# Given $h(\mathcal{O}_{\mathbb{Q}(\sqrt{d})}) = 1$, what is the longest possible run of inert primes in that ring?

Say $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is a unique factorization domain. A couple of primes will ramify, the rest will either split or be inert.

For example, in $\mathcal{O}_{\mathbb{Q}(\sqrt{-163})}$, the real primes from $2$ to $157$ are inert. $163$ ramifies, of course, and after that a prime can split or stay put. So there's at least one run of thirty-seven consecutive inert primes in this domain. Can there be a longer run? But there can't be runs of arbitrary length, right?

I might figure out that numbers congruent to such and such modulo this or that must split. But of course there can be consecutive primes such that none falls in that congruence class.

I'm also interested in this problem with $d$ positive, though I suspect that might be much more difficult.

• Undoubtedly you know that by Chebotaryev's density theorem the inert primes have asymptotic density $1/2$. That somehow suggests that in theory there should be arbitrarily long runs. OTOH we do get congruence conditions determining the set of split primes (this is an abelian extension). A good question! – Jyrki Lahtonen Jul 14 '15 at 21:19
• I'm going to have to read up on Chebotaryev's density theorem, as I currently know nothing about it. Thank you very much for telling me about it. – John-Luke Aug 19 '15 at 21:42

## 1 Answer

For positive $d$ you can construct arbitrarily long runs of nonresidues using the Chinese remainder theorem; just look for $d \equiv 5 \bmod 8$, $d \equiv 2 \bmod 3$, $d \equiv 2 \bmod 5$, $d \equiv 3 \bmod 7$ etc. Since the small primes are inert there is some hope that the corresponding field has class number $1$, but there is no guarantee.

The Lehmers actually studied similar questions here.