has order two(it is involution) and so complex conjugation acts trivially on the set of all roots. (cyclic group of order $3$ does not contain elements of order $2$)

So any roots is invariant under complex conjugation and so is real.

Galois group
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Let f be irreducible cubics in $\mathbb{Q}$. Any irreducible cubics in $\mathbb{Q}$ are automatically separable. So there are no repeating roots and there are only two cases:

Case 1. f has 1 irrational root and 2 complex roots

Case 2. f has 3 distinct irrational roots

As pointed out in one of the comments, $$|\mathrm{Gal}(f)| = [\mathbb Q_f : \mathbb Q] = [\mathbb Q_f : \mathbb Q(\alpha)][\mathbb Q(\alpha) : \mathbb Q]$$

where $\mathbb Q_f$ is the splitting field of f and $\alpha$ is a root of f.

As $[\mathbb Q(\alpha) : \mathbb Q]=deg(f,\alpha) =3$, $|\mathrm{Gal}(f)|$ is divisible by 3.

Since $\mathrm{Gal}(f)$ is a subgroup of $S_3$, $$\mathrm{Gal}(f) = S_3 \text{ or } A_3 \cong C_3$$

For Case 1, since the order of complex conjugation is 2 and there is no elements in $A_3$ with order 2, we must have $\mathrm{Gal}(f) = S_3$.

For Case 2, it is possible that $\mathrm{Gal}(f)$ be either $S_3$ or $A_3$ and it depends on whether the discriminant of f is a square. If the discriminant of f is a square, then $\mathrm{Gal}(f)$ is $A_3$. Otherwise, $\mathrm{Gal}(f)$ is $S_3$.

Examples:

(a) $f(X)=X^3 − 3X − 1$, $\Delta(f)=81$, $\mathrm{Gal}(f)=A_3$

(b) $f(X)=X^3 − 4X − 1$, $\Delta(f)=229 $, $\mathrm{Gal}(f)=S_3$

Now, suppose Galois group of f is cyclic. Since $S_3$ is not cyclic, case 1 cannot happen. So f must have 3 distinct irrational roots and we can also know that the discriminant of f should be a square.

Tim
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COMMENT.- (Just for information) It is known that any cubic having just one real root has Galois group $G=S_3$ which is not cyclic. However there are cubics having its three roots real whose Galois group is also $S_3$ such, for example the irreducible $f(x)=x^3-9x+3$ whose discriminant is equal to $2673$ not a rational square so also $G=S_3$.

Since $G$ is assumed cyclic and each $\sigma\in G$ is defined making correspond to a root a conjugate we can deduce without difficulty that the three roots are real.

For $f(x)=(x-1)(x-2)(x-3)$ $f$ is reducible and the $G$ is trivially cyclic.

For $f(x)=x^3-3x+1$ is irreducible whose discriminant is the square $81$ so $G=A_3$ and the roots are all real.

Piquito
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