In a previous post, I gave a short argument showing that, for odd primes p and N such that the p-class group of is non-trivial. This post is just to remark that the same argument works under weaker hypotheses, namely:

**Proposition:** Assume that N is p-power free and contains a prime factor of the form and that p is at least 5. Then the p-class group of is non-trivial.

The proof is pretty much the same. If N has a prime factor of the form then the genus field is non-trivial. Hence we may assume there are no such primes, from which it follows that has dimension one and is trivial, where S denotes the set of primes dividing Np. The prime q gives rise to a non-trivial class which is totally split at p (this requires that p be at least 5), and the field K itself gives rise to a class But now the vanishing of H^2 implies that and hence there exists a representation of G_S of the form:

where is the mod-p cyclotomic character. The class c gives the requisite extension (after possibly adjusting by a class in the one-dimensional space The main point is that the image of inertia at primes away from p is tame and so cyclic, but any unipotent element of has order p if p is at least three. This ensures c is unramified over K away from the primes above p. On the other hand, the class is totally split at p. This implies that the class c is locally a homomorphism of the Galois group of and so after modification by a multiple of the cyclotomic class in may also be assumed to be unmarried at p. The fact that ensures that and moreover the fact that p is at least 5 implies that the kernel of c is distinct from that of a, completing the proof. (This result was conjectured in the paper *Class numbers of pure quintic fields* by Hirotomo Kobayashi, which proves the claim for

Nice typo: “unmarried at p”…

WordPress continually auto-corrects “unramified” in this manner… this one must have slipped through!