Let K be a number field and suppose K contains no ppower roots of unity. Let P be a prime of K above the rational prime p. Can someone prove or disprove the assertion that the local field K_P will contain no ppower roots of unity?
Looks to me like this is false. Let K = Q(z)/(z^p1p^2). This is an extension of degree p, so it is disjoint from the pth cycloctomic field, and hence does not contain a pth root of 1. Thus, it also can not contain a p^kth root of 1. Now, let's see how z^p  1  p^2 factors in Q_{p}. There is already one pth root of 1+p^2 in Q_{p}; call this root a. (To see this, note that the power series (1+x)^{1/p} = 1+(1/p)x + (1/p choose 2)x^2 + ... converges for x=p^2.) Let P be a prime of K corresponding to a factor of z^p1p^2 other than za. (In fact, ( z^p1p^2)/(za) is irreducible over Q_{p}, but I don't need that.) So K_{P} contains a root b of z^p1p^2 other than a. But then b/a is in K_{p} and is a pth root of 1. This might be true if you ask K/Q to be Galois, but I would bet against it. 


Another counterexample, along the same lines as the one given by the other David S. but perhaps more "standard", is that Q_{p}(ζ_{p}) = Q_{p}((p)^{1/(p1)}); so K = Q((p)^{1/(p1)}) will do. As "unknown" has mentioned, Krasner's lemma explains why you would expect this to be false. 


There is a more general heuristic here, which I bet the number theorists can state more precisely. Questions about Galois groups tend to be locally constant in the padic setting. For example, consider the set of monic polynomials of degree d with coefficients in Q_{p}, topologized as Q_{p}^{d}. Then I believe that properties such as "has a root in Q_{p}", "splits completely in Q_{p}", "has abelian Galois group over Q_{p}" should be locally constant. This lead me to believe that I should be able to perturb z^p1 slightly to get a polynomial where the corresponding field still contained a root of (z^p1)/(z1). 

