Simultaneously principal open set.

For some reason, I have a hard time understanding the proof that in an intersection of two affine open subschemes, there exists an open set that is principal in both affine opens. Here is my attempt to make it slightly more clear.

The precise statement that I want to prove is the following. Let $X$ be a scheme and $U=\mathrm{Spec}~A$ and $V=\mathrm{Spec}~B$ be two affine open subschemes of $X$. For all $x \in U \cap V$, there exists $W = D_U(f) = D_V(g)$, a neighborhood of $x$, such that $f \in A$, $g \in B$, and $W \subset U \cap V$.

1) We would like to reduce to a simpler case. $U \cap V$ is an open subset of $V$. Hence there exists a principal open set $D(g) \subset U \cap V$ with $g \in B$ containing $x$. Then we may identify $D(g) = \mathrm{Spec}~B_g$. Therefore, we have $x \in \mathrm{Spec}~B_g \subset \mathrm{Spec}~A$. Suppose we found an open set $W$ containing $x$ that is principal inside both $\mathrm{Spec}~B_g$ and $\mathrm{Spec}~A$, then there exists some $h/g^n \in B_g$ and $f \in A$ such that $W = D_{\mathrm{Spec}~B_g}(h/g^n) = D_{\mathrm{Spec}~A}(f)$. As
$$(B_g)_{h/g^n} = B_{gh}$$
we see that $W = D_V(gh) = D_U(f)$, i.e. principal in both.

2) We may assume $x \in \mathrm{Spec}~B \subset \mathrm{Spec}~A$. Because $\mathrm{Spec}~B$ is an open subscheme of $\mathrm{Spec}~A$, we have an open immersion $i: \mathrm{Spec}~B \to \mathrm{Spec}~A$. This comes with a morphism of sheaves $\mathcal{O}_{\mathrm{Spec}~A} \to i_*\mathcal{O}_{\mathrm{Spec}~B}$ which is by definition the restriction map. Let $\rho$ denote the restrction map from $A$ to $B$. Then $i:\mathrm{Spec}~A$ to $\mathrm{Spec}~B$ is simply $\mathfrak{q} \mapsto \rho^{-1}(\mathfrak{q})$.

Pick $f \in A$ such that $D(f) \subset \mathrm{Spec}~B \subset \mathrm{Spec}~A$. Observe that $\rho(f) \in \mathfrak{q} \Rightarrow f \in \rho^{-1}(\mathfrak{q})$ by definition. Conversely, if $f \in \rho^{-1}\mathfrak{q}$, then $\rho(f) \in \rho(\rho^{-1}\mathfrak{q}) \subset \mathfrak{q}$. Therefore, we have $\rho(f) \not\in \mathfrak{q} \Leftrightarrow f \not\in \rho^{-1}\mathfrak{q}$.

Until here, we only know that $D_{\mathrm{Spec}~B}(\rho(f)) \subset D_{\mathrm{Spec}~A}(f)$ because $D(f)$ might not all be of the form $\rho^{-1} \mathfrak{q}$. But because $D(f) \subset \mathrm{Spec}~B$, we may conclude that $D_{\mathrm{Spec}~A}(f) \subset D_{\mathrm{Spec}~B}(\rho(f))$.


Warning (To my self): As it is written, $D_{\mathrm{Spec}~B}(\rho(f)) \subset D_{\mathrm{Spec}~A}(f)$ does not make sense. To be very precise, we have $U \subset \mathrm{Spec}~A$ with $U$ an open affine subscheme with an isomorphism $j: U \to \mathrm{Spec}~B$. Then $\rho$ is restriction to $U$ composed with the isomorphism $A \to \Gamma(U, \mathcal{O}_{\mathrm{Spec}~A}|_U)\to B$. And view $D_{\mathrm{Spec}~B}(\rho(f))$ as $D_U(j^b(\rho(f))$ where $j^b$ is the induced map of $B \to \Gamma(U, \mathcal{O}_{\mathrm{Spec}~A}|_U)$.