TITLE: If the n-th root of unity exists locally, does it exist globally?
QUESTION [8 upvotes]: My question is: is it true that $\zeta_n$ is in $K$ (a number field) iff $\zeta_n$ is in all but finitely many of the $K_{\mathfrak{p}}$?

REPLY [6 votes]: The answer is yes, and this follows (as BCnrd points out in the comments above) immediately from the Chebotarev Density Theorem.  See e.g. Corollary 9 here.
More generally, let $K$ be any global field and $f \in K[t]$ be any irreducible, separable polynomial of degree $d > 1$.  Then there are infinitely many places $v$ of $K$ such that $f$ does not split completely over $K_v$.  Indeed, let $\alpha$ be a root of $f$ in a separable closure of $K$, let $L = K[\alpha]$ and let $M$ be the Galois closure of $L/K$.  A finite place $v$ of $K$ splits completely in $L$ iff it splits completely in $M$ (Exercise 5.1.3), so by Chebotarev's Theorem, the set of primes which do not split completely in $L$ has density 
$1 - \frac{1}{[M:K]} > 0$.  
If $L/K$ is already Galois, the argument is simpler and the conclusion is stronger: there are infinitely many places $v$ of $K$ such that $f$ does not have any $K_v$-rational roots.  This is the case in the OP's question, which we recover by taking $f$ to be the minimal polynomial over $K$ of any one primitive $n$th root of unity.
On the other hand for every composite number $d$, there is a degree $d$ irreducible polynomial $f \in K[t]$ which is reducible in every completion $K_v$: this is a 2005 theorem of Guralnick, Schacher and Sonn.