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I am reading Hartshorne algebraic geometry. On page 70, the paragraph before Definition, it is said that the element 1 gives 1 in each $A_{\mathfrak{p}}$. What is the first "1"? Thank you very much.

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  • $\begingroup$ Maybe you could separate each question and contemplate more upon them, so that any of them might form an independent answer: this way you might get more attention, I suppose. $\endgroup$ – awllower Aug 25 '13 at 7:45
  • $\begingroup$ @awllower, thank you very much for your suggestion. $\endgroup$ – LJR Aug 25 '13 at 7:47
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The first 1 is the identity element of $\mathcal{O}(U)$. It is saying the ring identity in $\mathcal{O}(U)$, in other words the element 1, is the constant function $s: U \rightarrow \coprod_{\frak{p} \in \mathrm{U}} A_p$ given by $s(\frak{p})=1$ in each $A_\frak{p}$.

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  • $\begingroup$ thank you very much. Should $s(u)=1$ be $s(\mathfrak{p})=1$? $\endgroup$ – LJR Aug 25 '13 at 8:18
  • $\begingroup$ Yes, that's what I meant. I'll correct it. $\endgroup$ – Prometheus Aug 25 '13 at 8:21
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On page 70 Hartshorne constructs the structure sheaf on the spectrum of a commutative ring. The sections on an open subset are functions valued in the localizations which are given locally by fractions. Now one has to find a ring structure on this set. But this is easy using the ring structure of the localizations. For example, the multiplicative unit is the function which maps every point to the multiplicative unit in the corresponding localization. In fact, one easily checks that $\mathcal{O}_{\mathrm{Spec}(A)}(U)$ is a subring of the product $\prod_{\mathfrak{p} \in U} A_{\mathfrak{p}}$ which is a commutative ring with pointwise operations.

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