Quadratic field ideal find $\mathbb{Z}$-basis given a $\mathcal{O}_K$-basis

Question

Suppose we are working in an imaginary quadratic number field $\mathbb{Q}(\sqrt{d})$ (so $d$ is a fundamental discriminant with $d < 0$).

Now in the ring of integers $\mathcal{O}_K$ suppose we have an ideal $I$ where $(\alpha, \beta)$ represents an $\mathcal{O}_K$-basis for $I$: so we are saying $I = \{x\alpha + y\beta \, | \, x,y \in \mathcal{O}_K\}$. Now I want to do arithmetic with this ideal, but most computer programs (such as sage) assume you have a $\mathbb{Z}$-basis for an ideal.

So my question is either

1. Is it possible (in a program like sage) to work with an $\mathcal{O}_K$-basis?
2. If that doesnt work, is it possible to convert from an $\mathcal{O}_K$-basis to a $\mathbb{Z}$-basis?

Answer 1

1. In programs like Sage, you can certainly generate ideals with how many elements you want. After you do that, you can ask the program for a Z-basis of the ideal.

2. Suppose you have $\mathcal O_K$-generators $\alpha=u+v\sqrt{-d}$ and $\beta=w+t\sqrt{-d}$. Then it is easy to see that $\alpha,\beta,-dv+u\sqrt{-d},-td+w\sqrt{-d}$ all belong to the same ideal (just multiply $\alpha,\beta$ by $1$ and $\sqrt{-d}$). But for the very same reason, these are also $\mathbb Z$-generators for the ideal. Now you can use you favourite algorithm to compute a basis of that $\mathbb Z$-module.