Let $K$ be a number field and suppose $K$ contains no $p$power roots of unity. Let $\mathcal{P}$ be a prime of $K$ above the rational prime $p$. Can someone prove or disprove the assertion that the local field $K_{\mathcal{P}}$ will contain no $p$power roots of unity?
Looks to me like this is false. Let $K = \mathbb{Q}(z)/(z^p1p^2)$. This is an extension of degree $p$, so it is disjoint from the pth cycloctomic field, and hence does not contain a $p$th root of $1$. Thus, it also can not contain a $p^k$th 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 $(1+x)^{1/p} = 1+(1/p) x + \binom{1/p}{2} x^2 + ...$ converges for $x=p^2$.)
Let $\mathcal{P}$ 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 $\mathbb{Q}_p$, but I don't need that.) So $K_\mathcal{P}$ contains a root b of $z^p1p^2$ other than $a$. But then $b/a$ is in $K_\mathcal{P}$ and is a $p$th root of 1.
This might be true if you ask $K/\mathbb{Q}$ to be Galois, but I would bet against it.

2$\begingroup$ It's not even true if K/Q is Galois. Consider a quadratic extension K=Q(sqrt(n)) of Q and let p be 3 (this phenomenon has nothing to do with 3, it's just the easiest way to give a counterexample). If n is 3adically close to 3, but not 3, then K will be ramified at 3 and the completion of K at the prime above 3 will be Q_3(sqrt(3))=Q_3(zeta_3). $\endgroup$ – Kevin Buzzard Nov 3 '09 at 11:31
Another counterexample, along the same lines as the one given by the other David S. but perhaps more "standard", is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((p)^{1/(p1)})$; so $K = \mathbb{Q}((p)^{1/(p1)})$ will do. As "unknown" has mentioned, Krasner's lemma explains why you would expect this to be false.

3$\begingroup$ There is a criterion for a finite extension K of Q_p to contain a primitive pth root of 1. Let e be the ramification index of KQ_p, and k the residue field of K. The necessary and sufficient condition is that p1 should divide e and, upon writing p=u\pi^e (u unit in K and \pi a uniformiser of K), the reduction of u should be a (p1)th power in k. Cf. Chapter XV of Hasse's Number Theory, or prop. 25 of arXiv:0711.3878v1 [math.NT]. $\endgroup$ – Chandan Singh Dalawat Dec 31 '09 at 8:31
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 $p$adic setting. For example, consider the set of monic polynomials of degree $d$ with coefficients in $\mathbb{Q}_p$, topologized as $\mathbb{Q}_p^d$. Then I believe that properties such as "has a root in $\mathbb{Q}_p$", "splits completely in $\mathbb{Q}_p$", "has abelian Galois group over $\mathbb{Q}_p$" 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 $(z^p1)/(z1)$.

1$\begingroup$ Actually, if you perturb the coefficients of a polynomial over a local field slightly then the roots of the new polynomial generate the same field (this follows from "Krasner's lemma"; in particular, it follows that there are only finitely many extensions of a local field of a fixed degree. $\endgroup$ – naf Oct 23 '09 at 8:29