nt.number theory - Sum of reciprocals of primes modulo which a polynomial has a root

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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