I'm having troubles working on the exercise 7.2 of Lenstra - The Chebotarev Density Theorem (http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf). I have proved the first part, that is a group $G$ acting transitively on a finite set $X$ with at least two elements admits a $g$ that doesn't fix any $x\in X$. I want to show that

Let $f(X)\in\mathbb{Z}[X]$ an irreducible polynomial such that $f\pmod{p}$ admits a roots in $\mathbb{F}_p$ (the finite field with $p$ elements) for almost every prime $p$ (i.e. all but finitely many). Then $f(X)$ has degree 1.

I couldn't find help online and I'm stuck. I know that I have to use both the Chebotarev density theorem and the above group action fact, but I don't know how. I tried assuming the contrary, so $f(X)$ has $\ge 2$ roots and the Galois group of its splitting field over $\mathbb{Q}$ acts transitively on them, but I have no idea if this will lead me to a contradiction.

Any hint?


marked as duplicate by mercio, Dietrich Burde, drhab, Aditya Hase, Community Apr 30 '15 at 12:51

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  • $\begingroup$ By curiosity, how did you prove the first question? $\endgroup$ – Thibaut Dumont Apr 29 '15 at 21:16
  • 1
    $\begingroup$ Using Burnside's formula ad absurdum, using the fact that there is only one orbit and $|X|>1$. $\endgroup$ – Angelo Rendina Apr 29 '15 at 21:18
  • $\begingroup$ The contrapositive of what you need to prove is: if $f(X)$ has degree at least $2$, then there are infinitely many primes $p$ such that $f(X)$ has no root in $\Bbb F_p$. How does the Galois group act on the complex roots of $f$? How is that related to $f(X)$ not having a root in $\Bbb F_p$? (By the way, these Galois groups always act transitively on the roots of $f$, right?) $\endgroup$ – Greg Martin Apr 29 '15 at 21:23
  • $\begingroup$ @GregMartin I can't really get the hint, I'm sorry. I think the problem is that I fail to see the link between roots in $\mathbb{F}_p$ and roots in $\mathbb{Q}_p$. Except Hensel's lemma, but I don't think I can apply it because I'm not told that the root is simple. $\endgroup$ – Angelo Rendina Apr 30 '15 at 10:25
  • $\begingroup$ There are only finitely many primes (those dividing the discriminant) for which a given squarefree polynomial has non-simple roots. $\endgroup$ – Greg Martin Apr 30 '15 at 20:39

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