Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A_3$.
share|improve this question

1 Answer 1

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.

share|improve this answer
Thank you for this answer. With your method, further cases can be shown to have Galois group $S_3$ without using the discriminant. But it's not exactly what I hoped for: A deterministic and "elementary" way without using the discriminant to reliably distinguish $S_3$ from $A_3$. –  azimut Feb 17 '14 at 14:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.