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If $K:=\mathbb{Q}(\sqrt{-5})$, find a non-trivial extension $L$ such that no prime $\mathfrak{p}\subset\mathcal{O}_K$ ramifies in $L$.

I thought about using fact that every quadratic number field is contained in a cyclotomic field, in this case $K\subset \mathbb{Q}(\zeta_{20})=\mathbb{Q}(\zeta_5)\mathbb{Q}(i)$. Since $\mathbb{Q}(\zeta_{20})|\mathbb{Q}$ has nice propreties, I suppose $L=\mathbb{Q}(\zeta_{20})$ will do the job.

The thing is: I only know how to deal with the primes $\mathfrak{p}\subset \mathcal{O}_K$ individually (finding decomposition/inertia fields etc), but I don't know how to check whether or not every $\mathfrak{p}$ is unramified.

How should I approach this?

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    $\begingroup$ You can find the details of the computations around $L=Q(\sqrt 5,\sqrt {-1})$ in ex. 1 of chapter 6 of Samuel's ANT. In fact, $[L : Q(\sqrt {-5})]=2$, and you can check in any table that $Q(\sqrt {-5})$ has class number 2. This is a particular case of the "Hilbert class field" of a number field, which is its maximal abelian unramified extension. Here L is the Hilbert class field of $Q(\sqrt {-5})$ . $\endgroup$ – nguyen quang do May 21 at 8:03
  • $\begingroup$ @nguyenquangdo, in Samuel's exercise, he argues that $\mathbb{Z}\left[\sqrt{-1},\frac{1+\sqrt{5}}{2}\right]$ and that the absolute discriminant of $L=\mathbb{Q}(\sqrt{5}, \sqrt{-1})$ is only divisible by $2$ and $5$. I'm ok with that. But how does he know that the ramification indexes of the primes above $2$ and $5$ both equal $2$? $\endgroup$ – rmdmc89 May 22 at 21:45
  • $\begingroup$ My actual question is: how do I factor $2,5$ in $\mathbb{Q}(\sqrt{5},\sqrt{-1})$? $\endgroup$ – rmdmc89 May 22 at 21:46

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