This is a problem from Gortz and it does NOT assume that the underlying space is the spectrum of the ring or anything like that. Now I proved easily that given a clopen set of $X$ there is an idempotent element of $\mathcal{O}_X(X) $, using the "hint" which glued a section which was the identity on the clopen set $U$ and $0$ on $X-U$. I also know that, given a section of $\mathcal{O}_X(X) $ the set where it restricts to 1 and the set where it restricts to 0 will both be open and that a ring with idempotent $e$ splits as the direct product of $Re$ and $R(1-e)$. I'm pretty sure the answer will use both of these facts but I don't seem to be quite clever enough to put everything together. Any hints??
Edit: In fact I just noticed the bijection asked for is between idempotent elements of $\Gamma(X, \mathcal{O}_X) $ and clopen sets of $X$ and so I have an even stupider question : what is the difference between the ring $\mathcal{O}_X(X) $ and the ring $\Gamma(X, \mathcal{O}_X) $??
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Thanks Georges, have edited accordingly – R Mary Jun 10 '16 at 12:08
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2With the edit, when $\mathcal{F}$ is a sheaf then generally $\mathcal{F}(U)$ and $\Gamma (U,\mathcal{F})$ are notations for the same thing. – Andrew Dudzik Jun 10 '16 at 12:09
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Dear slade, am I right in thinking that they would be different for presheaves? (where I assume the second would be contained in the first?) – R Mary Jun 10 '16 at 12:16
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1@Slade is right. For presheaves it is best to use $\mathcal F(U)$. There is no distinction between $\mathcal F(U)$ and $\Gamma(X, \mathcal{O}_X) $ universally agreed upon by good authors. I would say that historically $\Gamma(X, \mathcal{O}_X) $ was used in the interpretation of sheaves as étalé spaces. Anyway, you should consider both as synonymous unless an author adopts a special convention, which is not the case in your problem. – Georges Elencwajg Jun 10 '16 at 12:32
1 Answers
The key tool is to introduce for any clopen subset $U\subset X$ the global section $e_U\in \mathcal O_X(X)$ defined by: $$e_U\vert _U =1_U\in \mathcal O_X(U),\quad e_U\vert _{X\setminus U} =0_{X\setminus U}\in \mathcal O_X(X\setminus U) $$
The sheaf axioms for $\mathcal O_X$ ensure that $e_U\in \mathcal O_X(X)$ is a well defined section.
That said, the required bijection is $$\operatorname {Clopen} (X)\stackrel \sim \to \operatorname {Idempot}(\mathcal O_X(X)):U\mapsto e_U$$
The inverse bijection being $$\operatorname {Idempot}(\mathcal O_X(X)) \stackrel \sim \to \operatorname {Clopen} (X) :e\mapsto U_e:=\{x\in X\vert e_x=1_x\in \mathcal O_{X,x}\}$$
Edit
Notice that for a fixed open $U$ we have $(e_U)_x=1_x\in \mathcal O_{X,x}$ for $x\in U$ and $(e_U)_x=0_x\in \mathcal O_{X,x}$ for $x\notin U$.
It is not allowable however to define $e_U$ by these properties: a section of a sheaf is not just a a random collection of germs in its stalks.
However this description of the germs of $e_U$ shows why $e_{U_e}=e$ for a given $e\in \operatorname {Idempot}(\mathcal O_X(X)) $, as requested by @Roland in his comment.
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Maybe you should write why these maps are inverse to each other. The fact that $e\mapsto U_e\mapsto e_{U_e}=e$ is not totally obvious and this is where we need the assumption that this is not only a ringed space but a locally ringed space. – Roland Jun 10 '16 at 13:22
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2Ah no wait I can see it now!! It is because local rings have no non trivial idempotent elements so on stalks e will map to either zero or one, correct? – R Mary Jun 10 '16 at 13:59
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Taking a section and mapping to its stalk at a point is a ring homomorphism, right? I apologize for asking such basic questions, I have so little algebraic intuition I get worried about saying anything – R Mary Jun 10 '16 at 14:02
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@GeorgesElencwajg Thx, I was under the impression that it was where the OP was stuck. But R Mary found out. – Roland Jun 10 '16 at 14:10
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