# 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 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?

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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 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 + (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 Qp, but I don't need that.) So KP contains a root b of z^p-1-p^2 other than a. But then b/a is in Kp 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.

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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 3-adically 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). –  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 Qpp) = Qp((-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.

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There is a criterion for a finite extension K of Q_p to contain a primitive p-th root of 1. Let e be the ramification index of K|Q_p, and k the residue field of K. The necessary and sufficient condition is that p-1 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 (p-1)-th power in k. Cf. Chapter XV of Hasse's Number Theory, or prop. 25 of arXiv:0711.3878v1 [math.NT]. –  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 Qp, topologized as Qpd. Then I believe that properties such as "has a root in Qp", "splits completely in Qp", "has abelian Galois group over Qp" 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).

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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. –  ulrich Oct 23 '09 at 8:29