How to find all finite fields in which a given polynomial has no root.

Question

Assume that a polynomial $p(x)$ is given. I am wondering if we may find all prime finite fields in which $p(x)$ has no root. For example if $p(x)=-x^6+3 x^5-6 x^4+7 x^3-6 x^2+3 x-1$, then $F_{11}$ is a such field. Thank you!

Answer 1

Your example polynomial $$p(x) = -x^6 + 3 x^5 - 6 x^4 + 7 x^3 - 6 x^2 + 3 x - 1=-(x^2-x+1)^3,$$ so for this particular polynomial it is easy to give a general answer. This is because the quadratic $x^2-x+1$ is the sixth cyclotomic polynomial. It has the obvious double root $x=-1$ modulo $p=3$. Otherwise it has zeros in the prime field $K=\Bbb{F}_p$ if and only $K$ has sixth roots of unity, which happens if and only if $p\equiv1\pmod 6$.

In general, for a random polynomial $f(x)$ with integer coefficients, this is a difficult question. If we know the Galois group $G$ of $f(x)$ as a group of permutations of roots of $f$, then Chebotarev's density theorem gives us a method for finding the fraction of primes for which there are modular zeros. But, in general the set of such primes won't be given by congruence conditions. That happens only (at least that's how I understand the situation) when the group $G$ is abelian. Class field theory has a lot to say about that.