# Hartshorne Proposition II.3.2

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

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

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 $$\langle f_1,\ldots,f_r\rangle=A$$.