A lemma about standard open sets of a scheme.

Question

I came across the following lemma in the Stacks project: Link.

I have two questions:

1. How do we obtain $D(g)=D(g_A)$?
2. How can we assume that $U\subset V$?

Answer 1

I start from question $2$!

$U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ are affine schemes, $\{D(s)\subseteq U\}_{s\in A}$ is a topological base for $U$, so there exists $f\in A$ such that $D(f)\subseteq U\cap V$ and $f\notin\mathfrak{p}_x$, that is $x\in D(f)$.

Without loss of generality, one can assume that $U=D(f)$; becasue they are both affine schemes!

By hypothesis, the previous examplification is equivalent to affirm $x\in U\subseteq V$; analogously to previous case there exists $g\in B$ such that $D(g)\subseteq U$ and $g\notin\mathfrak{q}_x$.

Passing to question $1$, by tag01HE, $i^{-1}\mathcal{O}_V\cong\mathcal{O}_U$ where $i:U\hookrightarrow V$ is the inclusion, in other words: $\forall u\in U,\,\mathcal{O}_{U,u}\cong\mathcal{O}_{V,i(u)}$; by universal property of stalks, one has a morphism $i^{\sharp}:B=\mathcal{O}_V(V)\to\mathcal{O}_U(U)=A$.

Because the stalks of $g$, viewed as global section of $\mathcal{O}_V$, coincide with the stalks of $i^{\sharp}(g)=g_A$ on $U$; that is for any $u\in U,\,g_u=g_{A,u}$ and then $D(g)=D(g_A)$.