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.
Asked
Active
Viewed 296 times
3
-
@ 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. – vgmath Jul 12 '17 at 20:56
-
Aren't the two events independent? – Igor Rivin Jul 13 '17 at 00:42
-
@vgmath: you're right, it's not the same question. I retracted my close vote. (For the record, the other question.) – R. van Dobben de Bruyn Jul 13 '17 at 16:24
2 Answers
4
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 $P(f)+P(g)=P(fg)+x.$
Aaron Meyerowitz
- 30,015
-
1I am not sure I understand why this answers the question, since Chebotarev only applies to irreducibles, I believe )(and @Aaron says as much). – Igor Rivin Jul 13 '17 at 04:09
-
It doesn't answer the question. But it does show it equivalent to a presumably hard problem: generalize Chebatorev to non-irreducibles (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. – Aaron Meyerowitz Jul 13 '17 at 04:58
-
@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? – vgmath Jul 19 '17 at 01:26
2
There are two cases: when $f, g $ are coprime polynomials and the other case is when they are not. In the first case, there are integer polynomials $f_1,g_1$ and a non-zero integer $c$ such that $f(x)f_1(x) + g(x) g_1(x)=c$, due to Euclid's algorithm. Therefore, if for a prime $p$ the polynomials $f, g $ have a common root then there exists $x\in \mathbb F_p$ such that $f(x)=0=g(x)\mod{p}$, hence, $c=0$ in $\mathbb F_p$. In other words, $p$ must divide $c$. However, $c\neq 0 $, hence, it has finitely many prime divisors. Therefore, the requested density is zero, simply because the list of primes is finite. The second case is when $f,g$ are not coprime-then since they are both irreducible, the primes $p$ for which there exists a common root modulo $p$ coincides with the primes $p$ for which the first polynomial has a root-apart (possibly) from at most a finite number of primes $p$. Hence, the second case is exactly what can be done via Chebotarev. Note that the same reasoning holds whenever you have any number of irreducible integer polynomials, not just $2$. Then the density will be zero except if all polynomials are multiples of the same.
Dr. Pi
- 2,949