6
$\begingroup$

Let $P(x)$ be an irreducible monic polynomial of degree $\ge4$ with integer coefficients. We all know that over a finite field $\mathbb F_p$, $P$ will often split, and I am interested in polynomials with the property that, for any given $p$, all splitting factors in $\mathbb F_p[x]$ have the same degree. Call such polynomials "uniformly splitting". (This doesn't exclude the possibility that for certain $p$, they may remain irreducible.)

E.g. it is not hard to see that $P(x)=x^4-x^2+1$ is uniformly splitting, because it will never split into two linear factors and one quadratic irreducible one.
I suspect the situation is similar for $P(x)=x^8-x^6+x^4-x^2+1$. (Is it really?)
On the other hand, e.g. $P(x)=x^8-x^6-x^4-x^2+1$ splits uniformly for all $p<43$, so far so good, but in $\mathbb F_{43}$, we have $$P(x)=(9+x) (19+x) (24+x) (34+x) (14+x^2) (40+x^2).$$

What are necessary or sufficient conditions for an irreducible polynomial to be uniformly splitting?

It seems to me like such polynomials are very rare, at least for degree $>4$. I conjecture that cyclotomic polynomials $\Phi_n(x^k)$, as long as they are irreducible, are uniformly splitting.

Given that the constant term must be $\pm1$, it seems reasonable to expect most if not all uniformly splitting polynomials to be symmetric (in the sense of having self-mirrored coefficients). Then all factors of a (uniform) split for a given $p$ look somewhat like "conjugates" of each other, in a yet-to-define broader way obviously involving the primitive elements of $\mathbb F_{p}$. Might Galois theory be of any help here? And:

Are there uniformly splitting polynomials of degree $>4$ which are not of the form $\Phi_n(x^k)$?

$\endgroup$
11
$\begingroup$

Yes, there are uniformly splitting polynomials of all degrees which are not of the form $\Phi_{n}(x^{k})$. (For example, $f(x) = x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$.)

The Chebotarev density theorem implies that if $f(x)$ is irreducible and $p$ does not divide the discriminant of $f$, then the factorization of $f(x)$ mod $p$ corresponds to a cycle type of an element of the Galois group of $f(x)$ (acting on the roots of $f(x)$). Moreover, every element of the Galois group shows up "equally often" in this way.

Modulo the small detail about primes dividing the discriminant of $f$, the question now is about transitive permutation groups $G$ with the property that every element consists of cycles of the same length. In such a permutation group, the only element that can have a fixed point is the identity, and this means the permutation group must be regular and indeed any regular permutation group arises from a group acting on itself by right multiplication. It is easy to see that in such a group, every element is a product of cycles of the same lengths. These are plentiful!

In particular, given any field $K/\mathbb{Q}$ which is Galois, there is an element $\alpha \in K$ so that $K = \mathbb{Q}[\alpha]$ and the minimal polynomial of $\alpha$ is a uniformly splitting polynomial.

$\endgroup$

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.