Why do no prime ideals ramify in the extension $\mathbb{Q}(\sqrt{p }, \sqrt{q})/\mathbb{Q}(\sqrt{pq })$?

Solution 1:

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