# 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.

-
Even the same cubic polynomial may not generate the same extension as itself: $x^3-2$ generated both a totally real extension, $\mathbb{Q}(\sqrt[3]{2})$, and an extension that is not contained in the real numbers, $\mathbb{Q}(\zeta\sqrt[3]{2})$, with $\zeta$ a root of $x^2+x+1$. Of course, they are isomorphic, but in the case of quadratics as you set them up you have equality, not merely isomorphism. You need to be much more precise in your question. –  Arturo Magidin Apr 22 '11 at 16:33
I'm looking for equality, not just isomorphisms. That is I'm essentially looking for a way to enumerate all degree 3 extensions of either $\mathbb{Q}$ or $\mathbb{Q}_p$. EDIT: and that would be inside some fixed algebraic closure. –  dstt Apr 22 '11 at 17:15
If you are looking for equality, then it is not enough to specify that the extension is given by "a cubic polynomial." As I point out above, this is not well defined, since the same cubic polynomial may give rise to two or three distinct (though isomorphic) extensions. You would also have to specify which root of the cubic polynomial you are taking. –  Arturo Magidin Apr 22 '11 at 17:16
"degree $3$ extension" is not the same thing as "extension generated by the roots of a cubic polynomial." The former are all abelian, hence classified by class field theory, but the latter includes degree $6$ extensions with Galois group $S_3$. –  Qiaochu Yuan Apr 22 '11 at 17:29
I guess the question could be rephrased to ask if there's a way to check that two cubic polynomials have the same splitting field in some fixed algebraic closure. This is essentially what we do in the quadratic case. –  dstt Apr 22 '11 at 17:39

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.

-
Alright, so $p$-adics are actually easier, because there's only one prime. –  Eric Apr 22 '11 at 18:41

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.

-
But while any quadratic field equals $\mathbb{Q}(\sqrt{a})$ for some $a$, not all cubic fields are of the form $\mathbb{Q}(\sqrt[3]{a})$. –  Zev Chonoles Apr 22 '11 at 16:10
Yes, I agree, I was just pointing out that it was not fair to demand a condition on two general roots of cubic polynomials that mirrors the condition for pure square roots. –  Phira Apr 22 '11 at 20:42