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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..

glS
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Vinny Chase
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  • @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 $b$ – Vinny Chase May 21 '18 at 03:42
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    $x^3-x$ has three distinct rational roots. Check out https://math.stackexchange.com/questions/38548/does-a-cubic-polynomial-with-3-real-roots-have-galois-group-c3 – Ashvin Swaminathan May 21 '18 at 03:50
  • @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 $\mathbb{Q}$ – Vinny Chase May 21 '18 at 04:16
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    If you write a nonzero rational number as a reduced fraction $\frac{p}{q}$, where $p \in \Bbb Z, q \in \Bbb N$ and $\gcd(p,q)=1$, then it is a square iff $p$ is positive and $p$ and $q$ are both squares. So it's not hard to figure out which rational numbers are squares. – Lukas Heger Oct 08 '23 at 20:19

1 Answers1

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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.

Qiaochu Yuan
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  • Hi there. I have one unanswered question. Would you like to see this question: https://math.stackexchange.com/questions/2531008/subextension-of-a-field-with-galois-series-of-subextensions-of-prime-degree Thank you very much for your help! – Hermi Nov 15 '23 at 03:35