Is every distinguished open of tensor product of rings covered by tensor product of distinguished open?

Question

Let $A,B$ be $R$-algebras and $f\in A\otimes_RB$. Denote $D(f)=\mathop{\mathrm{Spec}}(A\otimes_R B)_f$ the disguished open subset of $\mathop{\mathrm{Spec}}A\otimes_R B$. Can we cover $D(f)$ by open subsets of $D(f)$ of the form $D(a_i)\times_{\mathop{\mathrm{Spec}}R} D(b_i)$?

Answer 1

No. Let $k$ be a field, then $k(x)\otimes_k k(x)$ is not zero-dimensional, see this question. So its spectrum contains a proper closed subset of smaller dimension, thus contains a proper open subset and hence a proper distinguished open subset $D(f)$, which cannot be covered by opens $D(a_i)\times_{\mathop{\mathrm{Spec}}k}D(b_i)$ because each of them is just $\mathop{\mathrm{Spec}}k(x)\times_{\mathop{\mathrm{Spec}}k}\mathop{\mathrm{Spec}}k(x)$.