In a previous post, I gave a short argument showing that, for odd primes p and N such that \(N \equiv -1 \mod p,\) the p-class group of \(\mathbf{Q}(N^{1/p})\) 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 \(q \equiv -1 \mod p,\) and that p is at least 5. Then the p-class group of \(K = \mathbf{Q}(N^{1/p})\) is non-trivial.
The proof is pretty much the same. If N has a prime factor of the form \(1 \mod p,\) then the genus field is non-trivial. Hence we may assume there are no such primes, from which it follows that \(H^1_S(\mathbf{F}_p)\) has dimension one and \(H^2_S(\mathbf{F}_p)\) is trivial, where S denotes the set of primes dividing Np. The prime q gives rise to a non-trivial class \(b \in H^1_S(\mathbf{F}_p(-1))\) which is totally split at p (this requires that p be at least 5), and the field K itself gives rise to a class \(a \in H^1_S(\mathbf{F}_p(1)).\) But now the vanishing of H^2 implies that \(a \cup b = 0\) and hence there exists a representation of G_S of the form:
\(\rho: G_S \rightarrow \left( \begin{matrix} 1 & a & c \\ 0 & \chi^{-1} & b \\0 & 0 & 1 \end{matrix} \right),\)
where \(\chi\) is the mod-p cyclotomic character. The class c gives the requisite extension (after possibly adjusting by a class in the one-dimensional space \(H^1_S(\mathbf{F}_p)).\) The main point is that the image of inertia at primes away from p is tame and so cyclic, but any unipotent element of \(\mathrm{GL}_3(\mathbf{F}_p)\) 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 \(b\) is totally split at p. This implies that the class c is locally a homomorphism of the Galois group of \(\mathbf{Q}_p,\) and so after modification by a multiple of the cyclotomic class in \(H^1_S(\mathbf{F}_p)\) may also be assumed to be unmarried at p. The fact that \(b \ne 0\) ensures that \(c \ne 0,\) 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 \(p = 5.)\)
Nice typo: “unmarried at p”…
WordPress continually auto-corrects “unramified” in this manner… this one must have slipped through!
The proposition follows, as Iimura remarked in Acta Arith. 47 (1986), 153-166, from results due to Jaulent (Ann. Inst. Fourier 1981). I guess that it can be constructed classically by taking a subfield of the ray class group modulo q in the field of p-th roots of unity.
Thanks! It’s not surprising to find out this had been noted before (and even less surprising that my google search skills at the time didn’t find it).
I don’t quite get your second comment though; starting with ray class extensions over other fields will not (directly) give abelian extensions of K, so there has to be some back and forth between the class group of K and K(zeta_p) (say), which is certainly the type of game one can play. (And is more or less what is happening above, though certainly it could be described in more classical language since the cohomological formalism is just the classical one by another name…)