Is there a general method to find the size of the Galois group of a cubic polynomial over $\mathbb{Q}$?

Question

Given a monic cubic polynomial $f = x^3 + ax + b$, is there a general method for determining what the size of the Galois group will be over $\mathbb{Q}$?

I know:

• if $f$ has one real root and two imaginary then it will be isomorphic to the full group $S_3$.
• if $f$ has roots that are all rational, then the Galois Group is just the identity. Since the polynomial here is monic, the only possible rational roots are $\pm b$

The case I am trying to figure out is if $f$ has 3 real roots. I know that $f$ will be isomorphic to $S_3$ iff it has discriminant that is a nonsquare in $\mathbb{Q}, $ otherwise the Galois group is isomorphic to $A_3$, but its pretty hard (or at least I think it is...please let me know if theres some easy way) to find all the numbers that could be squares in $\mathbb{Q}$.

PS: In my head I just had the thought that a cubic polynomial can't have $3$ distinct rational roots. Is that true? I can't think of a counterexample. I don't need a proof just thinking out loud..

Answer 1

For the sake of getting this off the unanswered queue, yes, there is a general method:

1. First check for rational roots using the rational root theorem. If there are three of them then the polynomial splits so its Galois group is trivial. If there's only one then $f$ is the product of a linear and irreducible quadratic factor so its Galois group is $C_2$. If there aren't any, then $f$ is irreducible.

2. If $f$ is irreducible then, as you say, its Galois group is $S_3$ iff its discriminant $\Delta = -4a^3 - 27b^2$ is not a square, and $A_3 \cong C_3$ iff its discriminant is a square. As Lukas Heger says in the comments, it is easy to test if a rational number is a square: first, it has to be positive, and second, if you write it in lowest terms $q = \frac{a}{b}$ where $a, b \in \mathbb{N}$ and $\gcd(a, b) = 1$, then $q$ is a square iff $a$ and $b$ are both squares.

3. And a positive integer is a square iff every prime in its prime factorization has even exponent, although this test eventually becomes inefficient and it becomes more efficient for large positive integers to just numerically compute their square roots.