Is the relative discriminant a principal ideal?

Question

Let $F$ be a number field, $E$ a quadratic extension of $F$. we know $\mathfrak d_{E/F}$ is a integral ideal in $\mathfrak o_F$. Is it a principal ideal? How to prove it ?

Thanks?

Answer 1

The construction given in the comments works in principle, but not with fields with odd class number, as The Potter's Vessel pointed out.

Let $K$ be a quadratic field with cyclic class group of order $4$ and ${\mathfrak p}$ a prime ideal of odd norm $p$ in the class with order $2$. Find an element $\alpha \equiv 1 \bmod 4$ whose norm has the form $pa^2$; then the quadratic extension $K(\sqrt{\alpha}\,)$ will ramify exactly at ${\mathfrak p}$. Thus ${\mathfrak p}$ will be the relative discriminant, and it will not be principal.

As an example, take the prime $p = 89$ in $K = {\mathbb Q}(\sqrt{-17})$. Since $2p = N(5+3\sqrt{-17})$, genus theory tells us that the ideal class of the prime above $p$ is a square but not trivial. Now consider the element $\alpha = -23 + 4\sqrt{-17} \equiv 1 \bmod 4$. Its square root generates the extension $L/K$ defined by a root of $f(x) = x^4 + 46x^2 + 801$. The field discriminant is $\Delta = (4 \cdot 17)^2 \cdot 89$, hence exactly $89$ is ramified in the quadratic extension $L$, and thus the prime ideal above $89$ is the relatice discriminant.