Let $F$ be a number field and $S$ be a finite set of places of $F$ including archimedian places. Let $\zeta^S(s)$ be the partial L-function, that is the meromorphic continuation of the product of local zeta functions $\zeta_v(s)=\frac{1}{1-q^{-s}}$ outside the places in $S$.

Then we know that the complete zeta function $\zeta(s)$ has a simple pole at $s=0$.

Since $\zeta(s)=\zeta^S(s)\cdot \prod_{v \in S}\zeta_v(s)$ for $\Re(s) \gg0$ and each $\zeta_(s),\zeta^S(s),\zeta_v(s)$ has meromorphic continuation to $\mathbb{C}$, I suppose the order of $\zeta^S(s)$ at $s=0$ is (the order of $\zeta(s)$ at $s=0$) - $\sum_{v \in S}ord_{s=0}(\zeta_v(s))$.

So I think $\zeta^S(s)$ would have a pole or have zero at $s=0$ depending the set of $S$.

My reasoning is right?

  • $\begingroup$ Yes, indeed. Thus, for number fields other than the rationals and complex quadratic extensions thereof, even if $S$ is only the archimedean places, the finite-prime zeta will have a $0$ at $s=0$, since the Gamma functions will have at least a double pole there. $\endgroup$ – paul garrett Oct 7 '14 at 17:21
  • $\begingroup$ @garret, I really thank you for your answer. It helped me very much! $\endgroup$ – user29422 Oct 7 '14 at 18:21

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