# A question on totally splitting prime ideals

Let $$L/K$$ be a finite Galois extension of number fields, with $$\mathscr{O}_{L}$$ and $$\mathscr{O}_{K}$$ as the domains of algebraic integers respectively. Let $$\alpha \in \mathscr{O}_{L}$$ such that $$L=K(\alpha)$$, and $$f(X)\in \mathscr{O}_{K}[X]$$ to be the monic irreducible polynomial of $$\alpha$$ over $$K$$. Let $$\mathfrak{p}$$ be a nonzero prime ideal in $$\mathscr{O}_{K}$$. It is well known that if we do the assumption that $$f(X)$$ remains separable modulo $$\mathfrak{p}$$, which means $$f(X)$$ has no multiple roots in the residue field $$\mathscr{O}_{K}/\mathfrak{p}$$, then $$\mathfrak{p}$$ splits completely in $$\mathscr{O}_{L}$$ if and only if $$f(X)$$ has a solution modulo $$\mathfrak{p}$$. My question is that if we know $$\mathfrak{p}$$ splits completely in $$\mathscr{O}_{L}$$, can we have $$f(X)$$ remains separable modulo $$\mathfrak{p}$$ ? For example when $$L$$ to be the Hilbert class field of $$K$$, then $$\mathfrak{p}$$ splits completely is equivalent to $$\mathfrak{p}$$ is a principal ideal, and we can check some principal prime ideals by hand. So if the answer to my question is yes, then we can have $$h(\mathscr{O}_{K})=[L:K] < |\mathscr{O}_{K}/\mathfrak{p}|$$. Any help will be appreciated.

• I'm not sure what "remains separable" means here. Always $\mathcal{O}_K/\mathfrak{p}$ is a finite field, all of whose extensions are separable. Jun 15, 2020 at 10:17
• @Mindlack I think $\mathfrak{p}$ is ramified, but thanks anyway . Jun 15, 2020 at 11:20
• @OpenwinnerRay: Thank you for your correction. I thought of something else. Assume $[L:K]$ is larger than $|\mathcal{O}_K/\mathfrak{p}|$ and $\mathfrak{p}$ splits completely in $\mathcal{O}_L$. It is easy to check that there is inseparability mod $\mathfrak{p}$. So the question is: are there Galois field extensions of $K$ with arbitrarily large degree where $\mathfrak{p}$ splits completely? I can't think of anything right now, because my algebraic number theory is somewhat rusty... Jun 15, 2020 at 11:51
• @Mindlack Thank you for your idea, we may take $K$ to be $Q(\zeta_p)$ , where $p$ is a prime number and $\zeta_p$ is a $p$ -th primitive root of unity, and $L$ to be the Hilbert class field of $K$. Consider $\mathfrak{p}=(1-\zeta_p)\mathscr{O}_K$, then it is well known that $\mathfrak{p}$ is a prime, hence splits completely in $L$, and the residue field is just $\mathbb{F}_p$. So taking $p$ to be a irregular prime number will give us a counter example. Jun 15, 2020 at 12:09
• @Openwinner Ray: for my personal culture, why couldn’t we have, for some irregular prime, a smaller class number than $p$? OEIS that seems to confirm that these class numbers are large, but is there a reason? Jun 15, 2020 at 12:29

Let $$L=\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})$$, $$m_1=p^2+1$$, $$m_{i+1}=(pm_1\cdots m_i)^2+1$$, $$2^k \gt p \gt 2$$. Then $$p$$ splits completely in $$L$$, $$|L:\mathbb{Q}| = 2^k$$ and any $$f \in \mathbb{Z}[x]$$ generating $$L$$ has degree $$2^k \gt p$$, so $$f$$ must have multiple roots mod $$p$$. The ring of integers of $$L$$ is not monogenic over $$\mathbb{Z}$$.