On an separated variety, $\mathcal{O}_{X,Y_1}\subseteq \mathcal{O}_{X,Y_2}$ then $Y_2\subseteq Y_1$

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.

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 $$Y_1 \subset Y_2$$.
• 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 $O(U\cap V)\rightarrow O_{X,y}$? Apr 8 '20 at 18:48
• 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! Apr 8 '20 at 18:53