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Suppose $Y$ is a subvariety of a variety $X$ (according to Hartshorne this means if $X$ is quasi-affine or quasi projective then $Y$ is a locally closed subset of $X$, c.f. exercise 3.10, chapter 1). Now given $i : Y \to X$ the inclusion map, I am trying to figure out what the stalk at $x \in Y$ of $i_\ast(\mathcal{O}_Y) $ is. Now if I unwind the definitions, firstly for any open set $U \subseteq X$ we have $i_\ast(\mathcal{O}_Y)(U) = \mathcal{O}_Y(U \cap Y)$. I guess that the stalk at $x$ of $i_\ast (\mathcal{O}_Y) $ should consist of pairs $\langle U \cap Y,f \rangle$ where $f$ is a regular function on $U \cap Y$.

However why should it be the case that on stalks the map induced from restriction $(\mathcal{O}_X)_x \to (i_\ast \mathcal{O}_Y)_x$ is surjective? This seems to be saying to me that any regular function on an open subset $U \cap Y$ of $Y$ is the restriction of some regular function on an open subset of $X$, but is this true?

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up vote 4 down vote accepted

1) Yes, we have for the stalk of $i_\ast(\mathcal{O}_Y) $ at $x$ the equality $(i_\ast \mathcal{O}_Y)_x=\mathcal{O}_{Y,x} $ : as you write, this follows from the definitions.

2) Yes, the canonical morphism of local rings $\mathcal{O}_{X,x}\to \mathcal{O}_{Y,x}$ is an epimorphism.
Its kernel is $\mathcal I_x$, the stalk at $x$ of the sheaf of ideals $\mathcal I=\mathcal I_Y$ defining $Y$ in $X$.

These are not theorems but essentially definitions: the modern point of view is that closed subvarieties or, better, closed subschemes of a scheme $X$ correspond by definition to quasi-coherent ideals of $\mathcal O_X$: cf. Hartshorne, Proposition II 5.9 and this answer to Ravi Vakil's query, emphasizing that in the affine case closed subschemes of $Spec(A)$ are in perfect bijective correspondence with ideals of the ring $A$.
In my opinion elementary presentations of varieties tend to hide this simple and beautiful correspondence, replacing it by ad hoc, easy but not so transparent constructions.

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Dear Georges, thanks for your answer but I fail to see why the canonical morphism of local rings is an epimorphism. For example the claim amounts to saying that any regular function $f$ on a neigbourhood about $x$ open in $Y$ is the restriction of a regular function on a neighbourhood about $x$ that is open in $X$. Since $Y$ may be a closed subvariety how does this follow immediately? I am sorry if I am asking basic questions, but I am confused. – user38268 Jul 5 '13 at 7:09
I should state that in my mind, $\mathcal{O}_{Y,x}$ consists of pairs $\langle V \cap Y,f$ where $f$ is regular on $V \cap Y$ and $V$ is open in $X$. From this definition it is not at all clear to me why any $f$ like that comes from a regular function on an open subset of $X$. Certainly if $f$ were * a priori* regular on $V$ that it is clear for me. But certainly $f$ need not be at least from the definition of $\mathcal{O}_{Y,x}$. – user38268 Jul 5 '13 at 7:17
Dear @Benja, the difficulty might be in the notions of germ and stalk. Given your $f$ defined on an open neighbourhood $U$ of $x\in X$, the claim is that there exists an open neighbourhood $V$ of $x$ in $X$ and a function $F\in \mathcal O_X(V)$ such that $V\cap Y\subset U$ (note carefully that I didn't write an equality but an inclusion) and $F|V\cap Y=f|V\cap Y\subset U$. As to the definition of the stalk, it is constructed out of the set of pairs you mention by dividing out by the equivalence relation obtained from restrictions of functions. – Georges Elencwajg Jul 5 '13 at 7:32
Also, I'm not sure I understand your last sentence: only regular functions come up in this context. The elements of $\mathcal O_{Y,x}$ are constructed by restricting regular functions on open subsets of $X$ (and no other functions) , and then suitably identifying some of these restrictions. – Georges Elencwajg Jul 5 '13 at 7:39
Dear @Georges, thanks for your time to reply to my comments. Now when you say "the claim is that there exists an open neighbourhood $v$ of $x \in X$ and a function $F \in \mathcal{O}_X(V)$ such that $V \cap Y \subset U$ and $F|_{V \cap Y} = f|_{V\cap Y} \subseteq U$, may I know what is the justification for this claim?. It seems to me that this is automatic judging from your last comment: That elements of $\mathcal{O}_{Y,x}$ are constructed by restricting regular functions on open subsets of $X$. If I am understanding this statement correctly, have we rigged in the definition – user38268 Jul 5 '13 at 9:17

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