Classes of polynomials whose roots form a cyclic group

[burritoloco]

Hi, I'm currently doing a project and this topic has come up. Are there any known famous classes of polynomials (besides cyclotomic polynomials) that fit that description? In particular, I'm more interested in the case where the polynomials have odd degree. I know for example that the roots of cyclotomic polynomials form cyclic groups, but they have even degree except for Phi_1(x) and Phi_2(x). Many thanks!

 

[burritoloco]

Hmm, because all the finite multiplicative subgroups of the complex numbers are precisely the cyclic groups generated by primitive roots of unity, wouldn't this imply that the only polynomials with that characteristic (above) are the cyclotomic polynomials? How about when considering polynomials over finite fields instead?

 

[micromass]

Hi burritoloco!

Do you mean that the roots must form a cyclic group, or that the roots must generate a cyclic group? Judging from your talk about the cyclotomic polynomials, I guess the latter.

Well, when working over a finite field, then every polynomial $P(X)\in \mathbb{F}_p(X)$ with $P(X)\notin (X)$ will generate a cyclic group!

Also, in every field with characteristic p, the Artin-Schreier polynomial

$$X^p-X+a$$

with a nonzero, will generate a cyclic group.

 

[burritoloco]

Hi micromass and thanks for your reply!

I initially meant "form" but I now realize that 1 is not a root of cyclotomic polynomials except for Phi_1(x), so its roots cannot form a group! Thus "generate" seems to be the right word. So let me rephrase. Let's say we make a set with all the roots of a polynomial, and 1. For what polynomials will this set be a cyclic group?

 

[micromass]

Well, then I guess the only polynomials that satisfy this are some divisors of $X^n-1$...

 

[burritoloco]

I see :). What if instead of taking a cyclotomic polynomial as such a divisor, we take an irreducible factor of it (so over finite fields). Could it still satisfy the above?

 

[burritoloco]

By the above I meant my previous post...

 

[burritoloco]

Of course, the roots of an irreducible factor of a cyclotomic polynomial form a subset of the cyclic group generated by the corresponding primitive root of unity. The question is: when we include 1 in this subset, can it be a cyclic subgroup? When does this happen?

 

[burritoloco]

Hmm, I'm afraid not. The only way this subset is a proper subgroup of the cyclic group generated by an nth primitive root of unity, is if it's generated by a non-nth primitive root, say an mth primitive root of unity. So the corresponding polynomial would still be the mth-cyclotomic polynomial not necessarily irreducible over the finite field. Moreover, this polynomial would still have even degree (if it's not Phi_1(x), Phi_2(x)). So no luck!