Under Artin's holomorphy conjecture we can say, for any finite extension $K/F$ of number fields $\frac{\zeta_K(s)}{\zeta_F(s)}$ is entire. Here is a proof:
Let $\overline{K}$ be normal closure of $K$ over $F$. Say $G=\mbox{Gal}(\overline{K} /F)$ and $H=\mbox{Ga}l(\overline{K} /K)$. Clearly both of $G$ and $H$ are finite Galois extensions. Yes, we will take advantage of this. First of all note that, $$L(s, \mbox{Ind}_H^G(1_H),\overline{K}/K)=L(s, 1_{H}, \overline{K}/K)=\zeta_K(s)$$ \noindent We can write $$\mbox{Ind}_H^G(1_H)=1_G+\sum_{\chi \in \mbox{Irr}(G),\chi \neq 1} a_{\chi}\chi$$ \noindent From Frobenius reciprocity it follows, $1_G$ is coming with multiplicity one. And also $a_{\chi}$'s are non-negative as they are multiplicity of $\chi$ in $\mbox{Ind}_H^G(1_H)$. So we get $$L(s, \mbox{Ind}_H^G(1_H),\overline{K}/K)=\zeta_F(s)\prod_{\chi \neq 1}L(s,\chi,K)^{a_{\chi}} =\zeta_K(s)$$ where $a_{\chi}$'s are all non negative. Artin's conjecture does rest of the job.
A recent expository note by K. Murty and Foote (see here at page 474) says , under Artin conjecture $$\frac{\zeta_{K_1K_1}(s)\zeta_{K_1\cap K_2}(s)}{\zeta_{K_1}(s)\zeta_{K_2}(s)}$$ is entire. I am not understanding how to prove this. Could anyone please explain ?
Source of the paper: Foote, Richard; Ginsberg, Hy; Murty, Vijaya Kumar, Heilbronn characters., Bull. Am. Math. Soc., New Ser. 52, No. 3, 465-496 (2015). ZBL1326.20007.