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..

  • $\begingroup$ @pisco I know that, but how can I figure out all of the values for which the discriminant isn't a square in $\mathbb{Q}$. That seems hard to me since I don't just have to look at integer values for $a$ and $b$ $\endgroup$ – Vinny Chase May 21 '18 at 3:42
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    $\begingroup$ $x^3-x$ has three distinct rational roots. Check out math.stackexchange.com/questions/38548/… $\endgroup$ – Ashvin Swaminathan May 21 '18 at 3:50
  • $\begingroup$ @AshvinSwaminathan I feel silly for not seeing such an obvious counterexample. I actually saw that question earlier, but I guess here I am asking for an answer that more directly addresses what the conditions on $a$ and $b$ would have to be in order for a rational root to be the case or for the discriminant to be a square in $\mathbb{Q}$ $\endgroup$ – Vinny Chase May 21 '18 at 4:16

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