Why does the Dedekind zeta function of a number field have a pole at $s=1$? 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? 
 A: 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.)
A: 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
A: 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 $s \to 1^{+}$.
