Is there a characterization of irreducible polynomials over $\mathbb Q$ whose splitting field over $\mathbb Q$ are isomorphic to a rupture field?

In other words, of polynomials $P \in \mathbb Q(X)$ that are irreducible over $\mathbb Q$ and that split completely in $\mathbb Q(X) /(P)$.

Equivalently, if $\alpha$ is any root of $P$ then $\mathbb Q(\alpha)$ contains every root of $P$.

  • 2
    $\begingroup$ What is a rupture field? $\endgroup$ – lhf Dec 27 '18 at 15:52
  • $\begingroup$ It is a minimal extension containing at least one root. It only makes sense for $P$ irreducible, I will edit my question. Every ruture field is isomorphic to $\mathbb Q(X) / (P)$. $\endgroup$ – Régis Dec 27 '18 at 15:54
  • $\begingroup$ Maybe I am confused, but I think it's equivalent to the Galois group of the polynomial being cyclic. That's likely not the type of characterization you want but it might help with searching. (There may be no direct characterization.) $\endgroup$ – quid Dec 27 '18 at 16:11
  • $\begingroup$ Yes, I already saw that my condition is equivalent to $|Gal(P)|=deg(P)$. In particular it is the case if the Galois group is cyclic. I am not sure about the converse, though. $\endgroup$ – Régis Dec 27 '18 at 16:14
  • $\begingroup$ I thought that it follows from the transitivity of the action of the Galois group on the roots that with that cardinality restriction the group must me cyclic, but maybe that's not true. I have to confess I cannot give a clear argument now. At least in the case of prime degree it's true though. $\endgroup$ – quid Dec 27 '18 at 20:15

Here are the easy cases:

For quadratic polynomials, every rupture field is a splitting field.

For cubic polynomials, a rupture field is a splitting field iff the discriminant is a square.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.