# Determine $\operatorname{Aut}_{\mathbb{Q}}L$ for $L = \mathbb{Q}[u]$, where $u$ is a root of $x^3 - 3x^2 + 3$

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?

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

• $\Delta(x^3-3x^2+3) = 81 = 9^2$ Aug 3, 2020 at 1:05
• 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. Aug 3, 2020 at 2:00
• 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? Aug 3, 2020 at 2:10
• 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. Aug 3, 2020 at 12:15

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.