In the ring of integers of $\mathbb Q[\sqrt d]$, if every non-zero ideal $A$ is a lattice, then is every ideal generated by at most two elements?

Question

From Artin's Algebra:

He says every element of $A$ is an integer combination of the lattice basis. So, does this mean every ideal generated by at most two elements?

If we have an ideal $A=(a_1,a_2, \dots, a_n)$, then we can just write this as $(\beta_1, \beta_2)$ for $\beta_1, \beta_2$ are the lattice basis?

Answer 1

One property of Dedekind domains is that every nonzero ideal is generated by two elements.

Every ring of integers of a number field is a Dedekind domain.

Answer 2

Other people have told you that all ideals in Dedekind domains are generated by two elements, but I think there is a still a point to be made here.

For any number field $K$, the lattice basis of an ideal is a basis as a $\mathbf{Z}$-module. Since generators of an ideal $A \subset \mathcal{O}_K$ are generators of $A$ as an $\mathcal{O}_K$-module, the lattice basis is a fortiori a generating set of $A$. The problem is that the lattice basis consists of $[K:\mathbf{Q}]$ elements, while a minimal generating set for $A$ is typically much smaller (at most 2 by the result stated above). In your case, $K$ is a quadratic extension and so you get the result that any ideal is generated by two elements "for free", while there is some additional content to this result for higher degree number fields.