3
$\begingroup$

Suppose $f(x)\in\mathbf Z[x]$ is nonconstant. I would like to know if either of the following statements is true.

  1. If $a$ and $b$ are coprime integers (probably with some additional restriction), then there exist infinitely many primes $p$ such that $p\equiv a\bmod b$ and $f(x)$ has a root modulo $p$.

  2. It is possible choose $a$ and $b$ so that, for every prime $p\equiv a\bmod b$, $f(x)$ has a root modulo $p$.

$\endgroup$
1
  • 3
    $\begingroup$ Certainly some restrictions are needed --- there is no prime $p\equiv3\pmod4$ such that $x^2+1$ has a root modulo $p$. $\endgroup$ Apr 8, 2014 at 23:27

1 Answer 1

10
$\begingroup$

Here's a survey of the possible things that can happen. In regards to your first question, given any polynomial $f(x)$, there is a positive integer $M$ so that if $\gcd(b,M) = 1$, then there are infinitely many primes $p \equiv a \pmod{b}$ for which $f(x)$ has a root modulo $p$. (One can take for $M$ the modulus of the maximal abelian subextension $K/\mathbb{Q}$ of the splitting field of $f(x)$.) It is possible for $M$ to be $1$, in which case statement $1$ is true. On the other hand, there examples that show that some restraint on $b$ is necessary. The example Gerry gave is one. Another is $f(x) = x^{4} - 4x^{2} + 2$, which has a root modulo an odd prime $p$ if and only if $p \equiv 1 \pmod{8}$.

The answer to question 2 is no in general. Let $f(x) = x^5 + 20x + 16$. The splitting field of $f(x)$ over $\mathbb{Q}$ is a Galois extension $K$ with Galois group $A_{5}$. The Chebotarev density theorem applied to the compositum of $K$ with $\mathbb{Q}(e^{2 \pi i / b})$ implies that for any $a$ and $b$ with $\gcd(a,b) = 1$, it is possible to find a prime $p_{1} \equiv a \pmod{b}$ so that $f(x)$ has a root modulo $p_{1}$, and another prime $p_{2} \equiv a \pmod{b}$ so that $f(x)$ does not have a root modulo $p_{2}$.

On the other hand $f(x) = x^{3} - 2$ has a root modulo every prime $p \equiv 2 \pmod{3}$, and the polynomial $f(x) = (x^{2} - 2)(x^{2} - 3)(x^{2} - 6)$ has a root modulo every single prime number.

$\endgroup$
7
  • $\begingroup$ Could you suggest a reference where I can read a proof of your statement about the number $M$? Also, in the first paragraph, did you meant to write $p\equiv b\pmod M$ rather than $p\equiv a\pmod b$? $\endgroup$
    – 352506
    Apr 9, 2014 at 3:40
  • $\begingroup$ The statement I made about the number $M$ follows from the Chebotarev density theorem, together with some Galois theory (namely the "natural irrationalities theorem"). The best two sources I can think of for the Chebotarev density theorem are this article, and David Cox's book "Primes of the form $x^{2} + ny^{2}$", Chapter 2. $\endgroup$ Apr 9, 2014 at 12:14
  • $\begingroup$ In the first paragraph, I did mean to write $p \equiv a \pmod{b}$. Roughly speaking, if $\gcd(b,M) = 1$, the occurrence of $f(x)$ having a root modulo $p$, and the residue class of $p$ modulo $b$ are "independent". $\endgroup$ Apr 9, 2014 at 12:16
  • $\begingroup$ The link to the article appears to be broken. Is the article published in a journal that I could look up? $\endgroup$
    – 352506
    Apr 9, 2014 at 19:50
  • $\begingroup$ I typed the wrong URL earlier, sorry (and it seems I cannot edit it now). The article is "Chebotarev and his density theorem" written by Peter Stevenhagen, and Hendrick W. Lenstra. It was published in the Mathematical Intelligencer in 1996. You can find it from Lenstra's website. $\endgroup$ Apr 9, 2014 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.