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 $O_{K}$.
Best Answer
$\newcommand{\dQ}{\mathbb{Q}}\newcommand{\fp}{\mathfrak{p}}\newcommand{\fq}{\mathfrak{q}}\newcommand{\cO}{\mathcal{O}}$ 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.