How to determine the Galois group of irreducible polynomials of degree $3,4,5$

Question

Let $f$ be an irreducible (over $\mathbb{Q}$) polynomial in $\mathbb{Z}[x]$, $\deg (f)=3,4,5$. The Galois group of an irreducible polynomial $f\in \mathbb{Z}[x]$ acts transitively on distinct roots in $\mathbb{C}$ of $f$. Hence $Gal(f)\leq S_m$, and acts transitively on distinct $m$-roots, where $m\leq n$ for $n=\deg (f)$.

$m=3$, $S_3,A_3=\mathbb{Z}_3$

$m=4$, $S_4,A_4,D_4,\mathbb{Z}_4,K_4$ ($K_4$ is the Klein four group)

$m=5$, $S_5,A_5,D_5,\mathbb{Z}_5,Fr_5$ ($Fr_5$ is a Frobenius group)

For example, by Eisenstein criterion, the following polynomials are irreducible, but:

Determine over $\mathbb{Q}$, $Gal(x^3-3x+1)=\mathbb{Z}_3$, not $S_3$? Why?

Determine over $\mathbb{Q}$, $Gal(x^4+3x+3)=D_4$, not $A_4,K_4, S_4$? Why?

Also by determine coefficients, $x^4+8x+12$ is irreducible. Why $Gal(X^4+8x+12)=A_4$, over $\mathbb{Q}$, not $S_4,K_4, \mathbb{Z}_4$?

Is there any method to determine the Galois group?

Answer 1

The Galois group of an irreducible polynomial is transitive, so your list is the list of transitive permutation groups on $3, 4, 5$ elements.

And no, we cannot determine the Galois group if we have the number of real roots (in general), and no, there is no quick method to determine the Galois group, though for most polynomials of degree $n$ it is $S_n.$ Confirming or denying that it is $S_n$ is fast, as described at length in this quite recent preprint.