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Let $k$ be an algebraic closed field. Let $x$ be a point in $X=P_k^1$. What is $O_{X,x}$?

For example, if I have $x=(t-a)\in \text{Spec }k[t]$. Looking $x$ inside $P_k^1$, does $O_{X,x}=k[t]_{(t-a)}$? I'm confused when I have to deal with the sheaf of rings.

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    @Akhil: Dear Akhil, Tha$n$ks; I wondered if this was the case as I was posting my comment (hence my "Added" remark).2010-12-08

2 Answers 2

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For topological space $X$, open subset $U \subset X$ and point $x\in U$ we have for all sheaf $F$ on $X$ : $F_x = (F|U)_x$. Apply to $X=\mathbb P ^1, U=\mathbb A^1, x=(t-a)$

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    Note incidentally that if $R$ is a graded ring, and $f \in R$ a homogeneous element of degree $d$, then the homogeneous localization $R_{(f)}$ is isomorphic as a ring to $R^{(d)}/(f-1) R^{(d)}$ where $R^{(d)}$ consists of the fraction of $R$ that sits in degree $d$ or multiples thereof. This is an easy way of noting that $k[t,u]_{(u)} \simeq k[t]$. (Cf. EGA 2.2.)2010-11-08
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The stalk of the generic point is, of course, the rational function field $k(t)$. Now $\mathbb{P}^1$ is covered by two copies of $\mathbb{A}^1$. If a closed point $x$ corresponds to the maximal ideal generated by $(t-a)$ for some $a \in k$, then the stalk is

$O_{\mathbb{A}^1,x} = k[t]_{(t-a)}$

and this is isomorphic to $k[t]_{(t)}$.

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    @John: $\mathbb P^1$ is most easily *defined* as being two copies of $\mathbb A^1$ glued together in the usual way. So the formula for restrictions that you ask about is automatic, and is part of the construction.2010-12-08