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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