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Let $X$ be a separated integral variety and $Y_1,Y_2$ two irreducible closed subvarieties. Show that if $\mathcal{O}_{X,Y_1}\subseteq \mathcal{O}_{X,Y_2}$ then $Y_2\subseteq Y_1$.

Here $\mathcal{O}_{X,Y_i}$ is the local ring of $X$ at the generic point of $Y_i$. This is a stronger form of this question.

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For convenience we let $x,y$ be the generic points of $Y_1$ and $Y_2$.

Let $U$ be any affine open subscheme of $X$ containing $x$ and $V$ be some affine open subscheme of $V$ containing $y$.

Then $U \cap V$ is an affine scheme. Now $O(U \cap V)$ is the subring of $K(X)$ generated by $O(U)$ and $O(V)$, thus it has a natural morphism of $O(V)$-algebras $O(U \cap V) \rightarrow O_{X,y}$. This corresponds to the canonical map $\operatorname{Spec}\,O_{V,y} \rightarrow V$ factoring through the inclusion $U \cap V \rightarrow V$, thus implying $y \in U \cap V \subset U$.

So any open subset containing $x$ contains $y$, thus $Y_1 \subset Y_2$.

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  • $\begingroup$ I have a question. By the separated assumption I know we have a surjective map $$\begin{align*} O(U)\otimes_k O(V)&\rightarrow O(U\cap V)\\ f\otimes g&\mapsto fg\end{align*}$$ but how do you get from here the map $O(U\cap V)\rightarrow O_{X,y}$? $\endgroup$
    – Walter Simon
    Apr 8, 2020 at 18:48
  • $\begingroup$ Ooh I see know, every element in $O(U\cap V)$ is a sum of products $fg$, but as both $f,g$ are in $O_{X,y}$ ($f$ is there by the inclusion hypothesis) we get an inclusion $O(U\cap V)\hookrightarrow O_{X,y}$. Thanks! $\endgroup$
    – Walter Simon
    Apr 8, 2020 at 18:53

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