# Sum of reciprocals of primes modulo which a polynomial has a root

Dear all,

I am looking for a proof or a reference of the following statement:

Let $f$ be a non-constant polynomial with integer coefficients. Then the sum $\sum \{1/p \mid f \text{ has a root modulo } p\}$ diverges.

I am pretty sure that I saw it somewhere before but I cannot remember and I failed to find it in number theory books. A possible routes that has already been suggested to me is actually showing that the sums of reciprocals for which f even decomposes into linear factors has positive density which should stem from Galois theory. I am however not an expert in Galois theory so that I would prefer a direct proofs or a reference.

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This should follow from the Theorem of Frobenius mentioned on p. 7 (PDF numbering) of http://websites.math.leidenuniv.nl/algebra/chebotarev.pdf.

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Charles is completely right that this follows from Frobenius' theorem. Since you don't like Galois theory, here is a proof which does not explicitly mention Galois theory. (But it is hiding just out of sight.)

We may assume that $f$ is irreducible as, if $g$ divides $f$, then the set of primes for which $f$ has a root contains the set for which $g$ does.

Let $K$ be the field $\mathbb{Q}[x]/f(x)$. Let $R$ be the ring of integers of $K$, and let $S=\mathbb{Z}[x]/f(x)$. Note that $S$ is a finite index sublattice of $R$ and, if $p$ is a prime which does not divide $|R/S|$, then $R/p \cong S/p$. Also, for any prime $p$ which does not divide the discriminant of $f$, the polynomial $f$ factors into distinct factors in $\mathbb{F}_p[x]$.

Thus, if $p$ is large enough to not divide either $|R/S|$ or the discriminant of $f$, then $R/p \cong S/p \cong \mathbb{F}_p[x]/f(x) \cong \bigoplus \mathbb{F}_p[x]/(f_i(x))$ where $f_i$ are the irreducible factors of $f$ mod $p$. So, for such a prime $p$, prime ideals of $R$ which contain $(p)$ are in bijection with irreducible factors of $f$ mod $p$, and the norm of such a prime is $p^{\deg f_i}$.

So, if $f$ has a root modulo $p$, then $$\frac{1}{p} \leq \sum_{\pi \supseteq (p), \ \pi \ \mbox{prime}} \frac{1}{N(\pi)} \leq \frac{\deg f}{p}$$ and, if $f$ does not have a root modulo $p$, then $$\sum_{\pi \supseteq (p), \ \pi \ \mbox{prime}} \frac{1}{N(\pi)} \leq \frac{(\deg f)/2}{p^2}$$

We want to show that $$\sum_{p: \exists \pi \ \mbox{a prime of} \ R \ \mbox{with} \ N(\pi)=p} \frac{1}{p}$$ diverges. By the above inequalities, it is equivalent to show that $$\sum_{\pi \subset R, \ \pi \ \mbox{prime}} \frac{1}{N(\pi)}$$ diverges. (Note that the finitely many primes which divide $|R/S|$ or the discriminant of $f$ cannot change whether or not the sum converges.)

Now, we have unique factorization into prime ideals for $R$, so $$\sum_{I \subseteq R} \frac{1}{N(I)^s} = \prod_{\pi \subset R, \ \pi \ \mbox{prime}} \left( 1 - \frac{1}{N(\pi)^s} \right)^{-1}.$$

The left hand side is the $\zeta$ function of $K$. By the class number formula (see most books on algebraic number theory), $\zeta_K(s) = C/(s-1) + O(1)$ for some positive constant $C$, as $s \to 1^{+}$. So $$\log \zeta_K(s) = \log \frac{1}{s-1} + O(1) = \sum \log \left( \frac{1}{1-N(\pi)^{-s}} \right) = \sum \frac{1}{N(\pi)^s} + O(1/N(\pi)^{2s}).$$ We deduce that $$\sum \frac{1}{N(\pi)^s} = \log \frac{1}{s-1} + O(1)$$ so $$\sum \frac{1}{N(\pi)}$$ diverges.

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Thanks a lot guys. That was what i was looking for! --Alberto – Alberto Sep 3 '10 at 22:01