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Let $p,q $ be odd integer primes, $p \equiv 1 \pmod 4$ and $q \equiv 3 \pmod 4$.

$K = \mathbb{Q }[\sqrt{pq }]$, $L = \mathbb{Q}[\sqrt{p }, \sqrt{q} ]$. Why a prime ideal in $O_{K}$ doesn't ramify in $O_{L} $ ?

My thoughts : I calculated the field discriminants $d_{K } = 4pq $, $d_{L } = 16p^{2 }q^{2 }$, so a prime ideal in $\mathbb{Z }$ which ramify in $O_{L} $ already ramifies in $O_{K}$.

Zev Chonoles
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WLOG
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  • $ { 1, \sqrt{ pq} } $ is an integral basis for $O_{K}$, so we can calculate $d_{K}$ using this integral basis. Is it correct ? – WLOG Jul 11 '13 at 11:18
  • Sorry, I forgot to say that $p \equiv 1 $ ( mod $4$ ) and $q \equiv 3 $ ( mod $ 4$ ) – WLOG Jul 11 '13 at 11:21
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    It is, indeed, relatively rare that no primes ramify in an extension of number fields, but it does happen. Class field theory studies questions such as these. Here (when reinterpreted suitably) the infinite primes (real embeddings of a field) do "ramify". The maximal extension $M$ of a given field $K$ such that no primes (finite nor infinite) ramify in $M/K$ is called the Hilbert Class field. It is known that $[M:K]=h_K$, the class number of $K$. I am too ignorant to say much more about it. I did enjoy studying the topic from Milne's notes. – Jyrki Lahtonen Jul 11 '13 at 11:33
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    @JyrkiLahtonen: One correction: The Hilbert class field is only the maximal abelian such extension. It's quite possible to have unramified extensions even when $h_K=1$. – Cam McLeman Jul 12 '13 at 19:30
  • Oops. Thanks, @Cam. On a good day I might have remembered to add that bit. – Jyrki Lahtonen Jul 12 '13 at 19:32

1 Answers1

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$\newcommand{\dQ}{\mathbb{Q}}\newcommand{\fp}{\mathfrak{p}}\newcommand{\fq}{\mathfrak{q}}\newcommand{\cO}{\mathcal{O}}$ Note that if $K=\dQ(\sqrt{pq})$, then $L=\dQ(\sqrt{p},\sqrt{q})$ is just $K(\sqrt{p})=K(\sqrt{q})$. To see that $L/K$ is unramified, we just have to show that for each prime $\fp$ of $K$ and each prime $\fq\mid \fp$ of $L$, the extension $L_\fq/K_\fp$ is unramified. Since $L/\dQ$ is only ramified above $p$ and $q$, we only need to check those $\fp$ which lie above $p$ or $q$. Suppose $\fp\mid p$. Then $\cO_{L_\fq}=\cO_{K_\fp}[\sqrt{q}]$, but since $\cO_{K_\fp}$ has residue characteristic $p$, the extension $\cO_{K_\fp}[\sqrt{q}]/\cO_{K_\fp}$ is unramified. Similarly, if $\fp\mid q$, one writes $\cO_{L_{\fq}}=\cO_{K_\fp}[\sqrt{p}]$ to see that $L_\fq/K_\fp$ is unramified.

So essentially, one uses $L=K[\sqrt p]$ to see that $L/K$ is unramified away from $p$, and $L=K[\sqrt q]$ to see that $L/K$ is unramified away from $q$. This forces $L/K$ to be unramified.

Daniel Miller
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