A parametrization of quadratic fields is $\mathbb{Q}(\sqrt{m})$, where $m\ne1$ is a squarefree integer. That is, $\mathbb{Q}(\sqrt{m})$ is a quadratic field as $m$ varies, and all quadratic fields are included. There is another parametrization of quadratic fields, given by the 1-1 correspondence between $m$ and fundamental discriminants $d$. In particular, for each fundamental discriminant $d$ there is exactly one quadratic field of that discriminant.
Is there a similar parametrization for cubic fields?