Given $X=Spec(\mathcal O(X))$ is an affine scheme, the hyperplane-complement $D_f=Spec(\mathcal O(X)[\frac{1}{f}])$ is an open subscheme of $X$($D_f$ denotes the complement of $Z_f$). If I have an open subscheme of $D_f$, is it an open subscheme of $X$?
More generally, for any $D_I$ which is the complement of $Z_I\subseteq X$, for an open subschem of $D_I$, is it an open subscheme of $X$?
I am not sure for this. A proof or some explaination would be very appreciated.
To prove an open subscheme $D_I$ for $I\subseteq \mathcal O(X)[1/f]$ is an open subscheme of $X$, I think what I need to do is to conclude $D_I=D_{I'}$ for some $I'\subseteq \mathcal O(X)$. So may I please for an explaination which is elementary like this?
Your open subsubscheme $U \subset D_f$ is covered by principal open subschemes of $D_f$ because their underlying topological supports are a basis for the topology of $D_f$.
Moreover a principal open subscheme $(D_f)_g$ of a principal open subscheme $D_f$ of an affine scheme $S$ is also a principal open subscheme of $S$, you have in fact that $(D_f)_g=D_{fg}=Spec (O(X)[\frac{1}{fg}])$.
So $U$ is covered by open subschemes and so it is an open subscheme.
In the general case of $D_I$ you can reduce to the open principal subsets once again using the fact that principal open subsets are a basis for the topology.