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?



p-power roots of unity in local fields
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.


