Timeline for Affine scheme on spec(A) of a ring A as the sheafification of a pre-sheave on spec(A)?
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2019 at 19:25 | comment | added | bfhaha | Why "This shows that the associated sheaf of $\mathscr{O}'$ is indeed $\mathscr{O}_{\text{Spec }(A)}$"? I think it suffices to show that $\mathscr{O}'_{\mathfrak{p}}=\varinjlim\limits_{\mathfrak{p}\in U}\mathscr{O}'(U)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}\mathscr{O}'(X_f)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}A_f=A_{\mathfrak{p}}$. But I cannot prove $\varinjlim\limits_{\mathfrak{p}\in U}\mathscr{O}'(U)=\varinjlim\limits_{f\in A\backslash \mathfrak{p}}\mathscr{O}'(X_f)$. Could anyone help me? Thx! | |
| Apr 13, 2019 at 6:13 | comment | added | bfhaha | Interesting! Hartshorne verified (or expected readers to verify) $\mathscr{O}_{\text{Spec }A}$ is a sheaf directly. Then proved that $\mathscr{O}(D(f))\cong A_f$. You did that in a reverse way. | |
| Apr 13, 2019 at 6:07 | comment | added | bfhaha | $U=\text{Spec }(A_f)$ means the set of all prime ideals in $A$ which are correspondent with the prime ideals in $A_f$. That is, the set of all prime ideals in $A$ which are disjoint $\{1, f, f^2, ...\}$. | |
| May 13, 2017 at 21:45 | comment | added | Ingo Blechschmidt | The mapping $U \mapsto S_U$ is a sheaf on $\operatorname{Spec}(A)$, in fact a subsheaf of the constant sheaf $\underline{A}$. It is also called the "universal filter" or "generic filter" of $A$. The structure sheaf can then simply be constructed as the localization $\underline{A}[S^{-1}]$ (performed in the internal language of the topos of sheaves over $\operatorname{Spec}(A)$). | |
| Nov 11, 2011 at 12:20 | vote | accept | urelement | ||
| Nov 10, 2011 at 17:58 | comment | added | Martin Brandenburg | Probably this is the best definition of the structure sheaf on the specrum ... and I wonder why I haven't seen it yet :). | |
| Nov 10, 2011 at 6:27 | history | edited | user2035 | CC BY-SA 3.0 |
added 99 characters in body
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| Nov 10, 2011 at 6:17 | history | answered | user2035 | CC BY-SA 3.0 |