Lemma 33.41.4. Let $X$ be an integral separated scheme. Let $x_1, \ldots , x_ r \in X$ be a finite set of points such that $\mathcal{O}_{X, x_ i}$ is Noetherian of dimension $\leq 1$. Then there exists an affine open subscheme of $X$ containing all of $x_1, \ldots , x_ r$.

Proof. Let $K$ be the field of rational functions of $X$. Set $A_ i = \mathcal{O}_{X, x_ i}$. Then $A_ i \subset K$ and $K$ is the fraction field of $A_ i$. Since $X$ is separated, and $x_ i \not= x_ j$ there cannot be a valuation ring $\mathcal{O} \subset K$ dominating both $A_ i$ and $A_ j$. Namely, considering the diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A_1) \ar[d] \\ \mathop{\mathrm{Spec}}(A_2) \ar[r] & X }$

and applying the valuative criterion of separatedness (Schemes, Lemma 26.22.1) we would get $x_ i = x_ j$. Thus we see by Lemma 33.36.3 that $A_ i \otimes A_ j \to K$ is surjective for all $i \not= j$. By Lemma 33.36.7 we see that $A = A_1 \cap \ldots \cap A_ r$ is a Noetherian semi-local ring with exactly $r$ maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ r$ such that $A_ i = A_{\mathfrak m_ i}$. Moreover,

$\mathop{\mathrm{Spec}}(A) = \mathop{\mathrm{Spec}}(A_1) \cup \ldots \cup \mathop{\mathrm{Spec}}(A_ r)$

is an open covering and the intersection of any two pieces of this covering is $\mathop{\mathrm{Spec}}(K)$. Thus the given morphisms $\mathop{\mathrm{Spec}}(A_ i) \to X$ glue to a morphism of schemes

$\mathop{\mathrm{Spec}}(A) \longrightarrow X$

mapping $\mathfrak m_ i$ to $x_ i$ and inducing isomorphisms of local rings. Thus the result follows from Lemma 33.41.3. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).