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If $X$ is a projective variety (irreducible algebraic subset of a projective space), $U$ is an open subset of $X$, $x\in U\subset X$, is that true that $\mathcal{O}_{X,x}=\mathcal{O}_{U,x}$ ?

I think it is true but I do not know how to prove it. My idea is following: $\mathcal{O}_{X,x}=\lim_{x\in V}\mathcal{O}(V)$, so we can take the intersection with the given $U$ above and then taking the limit, we get the result $\mathcal{O}_{U,x}$. But I am not sure about my argument.

Please help me to correct my idea and prove/disprove that proposition.

Arsenaler
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    You have the right idea, but your argument is somewhat imprecise. What is your working definition of $\displaystyle \lim_{x\in V}\mathcal{O}(V)$? Are you familiar with the universal property of limits? – bradhd Jul 11 '13 at 17:01

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BenjaLim's answer is great and constitutes a direct proof of the claim in which you're interested (+1). An alternate way of looking at it is via direct limits as Brad suggests. On the one hand, $O_{X,x}=\lim_{x\in V} O(V)$ and on the other hand, $O_{U,x}=\lim_{x\in V\subseteq U} O(V)$. Can you use the universal property of direct limits to construct inverse maps $O_{X,x}\to O_{U,x}$ and $O_{U,x}\to O_{X,x}$? Hint: You've got a natural (in $V$) family of maps $O(V)\to O(U\cap V)$ for $V\ni x$ that defines $O_{X,x}\to O_{U,x}$ and conversely (the easy part, I suppose!), you've got a natural family of maps $O(V)\to O(V)$ for $U\supseteq V\ni x$ that defines $O_{U,x}\to O_{X,x}$.

Alternatively, if you're familiar with the basic theory of direct limits, just note that the index set for the direct limit defining $O_{U,x}$ is cofinal in the index set for the direct limit defining $O_{X,x}$ (according to the definitions I've given above). The proof of this is equivalent to the one suggested in the first paragraph if you unwind the logic. I'm happy to provide you with relevant definitions/more details if you'd like!

I hope this helps!

Amitesh Datta
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  • Dear @Amitesh, my answer is simplified version of your answer above, +1 for yours too! I believe that the OP's hypothesis of $X$ being projective is superfluous. –  Jul 11 '13 at 17:26
  • Dear @BenjaLim, Thanks! I think that it's great that we've got complementing answers to the OP's question. I completely agree with you that the OP's hypothesis of $X$ being projective is superfluous (so superfluous, in fact, that I didn't notice it until you pointed it out!). I suspect that the OP is learning about these ideas in the context of projective algebraic geometry hence the hypothesis ... – Amitesh Datta Jul 11 '13 at 17:32
  • Dear @Amitesh, the reason I noticed this is because I have been thinking about these ideals recently (more generally in the context of sheaves). I actually have a question which I posted here, would you be able to answer it? Regards, –  Jul 11 '13 at 17:34
  • Thank you very much Amistesh Datta, you are right, I faced this type of problem in the context of projective algebraic geometry :). Anyway, thanks for the great answer :) – Arsenaler Jul 11 '13 at 17:54
  • Dear @BenjaLim, sure I'll try to answer it. It's early morning in Perth at the moment, where I am, and waaay past my bedtime but I'll definitely look at it in the morning! – Amitesh Datta Jul 11 '13 at 17:55
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    @Arsenaler I'm glad it helped! If there's anything else I can help you with, then please don't hesitate to ask! – Amitesh Datta Jul 11 '13 at 17:56
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An element in $\mathcal{O}_{X,x}$ is an equivalence class of pairs $\langle U,s \rangle$ with $s$ a section of $\mathcal{O}_X(U)$ and $U$ open about $x$. The equivalence relation that we have is that two pairs $\langle U,s\rangle = \langle V,t\rangle$ if $s$ is "eventually equal" to $t$. That is we can find an open neighbourhood $W \subseteq U \cap V$ for which $s|_W = t|_W$. From this it is clear that $\mathcal{O}_{X,x} = \mathcal{O}_{U,x}$ because any pair on the left $\langle V,s \rangle $ is equal to a pair $ \langle U \cap V,s|_{U \cap V} \rangle$ which is an element of $\mathcal{O}_{U,x}$. Note $U \cap V \neq \emptyset$ because $x$ is in both sets. The other containment is also clear because anything open in $U$ about $x$ is also open in $X$ about $x$, since being open is a transitive property.