TITLE: Why does the Dedekind zeta function of a number field have a pole at $s=1$?
QUESTION [12 upvotes]: 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?

REPLY [3 votes]: 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 $s \to 1^{+}$.