Zeta function of number field and linear factors Suppose that $f(x) \in \mathbb Z[x]$ factors in linears over all $\mathbb F_p$ except for a finite number. How can I use the fact that $\lim_{s \to 1} (s-1)\zeta_K(s) \neq 0$ for any finite extension $K:\mathbb Q$ to prove that $f(x)$ factors in linears over $\mathbb Q[x]$? Here, zeta function is $\zeta_K(s):=\prod (1-\frac{1}{N(p)^s})^{-1}$ for all prime ideals $p$ in integers ring $\mathcal O_K$.
My idea is to take $K$ to be a splitting field of $f(x)$ and to prove that $K=\mathbb Q$. Unfortunately, I do not understand a connection between integers ring and $f(x)$: neither $\mathcal O_K \neq \mathbb Z[x]/f(x)$, nor I see why a ramification of ideals in $\mathcal O_K$ is connected with $\zeta_K(s)$. Could you tell me a right direction or a hint?
 A: I don't know how to prove everything, but there is how I understand it

$K$ is the splitting field of a monic polynomial $F \in \mathbb{Z}[X]$, then $K/\mathbb{Q}$ is a Galois extension. $n = [K:\mathbb{Q}]$. 
A prime $p$ factors in $\mathcal{O}_K$ as $(p) = \prod_{j=1}^{g(p)} \mathfrak{p}_j^{e(p)}$ with $N(\mathfrak{p}_j) = p^{f(p)}$, $g(p) = \frac{n}{e(p)f(p)}$ and $e(p) \ne 1$ iff $p | \Delta$ so
$$\zeta_K(s) = \sum_{I \subset \mathcal{O}_K} N(I)^{-s} = \prod_{\mathfrak{p}} \frac{1}{1-N(\mathfrak{p})^{-s}} = \prod_{p | \Delta} \frac{1}{1-N(\mathfrak{p})^{-s})^{ \frac{n}{e(p)f(p)}}}\prod_{p \nmid \Delta}\frac{1}{(1-p^{-sf(p)})^{ \frac{n}{f(p)}}}$$
Now if for almost every $p$, $F(X)$ splits completely in $\mathbb{F}_p$ then (assumed you proved the answer to your other question) $f(p) = 1$ and with $\nabla = \prod_{p, e(p) = 1,f(p) \ne 1}p$ you get
$$\zeta_K(s) =   \prod_{p | \Delta} \frac{1}{(1-p^{-sf(p)})^{ \frac{n}{e(p)f(p)}}}\prod_{p |\nabla }\frac{1}{(1-p^{-sf(p)})^{ \frac{n}{f(p)}}}\prod_{ p \nmid \Delta,p \nmid \nabla}\frac{1}{(1-p^{-s})^{ n}}$$
i.e. $\displaystyle\zeta_K(s) =\zeta(s)^n \prod_{p | \Delta \nabla}  \frac{(1-p^{-s})^n}{(1-p^{-sf(p)})^{ \frac{n}{e(p)f(p)}}}$ has a pole of order $n$ at $s=1$.
But from the class number formula this is a contradiction if $n \ne 1$
