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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 $A_3$.
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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.

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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

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