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  1. The proposition seems to use the following fact: $\operatorname{Spec}(A)=\operatorname{Spec}(A_{f_1})\cup\ldots\cup\operatorname{Spec}(A_{f_r}) \iff \langle f_1,\ldots, f_r\rangle=A$, where $\langle f_1, \ldots, f_r\rangle$ is the ideal generated by the elements $f_1, \ldots, f_r\in A$. Why is this true?

  2. If $U=\operatorname{Spec} B\subset\operatorname{Spec} A, f\in A$, and $D(f)\subset \operatorname{Spec} B$, then the proposition refers to the image $\bar{f}$ of $f$. Is the following a valid way to think of the map $A\to B$ implied here? Since $\operatorname{Spec} B$ is a subscheme, we must have $\mathcal{O}_{A}(\operatorname{Spec} B)\cong\mathcal{O}_{B}(\operatorname{Spec} B)\cong B$; at the same time there is a restriction map $A\cong\mathcal{O}_{A}(\operatorname{Spec} A)\to \mathcal{O}_{A}(\operatorname{Spec} B)$. The composition of these is the implied map $A\to B$.

P. S.

Part 2 of my question had already been asked here: Proposition II.$3.2$ in Hartshorne

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    $\begingroup$ Please use $\operatorname{Spec}$ to format $\operatorname{Spec}$: this produces better spacing. I've made the upgrade for you this time. $\endgroup$
    – KReiser
    Jan 9, 2023 at 16:47

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For the first question : note that ${\rm Spec}(A_f)$ is canonically isomorphic to the principal open $D(f)$ and $D(f_1)\cup\ldots\cup D(f_r)={\rm Spec}(A)\setminus V(\langle f_1,\ldots,f_r\rangle)$ so this union is ${\rm Spec}(A)$ in its entirety if and only if $V(\langle f_1,\ldots, f_r\rangle)=\emptyset$ if and only if $\langle f_1,\ldots,f_r\rangle=A$.

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