Convergence of the series for the Dedekind zeta function

Question

The Dedekind zeta function of an algebraic number field $K$ is defined as $\zeta_K(s)\mathrel{\stackrel{\rm def}=} \sum\limits_{I \subset\mathcal O_K} \frac{1}{(N_{K/\mathbb Q}(I))^s}$, where $N_{K/\mathbb Q} (I)$ is the norm of the ideal $I$, i.e., the cardinality of $\mathcal O_K/I$. The first question I have about this function is why the series converges for $s \in \mathbb C$ such that ${\rm Re}(s)>1$ and diverges otherwise. To answer this question should I somehow describe ideals in algebraic number fields? Is there a general way to do this?

Answer 1

The convergence for ${\rm Re}(s) > 1$ is discussed in textbooks on algebraic number theory, e.g., Borevich and Shafarevich's "Number Theory" or Lang's "Algebraic Number Theory." Have you looked in any books that discuss zeta-functions of number fields? If not, where are you finding the definition of $\zeta_K(s)$?

To show the series converges when ${\rm Re}(s) > 1$, we can show first that it converges for real $s > 1$. Then it will converge absolutely for $s \in \mathbf C$ with real part greater than $1$, since $|1/{\rm N}(I)^s| = 1/{\rm N}(I)^{{\rm Re}(s)}$, and thus the series converges for $s$ with ${\rm Re}(s) > 1$.

When $s$ is real and greater than $1$, consider the Euler product $\prod_{{\mathfrak p}} 1/(1 - 1/{\rm N}(\mathfrak p)^s)$. Convergence of this product is equivalent to convergence of $\prod_{{\mathfrak p}} (1 - 1/{\rm N}(\mathfrak p)^s)$. A theorem about infinite products is that a product $\prod_{n \geq 1} (1 + x_n)$ with $|x_n| < 1$ converges absolutely (implying it can be expanded out into a convergent series by multiplying out all the terms) if the series $\sum_{n \geq 1} x_n$ converges. Therefore we consider the series $\sum_{\mathfrak p} 1/{\rm N}(\mathfrak p)^s$. If $\mathfrak p$ divides $p\mathcal O_K$ then ${\rm N}(\mathfrak p) \geq p$ and there are at most $[K:\mathbf Q]$ primes $\mathfrak p$ dividing each $p\mathcal O_K$. Therefore $$ \sum_{\mathfrak p} \frac{1}{{\rm N}(\mathfrak p)^s} \leq \sum_p \frac{[K:\mathbf Q]}{p^s} < \infty $$ by convergence of the Dirichlet series for the Riemann zeta-function. This proves $\prod_{\mathfrak p} 1/(1 - 1/{\rm N}(\mathfrak p)^s)$ converges and is equal to $\sum_I 1/{\rm N}(I)^s$.

To show failure of convergence of $\sum 1/{\rm N}(I)^s$ with ${\rm Re}(s) \leq 1$, the case ${\rm Re}(s) < 1$ follows from divergence at $s = 1$ because a basic theorem about Dirichlet series says convergence at a point $s_0$ implies convergence at all $s$ with ${\rm Re}(s) > {\rm Re}(s_0)$. Therefore divergence at $s = 1$ implies divergence if ${\rm Re}(s) < 1$. The divergence at $s = 1$ is related to the existence of a pole at $s = 1$; see any textbook discussing the zeta-function of a number field.

Divergence if ${\rm Re}(s) = 1$ with $s \not= 1$ is more complicated. Have you ever seen a proof that $\sum_{n \geq 1} 1/n^s$ does not converge if ${\rm Re}(s) = 1$ and $s \not= 1$?