Determining the Galois group of a cubic without using discriminant

abstract-algebrafield-theorygalois-theorygroup-theory

Question: Suppose I want to calculate the Galois group of a cubic polynomial. If that cubic is irreducible over $\mathbb{Q}$, then we can use the discriminant to determine if the Galois group is $A_3$ or $S_3$. That is, if the discriminant is a square, the Galois group is $A_3$, otherwise, the Galois group is $S_3$. BUT, the discriminant of a cubic is $18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$. Suppose this isn't memorized and the derivation is quite tedious… how else could we approach a cubic? In particular, I'm interested in irreducible cubics over $\mathbb{Q}$ (an example would be nice), but if you'd also like to say something about reducible cubics over $\mathbb{Q}$, that would be welcomed as well!

Thank you.

After typing this question, I see that it is asked here: Galois group of an irreducible cubic polynomial without using the discriminant, with an answer, but, much like the OP, I am wondering if there are "elementary methods", such that you may find in a typical course in Abstract Algebra, but may not usually be taught for whatever reasons.

Best Answer

Perhaps not quite basic Galois theory, but definitely a standard part of basic algebraic number theory:

Assuming we've rearranged so that the (irreducible) cubic $f(x)$ is monic and has integer coefficients, if we can find a prime $p$ such that the reduction of $f$ mod $p$ factors as linear polynomial and irreducible quadratic, then there is a $2$-cycle in the global Galois group. The irreducibility implies presence of a $3$-cycle, so there's no choice left but that the Galois group is $S_3$.

Similarly, if the cubic has exactly one real root (and two complex), we also conclude that there's a $2$-cycle in the Galois group.

The same line of thinking applies to irreducible prime degree $q$ polynomials: the irreducibility implies existence of a $q$-cycle in the Galois group. If not all roots are real, and/or if there is some prime modulo which there is an irreducible quadratic factor, then the Galois group has to be $S_q$, since any $q$-cycle and a $2$-cycle generate the whole thing.

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