Let $f(x)$ and $g(x)$ be irreducible polynomials over rationals. I know that to find the density of the primes $p$ such that (say) $f(x)$ has a root mod $p$ can be found using Chebotarev density theorem, looking at the density of elements with fixed points in the Galois group of $f(x)$. (Thinking of the Galois group embedded in $S_d$, where d is the degree of $f$) I was wondering what is the general way of doing this when we're given two polynomials $f$ and $g$, and if we try to find the density of primes such that both $f$ and $g$ have roots mod $p$. Should we somehow look at the compositum of the splitting fields of $f$ and $g$? That's my feeling, but I don't know how to do that. Thanks.

$\begingroup$ @ R. van Dobben de Bruyn : That's slightly different. There, it finds density of primes which are split in the Galois closure, but prime doesn't necessarily have to split for these polynomials to have root mod $p$, so that answer only gives lower bound for the density. $\endgroup$ – vgmath Jul 12 '17 at 20:56

$\begingroup$ Aren't the two events independent? $\endgroup$ – Igor Rivin Jul 13 '17 at 0:42

$\begingroup$ @vgmath: you're right, it's not the same question. I retracted my close vote. (For the record, the other question.) $\endgroup$ – R. van Dobben de Bruyn Jul 13 '17 at 16:24
Let $P(h)$ be the proportion of primes $p$ such that the monic polynomial $h(t)$ has a root mod $p.$ You say that you can compute this when $h$ is irreducible. Your question is equivalent to asking if you can do it when $h$ is a product of two irreducibles:
You ask about finding $x,$ the proportion of primes where $f$ and $g$ both have roots. But $P(f)+P(g)=P(fg)+x.$

1$\begingroup$ I am not sure I understand why this answers the question, since Chebotarev only applies to irreducibles, I believe )(and @Aaron says as much). $\endgroup$ – Igor Rivin Jul 13 '17 at 4:09

$\begingroup$ It doesn't answer the question. But it does show it equivalent to a presumably hard problem: generalize Chebatorev to nonirreducibles (or at least those with two factors.) So something like computing two splitting fields $F,G$ and their composite $FG$ is not going to solve the given problem unless it solves the harder one. $\endgroup$ – Aaron Meyerowitz Jul 13 '17 at 4:58

$\begingroup$ @Aaron Meyerowitz , @ Igor Rivin : Sorry for my ignorance if I'm wrong, but I couldn't see why Chebatorev requires irreduciblity? Doesn't it work for splitting field of any polynomial? Is there any quick counter example when it doesn't work? $\endgroup$ – vgmath Jul 19 '17 at 1:26