Now, I broke this down into two cases:

Case 1 - the other roots (let's call them $u_2$ and $u_3$) of the polynomial (let's call it $h(x)$) are in $\mathbb{Q}[u]$. In this case, $L = \operatorname{Gal}(h, \mathbb{Q})$, which implies $|\operatorname{Aut}_{\mathbb{Q}}L| = 3$, so $\operatorname{Aut}_{\mathbb{Q}}L \simeq A_3$, the set of even permutations of 3 elements.

Case 2 is where I got stuck. If the other roots are not in $L$, then $\operatorname{Gal}(h, \mathbb{Q}) = \mathbb{Q}[u, u_2]$, since $u + u_2 + u_3 = 3$. Then, I determined $|\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}[u, u_2]| = 6$, making $\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}[u, u_2] \simeq S_3$. But that didn't give me what I needed for $\operatorname{Aut}_{\mathbb{Q}}L$. My intuition actually tells me this case can't happen, but I don't know how to prove it...

Is everything correct so far? What can I do to finish the problem?

Thanks in advance!

EDIT: I forgot to mention I do know all three of the roots are real, and that is what leads me to believe Case 1 is always true

  • 1
    $\begingroup$ $\Delta(x^3-3x^2+3) = 81 = 9^2$ $\endgroup$
    – Kenny Lau
    Aug 3, 2020 at 1:05
  • $\begingroup$ For most cubics, e.g., $x^3-2$, the automorphism group you're after is the one element group – the identity is the only automorphism. But your cubic is what's called a cyclic cubic, you are in your Case 1, and the automorphism group is cyclic of order three. As @Kenny comments, the discriminant is a square, that's the tell-tale sign. $\endgroup$ Aug 3, 2020 at 2:00
  • $\begingroup$ Thank you for the answer! In this case, I'm going to need a little more info. How is the discriminant of a cubic defined? And how can I prove this fact, $\Delta$ square number $\implies$ the cubic is cyclic? I'm sorry for the lack of information, I had never seen this concept before. In case it's too much to answer, can you point me to some references? $\endgroup$
    – Gauss
    Aug 3, 2020 at 2:10
  • $\begingroup$ If you want to be sure I see a comment, Gauss, you have to put @Gerry in it. Fortunately, another user was here to help. $\endgroup$ Aug 3, 2020 at 12:15

1 Answer 1


All 3 roots being real does not imply cyclic Galois group. For example $x^3 - 3x - 1$.

The discriminant is the square of the expression $$(u_1 - u_2)(u_1 - u_3)(u_2 - u_3);$$

this expression is invariant when acted on by $A_3$ and flips sign when acted on by $S_3$ permutations outside of $A_3$. Its square is invariant under all transformations and therefore can always be expressed with rational coefficients.

If the Galois group of the cubic is $A_3$, then the discriminant will be a square, since the expression above will be expressible with rational coefficients.

More details here: https://kconrad.math.uconn.edu/blurbs/galoistheory/cubicquartic.pdf


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