Question

Let K be a number field and suppose K contains no p-power 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 p-power roots of unity?

Answer 1

Looks to me like this is false. Let K = 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 Q_p. There is already one p-th 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^p-1-p^2 other than z-a. (In fact, ( z^p-1-p^2)/(z-a) is irreducible over Q_p, but I don't need that.) So K_P contains a root b of z^p-1-p^2 other than a. But then b/a is in K_p and is a p-th root of 1.

This might be true if you ask K/Q to be Galois, but I would bet against it.

Answer 2

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/(p-1)); so K = Q((-p)^1/(p-1)) will do. As "unknown" has mentioned, Krasner's lemma explains why you would expect this to be false.

Answer 3

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 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^p-1 slightly to get a polynomial where the corresponding field still contained a root of (z^p-1)/(z-1).