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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?

2 Answers2

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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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Georges Elencwajg
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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. –  Jul 05 '13 at 07:09
  • 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 $\mathcal{O}{Y,x}$. –  Jul 05 '13 at 07:17
  • 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 05 '13 at 07:32
  • 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 05 '13 at 07:39
  • 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* –  Jul 05 '13 at 09:17
  • of $\mathcal{O}_{Y,x}$ that its elements are restrictions of regular functions of open subsets $U\subseteq X$ to $U \cap Y$? –  Jul 05 '13 at 09:18
  • Dear @Benja,yes, an element of $\mathcal O_{Y,x}$ is always the germ of the restriction of a regular function $F$ defined on some open neighbourhood of $x$ in $X$. This is true by definition, although depending on the presentation it may be somewhat hidden in some books. (In your last one-line comment, do not forget that you have to take the germ of the restriction in order to land in $\mathcal O_{Y,x}$) – Georges Elencwajg Jul 05 '13 at 09:27
  • Dear @Georges, if indeed that is the definition of $\mathcal{O}{Y,x}$ then isn't what I am asking tautologically clear from the definition? I was confused for nothing! Though something still bothers me, if $$\begin{eqnarray*} \mathcal{O}{Y,x} &:=& \lim_{U\ni x} (i_\ast \mathcal{O}{Y})(U) \&:=& \lim{U \ni x} \mathcal{O}Y(Y \cap U) \end{eqnarray*}$$ then it is not so clear that an element of $\mathcal{O}{Y,x}$ is the germ of the restriction of a regular function $F$ on some open neighbourhood of $x \in X$. –  Jul 05 '13 at 09:31
  • Dear Benja, your last definition contains, I think, a vicious circle since you can't define $\mathcal O_{Y,x}$ in terms of $\mathcal O_Y$. Once again, I think it is best to take germs of restrictions of regular functions defined on open neighbourhoods of $x$ in $X$. I suggest we leave these foundational questions at that for the time being. – Georges Elencwajg Jul 05 '13 at 09:53
  • Dear @Georges, I agree with leaving this question as it stands. Even though I still don't understand some things, I am moving on as I do not want just one thing to bog me down and keep me depressed :D In fact, this has given me more energy to learn scheme theory, I want to see how subschemes are defined! Regards, –  Jul 05 '13 at 14:31
  • Yes, learning schemes is an excellent idea. I'm sure you will succeed very well: good luck! – Georges Elencwajg Jul 05 '13 at 17:25
  • I find the statement ‘you can't define $(\mathscr O_Y)x$ in terms of $\mathscr O_Y$’ to be perplexing, in view of the general fact that if $Z\subset X$ is a closed subset of a topological space and $\mathscr F$ is a sheaf on $Z$, then the stalk $(i* \mathscr F)_P$ (where $i$ is the inclusion) is $\mathscr F_P$ if $P\in Z$, $0$ otherwise. – Tomo Nov 12 '15 at 04:45
  • I suppose the original poster is gone, but for those returning to this who are still confused, when Georges says ‘This is true by definition’ a few comments above, and then claims one cannot define the sheaf $\mathscr O_Y$ without mention of $X$, I believe he is pointing out that in the definition of a subvariety, the structure is induced by the structure of $X$ (the induced structure mentioned in ex. I.3.10 of Hartshorne, since it seems like the poster was thinking about ex. II.1.21b in the same book). – Tomo Nov 12 '15 at 05:05
  • The original poster was asking, essentially, if you could define a structure on Y as a variety without mentioning X, and then compare that structure to the induced structure induced by X, why do the two have to coincide. And the simple answer is ‘don’t worry about that, the structure on Y is the induced structure,’ which is the only answer that makes sense, since Y need not even be a closed subvariety, etc. – Tomo Nov 12 '15 at 05:10
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Let me illustrate how to find the lift of regular functions. (And I agree the comments that the former answer didn't reveal it clearly)

In short: to see the surjectivity, we need to use the definition of regular functions, in the affine case!

Here I adopt the definitions in Hartshorne, Chapter 1.

First we can find an affine cover of $X$. So $\forall (U,f)\in (i_{\ast}\mathcal{O}_{Y})_{x}$, we may assume that $U$ is an open subset of an affine space. Also now $Y\cap U$ is a quasi-affine variety since it is a subvariety of a quasi-affine variety $U$. Since $f$ is regular at $x$, we can find an open neighborhood of $x$ in $Y\cap U$, on which $f$ is a quotient of two polynomials. Note that these two polynomials are lived in the affine space. So we can view the quotient in the affine space and find a open neighborhood of $x$ in $U$, on which the quotient is still regular, since the zero set of the denominator is closed. We denote the quotient as $\bar{f}$ and find that $(V,\bar{f})\mapsto (U,f)$.

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