Consider a proper and integral scheme $X\rightarrow\textrm{Spec}(A)$$X\rightarrow\operatorname{Spec}(A)$ over a Noetherian ring $A$ and $D\in\textrm{Div}(X)$$D\in\operatorname{Div}(X)$ an effective ample Cartier divisor on $X$. Is it true that its complement $X\setminus\textrm{Supp}(D)$$X\setminus\operatorname{Supp}(D)$ is affine? If not, then what would be the necessary conditions on $A$ that would make it true?
If $A$ is an algebraically closed field, then this is a classical result, whose proof can be found (for example) here:
https://math.stackexchange.com/questions/264131/is-the-complement-of-an-ample-divisor-always-affine
and some relevant stuff is here: https://math.stackexchange.com/questions/1197197/relationship-between-very-ample-divisors-and-hyperplane-sections and here: https://math.stackexchange.com/questions/2450746/a-definition-of-a-very-ample-divisor.
Even with the ideas from these links, I can't really solve the problem. I mostly tried using the property that a very ample divisor induces a closed embedding, but I suspect my arguments might be too elementary. I would be interested, in particular, in the case where $A$ is a Dedekind ring (or, if not, a DVR).
