Let $f\in\mathbb Q[x]$ be irreducible of degree $3$.
Since the Galois group $G$ of $f$ is a transitive subgroup of $S_3$, it is either $S_3$ or $A_3$. Those two possibilities are easily distinguished by computing the discriminant of $f$.
My question is:
Is there is an "elementary" and deterministic way to determine the Galois group of $f$ without using (or replicating) the discriminant.
One way to settle a part of the cases is to check if $f$ has roots in $\mathbb C\setminus\mathbb R$, which can be done by standard calulus methods. In that case, the complex conjugation provides an $\mathbb Q$-automorphism of order $2$ and hence, $G$ must be $S_3$.
But how to proceed if $f$ has three real roots?
Test cases of irreducible cubic polynomials with three real roots:
- The polynomial $2X^3 + X^2 - 3X - 1\in\mathbb Q[X]$ has Galois group $S_3$.
- The polynomial $X^3 + X^2 - 2X - 1\in\mathbb{Q}[X]$ has Galois group $A_3$.
Similarly, but slightly more sophisticated-ly, if for some prime $p$ the cubic can be shown to have a root in a quadratic extension of the $p$-adics $\mathbb Q_p$, the same conclusion is reached, namely, that the Galois group is the full symmetric group.
This can be tested reasonably via Hensel's lemma: for example, suppose that the cubic is rearranged to be monic with integer coefficients, and mod $p$ (not $2,3$, maybe) factors as a linear factor and an irreducible quadratic factor. Then Hensel's lemma guarantees a root in $\mathbb Q$, but also an irreducible quadratic factor.