resultants over finite fields vs resultants over their closure I'm having a little trouble grokking the following line of argument. I have two monic, univariate polynomials $p(x)$ and $q(x)$ over $\mathbb{Z}/p\mathbb{Z}$ that share a common root, where $p$ is prime. I want to imply that they must therefore have a common root over $\mathbb{C}$. I am told that the proof of this involves resultants and minimal polynomials, but I am missing the nub of the argument. Why is it so? Any help shedding some light on this would be much appreciated.
 A: So we are given two polynomials $p(x),q(x)\in\Bbb{Z}[x]$ such that when reduced modulo any prime $\ell$ they share a common root in $\Bbb{Z}/\ell\Bbb{Z}$. The claim is that they must have a common complex zero.
Assume contrariwise that $p(x)$ and $q(x)$ do not share a common complex zero. Because $\Bbb{C}$ is algebraically closed, this means that the greatest common divisor
$m(x)=\gcd(p(x),q(x)$ must be constant $1$.
By Bezout's identity there exists polynomials $u(x),v(x)\in\Bbb{Q}[x]$ such that
$$
u(x)p(x)+v(x)q(x)=m(x)=1.
$$
(If you have not heard Bezout's name associated with this fact you can alternatively recall that the ring $\Bbb{Q}[x]$ is a principal ideal domain, and $m(x)$ is a generator of the ideal $\langle p(x),q(x)\rangle$).
In general the polynomials $u(x),v(x)$ have rational coefficients. But they have only finitely many terms, so the denominators of those coefficients have a least common multiple $d$. So multiplying the above equation by $d$ we get polynomials
$U(x)=du(x),V(x)=dv(x)\in\Bbb{Z}[x]$ such that
$$
U(x)p(x)+V(x)q(x)=d.\qquad(*)
$$
Now let us choose the prime $\ell$ such that $\ell\nmid d$. It follows that $p(x)$ and $q(x)$ cannot have a common zero in $\Bbb{Z}/\ell\Bbb{Z}$. For if the residue class $\overline{a}$ where such a zero, then the integers $p(a)$ and $q(a)$ would both be divisible by $\ell$. This contradicts the equation $(*)$, because its left hand side
$U(a)p(a)+V(a)q(a)$ would then also be divisible by $\ell$, but the right hand side is not.
This is the desired contradiction, and the claim follows.

Initially it was unclear whether the assumption should hold for all primes $\ell$. This cannot be relaxed essentially (well, it is sufficient if this is known to hold for infinitely many primes $\ell$). Most certainly a single prime won't do. For example $p(x)=x+1$ and $q(x)=x+4$ share the root $x=2$ modulo $\ell=3$, but the obviously do not share any complex zeros. I leave it as an exercise to generalize this example to one that
works for a given finite set of primes.
