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in proofing Noetherian property as a local property,he proof as follow:enter image description here

but I did not find this isomorphism so easy digested:enter image description here

and whats more ,spec (Aj)fj shall be D(fj),and under different underlying topology spaces ,does the same character D make any sense?If not ,how should I understand D(f)=D(fj),plz give me a detail proof or explain about this isomorphism,since I am lack of exp in this course,thanks a lot,yours Yoshinobu~

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Is this some old version of Hartshorne's book? It looks different. Chattier, really! – Dylan Moreland Aug 5 '11 at 16:45
I mistake it!It is Ueno's book!LoL~ – Yoshinobu Osawa Aug 6 '11 at 5:10
up vote 1 down vote accepted

$\DeclareMathOperator{\Spec}{Spec}$Let's go through this carefully. We have $X = \Spec R$, an open subset $U \subset X$, and a morphism of schemes $g\colon \Spec A_j \to X$ which induces an isomorphism $(\Spec A_j, \mathscr{O}_{\Spec A_j}) \to (U, \mathscr{O}_X|_U)$. We find an element $f \in R$ such that $D(f) \subset U$. Now, Hartshorne proves earlier that $g$ corresponds to a homomorphism $\varphi\colon R \to A_j$, so that $g(\mathfrak{p}) = \varphi^{-1}(\mathfrak{p})$. And you can check that $g^{-1}(D(f)) = D(\varphi(f)) = D(f_j)$. (This is one way of proving that $g$ so induced are continuous.)

So the isomorphism will restrict to an isomorphism $(D(f_j), \mathscr{O}_{\Spec A_j}|_{D(f_j)}) \to (D(f), \mathscr{O}_X|_{D(f)})$. Replacing the source and target of this by isomorphic schemes, we get an isomorphism $\Spec (A_j)_{f_j} \to \Spec R_f$. Perhaps using an equals sign is a bit cavalier, but you'll have to get used to it.

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thanks alot ,I am interested in arithemetic algebraic geomwtry,but find it so hard to start.I learn some algebraic number theory before,like neukirch's book,but non algebraic geometry,how to get started? – Yoshinobu Osawa Aug 6 '11 at 5:22
@Yoshinobu I wish I knew! There have been a few similar questions on this site and MathOverflow. – Dylan Moreland Aug 6 '11 at 21:29

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