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There is a post about this exact question here I am having the exact same issue and don't find the solution there complete.

Let $Y\subseteq X$ be a subvariety. Let $\mathcal{O}_{Y,X}$ be the set of equivalence classes $\langle U,f\rangle$, where $U\subseteq X$ is open, $U\cap Y\neq \emptyset$, and $f$ is a regular function on $U$. We say that $\langle U,f\rangle$ is equivalent to $\langle V,g\rangle$, if $f=g$ on $U\cap V$. Show that $\mathcal{O}_{Y,X}$ is a local ring, with residue field $K(Y)$ and dimension = $\dim X-\dim Y$.

Let $m=\{\langle U,f\rangle\in \mathcal{O}_{Y,X}|f(P)=0,\ \forall P\in U\cap Y\}$. I have proved that $m$ is the unique maximal ideal of $\mathcal{O}_{Y,X}$. The part that I am stuck on is trying to show that $\mathcal{O}_{Y,X}/m\cong K(Y)$. Many places I've seen just say it's obvious like the solution posted there. I defined a map $\varphi:\mathcal{O}_{Y,X}\rightarrow K(Y)$ by $\langle U,f\rangle\mapsto\langle U\cap Y,f\rangle$. It is clear this map is well defined, and has kernel $m$. The only thing is now I can't show that it is onto. So let $\langle U,f\rangle\in K(Y)$. Since $U$ is open in $Y$ there exists a $V$ such that $U=V\cap Y$. the hope is to extend $f$ to a $g$ such that $g$ is regular on $V$ and $G|_{U}=f$.

I am most interested in a construction of this $g$. With this the problem is complete. Any help would be appreciated. Thanks.

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TheNumber23
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You can reduce to the affine case as follows. Let $\left<U, f\right> \in K(Y)$. Choose $V$ open in $X$ such that $U = V \cap Y$. Since $V$ has an open cover by affine varieties, we can replace it by an affine open $V'$ such that $U \supseteq V' \cap Y \neq \emptyset$, since at least one of these affine opens must intersect $Y$. Thus we can suppose that both $V$ and $U$ are affine, so that the $K$-algebra homomorphism corresponding to the closed immersion $U \hookrightarrow V$ is a surjective map of $K$-algebras $A(V) \to A(U)$. The rational function $\left<U, f\right>$ can be viewed as an element of $A(U)$, and it lifts to an element $\left<g, V\right>$ of $A(V)$, which we can view as an element of $\mathcal O_{X,Y}$.

Bruno Joyal
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  • What is meant by the closed immersion? – TheNumber23 Dec 02 '13 at 20:50
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    @TheNumber23 It's just a fancy word for the inclusion of a closed subvariety. – Bruno Joyal Dec 02 '13 at 21:11
  • Ok that's what I though. the map is defined by $x\mapsto x$? Then how is the map onto? – TheNumber23 Dec 02 '13 at 22:29
  • @TheNumber23 I suggest you try to prove the following: let $h : A\to A'$ be a map of affine $K$-algebras (reduced, finite-type $K$-algebras), and let $­f: V' \to V$ be the morphism of affine varieties which it determines. Then $f$ is an isomorphism of $V'$ with a closed subvariety of $V$ if and only if $a$ is a surjective homomorphism of rings. (Hint: you know what the closed subsets of $V$ look like: they look like $V(I)$ for an ideal $I$ of $A$. Try to show that $A'=A/I$ and that $h$ identifies with the quotient map $A \to A/I$.) – Bruno Joyal Dec 02 '13 at 22:37
  • This statement is saying if $f$ is injective if and only if it's corresponding map $h$ is onto? – TheNumber23 Dec 04 '13 at 18:41
  • @TheNumber23 No, $f$ must also be a closed map, and an isomorphism onto the image. – Bruno Joyal Dec 04 '13 at 20:25
  • What allows you to suppose that $U$ is affine? – JDZ Dec 03 '19 at 02:24