Affine Communication Lemma

Definition. A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$. 
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Affine Communication Lemma

Lemma. Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Proof. Let $U = \text{Spec} A$ be an affine open of $X$. Cover $U_i = \text{Spec} A_i \cap U$ by simultaneous distinguished open sets of both $U$ and $\text{Spec} A_i$ (possible by Lemma below). By the first property of affine-local $P$, these distinguished open sets all have property $P$. But together they cover $U$, so by the second property of affine-local $P$, $U$ also has $P$. QED.
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Examples of Affine-Local Properties:

• reduced-ness
• Noetherian
• finite type $B$-scheme

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Covering by Simultaneous Distinguished Open Sets

Lemma. Let $\text{Spec} A$ and $\text{Spec} B$ be affine open subschemes of a scheme $X$. Then $\text{Spec} A \cap \text{Spec} B$  can be covered by open sets that are simultaneously distinguished open subschemes of both $\text{Spec} A$ and $\text{Spec} B$.
Proof.
In the proof below we need to use the following observation: Suppose $g \in O_X(U)$ and $V \subset U$. Let $U_g$ denote the non-zeros of $g$, i.e. $x \in U$ such that $g$ is not zero in $\kappa(x) = O_x/\mathfrak{m}_x.$. Then $$U_g \cap V = V_{g|_V}. $$

Let $p \in \text{Spec} A \cap \text{Spec} B$. We claim that there is a simultaneous distinguished open of both $\text{Spec} A$ and $\text{Spec} B$ containing $x$.

Let $\text{Spec} A \supset \text{Spec} A_f \ni p$ and let $p \in \text{Spec} B_g \subset \text{Spec} A_f$. We claim that $\text{Spec} B_g$ is also a distinguished open of $\text{Spec} A$.

Thus the idea is to get rethink of $\text{Spec}B_g$, the non-zeros of a section $g$ of $B$, first as non-zeros of a section of $\text{Spec} A_f$ (restriction of $g$), then as the non-zeros of a section of $A$.  

Indeed $ \text{Spec} B_g$ are just the non-zero locus of $g \in B= O_X(V)$.  The problem is that $g$ is not a section on $\text{Spec} A$. However, as $\text{Spec} A_f \subset \text{Spec} B$, we can restrict $g$ to  $\text{Spec} A_f $. Let $g'$ denote this restriction. Then $ \text{Spec} B_g$ are exactly the nonzeros of $g'$. Thus it is $\text{Spec} (A_f)_g' = \text{Spec} A_{fh}$ if $g' = h/f^n$ for some $h \in A$.