TITLE: unramified quadratic extension of number field
QUESTION [7 upvotes]: I try to understand the following statement
There are only finitely many quadratic unramified extension of a number field $K$
I know by Kummer theory that such extensions are of the form $K(\sqrt{a})$ for some $a \in K$ such that $a$ generates $K^*/(K^*)^2$. But I can't figure out how to use this to show the claim.

REPLY [3 votes]: Here is a proof similar to tracing's but phrased in slightly different language, and which shows that, in fact, $K$ has only finitely many quadratic extensions unramified outside of primes over $2$. 
Namely, let $U$ be $\mathrm{Spec}(\mathcal{O}_K)-S$, where $S$ is the set of primes lying over $2$. In other words, $U=D(2)$, and so is just $\mathrm{Spec}(\mathcal{O}_K[\frac{1}{2}])$. Note then that we have a surjection
$$\mathrm{Cl}(\mathcal{O}_K)\to \mathrm{Cl}(U)$$
and so $\mathrm{Cl}(U)$ is finite (this also just follows since it's a ring of $S$-integers, and so one can use the generalized version of the finiteness of the class number). In particular, we se that $\mathrm{Cl}(U)[2]$ is finite. 
So then, note that from Kummer theory we have a short exact sequence
$$1\to \mathbf{G}_m(U)/\mathbf{G}_m(U)^2\to \mathrm{Hom}_\mathrm{cont.}(\pi_1(U),\mathbb{Z}/2\mathbb{Z})\to \mathrm{Pic}(U)[2]\to 1$$
But, $\mathrm{Pic}(U)[2]=\mathrm{Cl}(U)[2]$ is finite, and $\mathbf{G}_m(U)/\mathbf{G}_m(U)^2$ is finite by the generalized version of Dirichlet's unit theorem (again, it's just a ring $S$-integers). 
Thus, we see that $\mathrm{Hom}_{\mathrm{cont.}}(\pi_1(U),\mathbb{Z}/2\mathbb{Z})$ is finite. But, this is precisely the set of quadratic extensions $L/K$ unramified outside of $S$ (i.e. unramified outside of primes dividing $2$). So, evidently the set of quadratic extensions unramified everywhere is finite. 
Of course, note that if $K\supseteq\mu_n(\overline{K})$ the above proof goes through to work for degree $n$ unramified abelian extensions.