I think I have this right but I want to make sure. The roots of this polynomial are $-\sqrt[3]3 = \alpha$ and $\frac{\sqrt[3]3(1 \pm i\sqrt3)}{2} = \beta_1, \beta_2$.

The possible automorphisms will send these roots to other roots. We have the identity (of order 1) and the automorphisms (of order 2):

$\pi_1 : \alpha \rightarrow \beta_1$

$\pi_2 : \alpha \rightarrow \beta_2$

$\pi_3 : \beta_1 \rightarrow \beta_2$

As well as the following (of order 3)

$\pi_4: \alpha \rightarrow \beta_1, \beta_1 \rightarrow \beta_2, \beta_2 \rightarrow \alpha$

$\pi_5: \alpha \rightarrow \beta_2, \beta_2 \rightarrow \beta_1, \beta_1 \rightarrow \alpha$

Is this correct?

  • 1
    $\begingroup$ Your reasoning makes it sound like the Galois group of any n-th degree polynomial is the full permutation group $S_n$. (And for clarity you should probably write e.g. $\alpha \leftrightarrow \beta_1$ for the order two permutations, but that's pretty minor.) $\endgroup$ – spaceisdarkgreen Jul 28 '17 at 0:05

Just to extend on the comment: Yes, this is correct, but you should probably give a reason, why all of these are actually automorphisms and why there aren't more. For example you could argue with the degree of the extension $F/\mathbb{Q}$, where $F$ is the splitting field of the polynomial over $\mathbb{Q}$.


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