Why does the Dedekind zeta function of a number field have a pole at $s=1$?

Question

The analytic class number formula tells us that the Dedekind zeta function $\zeta_K$ of a number field $K$ has a pole at $s=1$ with residue $$\frac{2^{r_1}(2\pi)^{r_2}\text{Reg}_Kh_K}{w_K\sqrt{|\Delta_K|}}.$$ Is there a quick way to see that $\zeta_K$ should have a pole at $s=1$ without explicitly computing the residue? Is there some theorem concerning $L$-functions that yields this fact immediately?

Answer 1

It is not too hard if $K/\mathbb Q$ is an abelian extension. Put $G$ to be the Galois group of $K/\mathbb Q$. Then we have the factorization $$ \zeta_K(s) = \zeta(s) \prod L(s, \chi) $$ where $\chi$ runs over nontrivial characters of $G$. These are Dirichlet characters, and by Dirichlet we know these are nonvanishing at $s=1$, so you get a (simple) pole for $\zeta_K$ at $s=1$.

If $K/\mathbb Q$ is Galois, one has an analogous factorization in terms of Artin $L$-functions, $L(s,\rho)$, where $\rho$ runs over nontrivial irreducible representations of $G$. By a theorem of Brauer, one again knows $L(1, \rho) \ne 0$ and we get the desired result.

Finally, for $K/\mathbb Q$ non-Galois, one can look at the Galois closure $L$ and compare the Artin factorizations for $L/\mathbb Q$ and $L/K$, that as I recall gives you what you want.

However, I haven't thought about the details of either of these approaches (Artin $L$-function or analytic class number formula) recently enough to have an opinion about which approach should be considered "quicker." (One also needs to decide where to start the race from.)

Answer 2

For an arbitrary number field, proving the analytic continuation of $\zeta_K(s)$ is non-trivial. For a totally complex field (no real embeddings) it needs looking at $$\zeta_K(s)(2\pi)^{-n s} \Gamma(s)^n =\int_{\Bbb{R}^n / L} (\Theta(e^{t_1},\ldots,e^{t_n})-|C_K|)\exp(\sum_{j=1}^n e^{s t_j})d^n t$$ where

• $\Theta(x_1,\ldots,x_n) =\sum_{c \in C_K} \sum_{a \in c^{-1}} \exp(-2\pi \frac{N(c)}{N(cc^{-1})}\sum_{j=1}^n x_j|\sigma_j(a)|^2)$

• $\sigma_j,\overline{\sigma_j}$ are the $n = \frac{[K:\Bbb{Q}]}{2}$ pairs of complex embeddings $K \to \Bbb{C}$

• $c,c^{-1}$ are ideals of $O_K$ representative of the ideal class group $C_K$

• $L$ is the lattice of $\Bbb{R}^n$ generated by the $(\log |\sigma_1(u_r)|^2,\ldots,\log |\sigma_n(u_r)|^2)$ with $O_K^\times = \prod u_r^\Bbb{Z}$.

• $d^nt$ is the usual measure of $\Bbb{R}^n$ and $\int_{\Bbb{R}^n / L}$ means integrating over a fundamental domain

From there the analytic continuation, the unique simple pole at $s=1$, the class number formula, the functional equation (*) are easily obtained.

(*) $\sum_{a \in c^{-1}}$ reduces to summing over a lattice, the functional equation relates it with its different ideal

Answer 3

If the goal is just to show that $\lim_{s \to 1^+} \zeta_K(s) = \infty$, then we can take the standard proof of the class number formula and strip it down to something quite short.

Let $n = [K: \mathbb{Q}]$. Let $A$ be the ring of integers of $K$ and let $V = A \otimes \mathbb{R}$. So $A$ is a discrete lattice in the $n$-dimensional vector space $V$. Addition and multiplication as maps $A \times A \to A$ extend to polynomial maps $V \times V \to V$, making $V$ into a ring. The norm map $N : K \to \mathbb{Q}$ similarly extends to a polynomial $N:V \to \mathbb{R}$. On the dense open set $N \neq 0$, we can extend $a \mapsto a^{-1}$ to a continuous map $\{ N \neq 0 \} \to \{ N \neq 0 \}$. We will denote all of these extended operations $+$, $\times$, $N$, ${a}^{-1}$ on $V$ by the same symbols as the original ones.

The $\zeta$-function is a sum over ideals $\sum_{I \subseteq A} |I|^{-s}$, so it is bounded below by the sum over principal ideals $\sum_{I \subseteq A,\ I \ \mbox{principal}} |I|^{-s}$. Suppose we can construct some subset $\Delta$ of $V$ such that no two elements of $\Delta \cap A$ generate the same ideal, then we will have $$\zeta_K(s) \geq \sum_{a \in \Delta} N(a)^{-s}.$$ If $\Delta$ is reasonably nice, and bounded away from $0$, then we can approximate the sum as an integral $$\tfrac{1}{\mathrm{Vol}(A)} \int_{x \in \Delta} N(x)^{-s} \qquad (\ast)$$ where $\mathrm{Vol}(A)$ is the volume of a fundamental domain of the lattice $A$.

So the goal is to construct $\Delta$ small enough that no two elements of $\Delta \cap A$ generate the same ideal, but large enough that we can show $(\ast)$ diverges as $s \to 1^{+}$.

The lattice $A$ is discrete. So we can find $\epsilon$ small enough that the ball $B_{\epsilon}(1)$ of radius $\epsilon$ around $1$ contains no element of $A$ other than $1$. We can find $\delta$ small enough that, if $a$ and $b$ lie in $B_{\delta}(1)$ then $a^{-1}b$ lies in $B_{\epsilon}(1)$. Thus, for $u_1$ and $u_2 \in B_{\delta}(1)$, either $u_1=u_2$ or $u_1^{-1} u_2 \not \in A$.

Let $X$ be the hypersurface $N^{-1}(1)$ in $V$. This is an $n-1$ dimensional manifold. Let $Y = X \cap B_{\delta}(1)$. Let $\Delta$ be those elements that can be written as $tu$ with $t \in [1,\infty)$ and $u \in Y$.

We claim that, if $a_1 = t_1 u_1$ and $a_2 = t_2 u_2$ in $\Delta \cap A$ generate the same ideal then $a_1 = a_2$. Since $(a_1) = (a_2)$, we must have $N(a_1) = N(a_2)$. We have $N(a_1) = t_1^n N(u_1) = t_1^n$ and likewise for $a_2$, so $t_1 = t_2$. The condition that $(a_1) = (a_2)$ further means that $a_1^{-1} a_2$ is a unit of $A$. But $a_1^{-1} a_2 = u_1^{-1} u_2$ and we selected $\delta$ so that $u_1^{-1} u_2 \in A$ implies $u_1 = u_2$.

So our domain $\Delta$ has the required property, and it remains to estimate the integral $(\ast)$. We make the change of variables $x = tu$, and note that the determinant of the Jacobian is $t^{n-1}$. We have $N(tu) = t^n N(u) = t^n$. So $(\ast)$ is $\int_{t=1}^{\infty} \int_{u \in Y} t^{n-1} t^{-ns} = \mathrm{Vol}(Y) \int_{t=1}^{\infty} t^{n-1-ns} dt$. Sure enough, this diverges as $s \to 1^{+}$.