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 $K\_{\mathfrak{p}}$?



If the n-th root of unity exists locally, does it exist globally?
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](http://alpha.math.uga.edu/%7Epete/8430notes5.pdf).


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](http://alpha.math.uga.edu/%7Epete/8430notes5.pdf)), 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.


