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 $\Phi_n(x^k)$?


1 Answer 1


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.


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