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.
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$\begingroup$ 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. $\endgroup$– Angina SengJun 15, 2020 at 10:17
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$\begingroup$ @Mindlack I think $\mathfrak{p}$ is ramified, but thanks anyway . $\endgroup$– Openwinner RayJun 15, 2020 at 11:20
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$\begingroup$ @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... $\endgroup$– AphelliJun 15, 2020 at 11:51
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$\begingroup$ @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. $\endgroup$– Openwinner RayJun 15, 2020 at 12:09
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$\begingroup$ @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? $\endgroup$– AphelliJun 15, 2020 at 12:29
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1 Answer
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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 $\mathbb{Z}$.
