Lemma 29.49.7. Let $X$ be an integral separated scheme. Let $Z_1$, $Z_2$ be distinct irreducible closed subsets of $X$. Let $\eta _ i$ be the generic point of $Z_ i$. If $Z_1 \not\subset Z_2$, then $\mathcal{O}_{X, \eta _1} \not\subset \mathcal{O}_{X, \eta _2}$ as subrings of $R(X)$. In particular, if $Z_1 = \{ x\} $ consists of one closed point $x$, there exists a function regular in a neighborhood of $x$ which is not in $\mathcal{O}_{X, \eta _{2}}$.

**Proof.**
First observe that under the assumption of $X$ being separated, there is a unique map of schemes $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2}) \to X$ over $X$ such that the composition

is the canonical map $\mathop{\mathrm{Spec}}(R(X)) \to X$. Namely, there is the canonical map $can : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2}) \to X$, see Schemes, Equation (26.13.1.1). Given a second morphism $a$ to $X$, we have that $a$ agrees with $can$ on the generic point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$ by assumption. Now $X$ being separated guarantees that the subset in $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$ where these two maps agree is closed, see Schemes, Lemma 26.21.5. Hence $a = can$ on all of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta _2})$.

Assume $Z_1 \not\subset Z_2$ and assume on the contrary that $\mathcal{O}_{X, \eta _{1}} \subset \mathcal{O}_{X, \eta _{2}}$ as subrings of $R(X)$. Then we would obtain a second morphism

By the above this composition would have to be equal to $can$. This implies that $\eta _2$ specializes to $\eta _1$ (see Schemes, Lemma 26.13.2). But this contradicts our assumption $Z_1 \not\subset Z_2$. $\square$

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## Comments (2)

Comment #3209 by Dario Weißmann on

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