Explicit description of the extension generated by the square root of a fundamental unit of a real quadratic field

Question

Suppose $K={\mathbb{Q}}(\sqrt{m})$ is a real quadratic field which contains a totally positive fundamental unit $\epsilon$ (hence its norm is $1$). Suppose that $\epsilon \equiv 1 \pmod 8$, so that the quadratic extension $K(\sqrt{\epsilon})$ is unramified over $K$. In particular, it is contained in the genus field; i.e., there exists $d \mid m$ such that $K(\sqrt{\epsilon}) = K(\sqrt{d})$. Is there a recipe for $d$ in terms of $\epsilon$?

A guess: suppose $m$ is odd; then for each odd prime $p \mid m$ let $\tilde{p}$ denote the unique ideal in $K$ dividing $p$, so in particular $\epsilon \equiv \pm 1 \pmod{\tilde{p}}$. Is it true that $d = \prod p$ where the product is over the primes dividing $m$ where $\epsilon \equiv 1$ (or $-1$ as $d$ and $m/d$ give the same extension)?

Answer 1

This has nothing to do with ramification. Write your unit in the form $\varepsilon = t + u \sqrt{m}$; then $t^2 - mu^{2} = 1$, hence $t^2 - 1 = (t-1)(t+1) = mu^2$. If $t$ is odd, then unique factorization implies $t-1 = 2ar^2$ and $t+1 = 2bs^2$, where $ab = m$, hence $t = ar^2 + bs^2$. It is then easily verified that $(r\sqrt{a} + s \sqrt{b})^2 = t + u \sqrt{m}$, and $k (\sqrt{\varepsilon}) = k(\sqrt{a})$. The case where $t$ is even is taken care of similarly.