Cubic polynomials that generate the same extension?

For quadratic extensions we can easily determine when $\mathbb{Q}(\sqrt{a})=\mathbb{Q}(\sqrt{b})$ by checking if $a/b$ is a square and this is easy to prove. I was wondering if there are any good rules for extensions generated by roots of cubic polynomials? Are there any other cases that are easy to work with? Does this simplify at all if we work in a local field e.g. the $p$-adics?

EDIT: To fix the ambiguity of the question, I'll change it as follows. In the quadratic case, we can write every polynomial in the form $X^2-a$ after a linear change of variables, so having two quadratic polynomials, we do the change of variables and check if the resulting polynomials satisfy the square test. If they do, then their splitting fields are the same.

For the cubic case, we can by a linear change of variables write any cubic as $X^3+aX+b$, so the question is then if there's an easy way to test if two such polynomials have the same splitting fields?

I guess this is somewhat equivalent to classifying all $C_3$ and $S_3$ extension of either $\mathbb{Q}$ or $\mathbb{Q}_p$. This depends on whether or not an $S_3$ extension is always the splitting field of a cubic.

Solution 1:

Suppose that f and g are monic irreducible polynomials in Q[X] and that they both have a root in the same cubic extension F. Then their discriminants differ from the discriminant of F by a square, so their discriminants have to differ by a square. Unlike in the quadratic case this is only a necessary condition. If f and g have the same discriminant, to see whether they have a root in some common cubic extension, you have to look at their factorizations modulo the primes not dividing their discriminants. If f and g have a different number of factors modulo such a p, then they don't.

Solution 2:

I think that there is something unfair about your question (unfair to the cubic polynomials).

If I ask if $\mathbb Q(\sqrt 5) = \mathbb Q(\varphi)$ where $\varphi$ is the golden ratio, you cannot use your test.

If you ask whether $\mathbb Q (\sqrt[3]a) = \mathbb Q(\sqrt[3]b)$ then the situation is analogous and you check whether $a/b$ or $a/b^2$ are cubes.