p-power roots of unity in local fields 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?
 A: Looks to me like this is false. Let $K = \mathbb{Q}(z)/(z^p-1-p^2)$. This is an extension of degree $p$, so it is disjoint from the p-th 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 Qp. There is already one p-th root of $1+p^2$ in Qp; 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 $z^p-1-p^2$ other than z-a. (In fact, $( z^p-1-p^2)/(z-a)$ is irreducible over $\mathbb{Q}_p$, but I don't need that.) So $K_\mathcal{P}$ contains a root b of $z^p-1-p^2$ 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.
A: Another counterexample, along the same lines as the one given by the other David S. but perhaps more "standard", is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((-p)^{1/(p-1)})$; so $K = \mathbb{Q}((-p)^{1/(p-1)})$ will do.  As "unknown" has mentioned, Krasner's lemma explains why you would expect this to be false.
A: 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 $z^p-1$ slightly to get a polynomial where the corresponding field still contained a root of $(z^p-1)/(z-1)$. 
