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For a graded ring $S$, $\mathrm{Proj}S$ is constructed in Hartshorne's Algebraic Geometry. Then Proposition 2.5 on the same page says that $$(D_+(f), \mathcal O|_{D_+(f)}) \cong \mathrm{Spec}S_{(f)}$$ for any homogeneous $f \in S_+$. The bijection between the underlying topological spaces is given by $$\varphi: \mathfrak a \mapsto (\mathfrak a S_f) \cap S_{(f)}.$$

I think I need some concrete examples to understand this, so I take $S = \mathbb Z[X_0,X_1]$ where the degree of $X_0$ and $X_1$ is $1$. Let $f=X_0$, $\mathfrak a$ be the ideal in $S$ generated by $X_1$, then it is homogeneous, prime and does not contain $f$. Through $\varphi$, it is mapped to $(\mathfrak a S_f) \cap S_{(f)}$. But what is $\mathfrak a S_f$? I think since $\frac{1}{X_1} \in S_f$ and $X_1 \in \mathfrak a$, $\mathfrak a S_f$ is $S_f$ itself. But $S_{(f)}$ is a subring of $S_f$, $S_f \cap S_{(f)} = S_{(f)}$, which is not a prime ideal of $S_{(f)}$. So where is wrong?

Thank you very much.

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  • $\begingroup$ Well, since $\mathfrak{a}$ doesn't contain $f$, you know that $\mathfrak{a}S_f$ is prime--so not everything. In particular, why is $\frac{1}{X_1}\in S_f$? It's not. $\endgroup$ Commented Sep 30, 2013 at 6:14
  • $\begingroup$ @AlexYoucis Thank you very much for the comment. I think $S_f$ consists of $g(X_0,X_1)/h(X_0,X_1)$ for $g(X_0,X_1),h(X_0,X_1) \in \mathbb Z[X_0,X_1]$, such that $h(0,X_1) \neq 0$. If this is wrong, they would you please tell me what $S_f$ is? Thanks again. $\endgroup$
    – sunkist
    Commented Sep 30, 2013 at 6:27
  • $\begingroup$ $S_f$ is just the localization of $S$ at $\{1,f,\ldots,\}$. In other words, the only things in the denominator are powers of $f$. $\endgroup$ Commented Sep 30, 2013 at 6:27
  • $\begingroup$ I am confused. Would that be denoted $(f)^{-1}S$? For an ideal $P$, $S_P$ is making $S-P$ invertible... $\endgroup$
    – sunkist
    Commented Sep 30, 2013 at 6:30
  • $\begingroup$ We're not localizing at a prime, we're localizing at $f$. You may be more familiar with the notation $T^{-1}S$ where $T=\{1,f,\ldots\}$ or something, but here $S_f$ means localizating at $\{1,f,\ldots\}$. If you've never seen this notation before, I can understand why you may think we're localizing at a prime though. $\endgroup$ Commented Sep 30, 2013 at 6:31

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The problem with what you wrote is the fact that $X_1$ is not invertible in $S_f$.

In particular, $\mathfrak{a}S_f=X_1 S_f\ne S_f$. Then,

$$\mathfrak{a}S_f\cap S_{(f)}=\frac{X_1}{X_0}S_{(f)}$$

This ideal IS prime. Namely, if you define the isomorphism $S_{(f)}\to \mathbb{Z}[T]$ in the natural way, $\displaystyle \frac{X_1}{X_0}\mapsto T$.

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  • $\begingroup$ Now I see where I was wrong. Thank you very much. $\endgroup$
    – sunkist
    Commented Sep 30, 2013 at 7:29

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