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Are there any reasonable ways to define zeta function of a maximal abelian extension $\mathbb{Q}^{ab}$ of the rationals, or at least zeta functions of some its infinite-dimensional subfields?

In particular, can we use zeroes of Dedekind zeta functions in the following way? Let $N$ be positive integer, let $\zeta_N$ be the Dedekind zeta function of the $N$-th cyclotomic extention of $\mathbb{Q}$, and let $Z_N:= \{s\in\mathbb{C}: \zeta_N(s)=0, 0< \mathrm{Re} s <1\}$ be its non-trivial zeroes. It is known that $Z_N \subset Z_M$ whenever $N$ divides $M$. But is $$Z:=\bigcup_{N\ge 1}{Z_N}$$ a discrete subset of $\mathbb{C}$, so that we could construct a meromorphic function having $Z$ as its zero set?

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The usual definition with "ideal" replaced by "cofinite ideal" (ideal such that the quotient is finite) should still work, although I don't know if it captures the same kind of information. –  Qiaochu Yuan Jun 30 '13 at 19:42
@Qiaochu The problem is that $\mathcal{O}_{\mathbb{Q}^{ab}}=\mathbb{Z}[\mu_\infty]$ has no ideals of finite index. –  Kevin Ventullo Jun 30 '13 at 20:08
By the way, there is no hope that the functions $\zeta_N$ will converge to a holomorphic function, since e.g. the order of vanishing at $s=0$ is getting larger and larger. I am no analytic number theorist, but my guess is that the set $Z$ will be dense on the line $Re(s)=1/2$. If this is true, one could still ask whether the sets $Z_N$ converge "in measure'' to some distribution, à la Sato-Tate. –  Kevin Ventullo Jun 30 '13 at 20:43
@ Kevin Ventullo : One can replace Dedekind zeta function with its completed version having simple poles at s=1 and s=0. Numerical simulation suggests that zeroes of Diriclet L-functions and Dedekind zetas might go closer as order N increases, but I don't know any analytic reason for such behaviour. –  Yauhen Radyna Jun 30 '13 at 21:46

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