# Direct limit of $\mathscr{F}(U)$ is the same as direct limit of $\mathscr{F}(X_f)$, where $P\in U$ and $f\notin P$

This question is from Mumford's The Red Book of Varieties and Schemes (Section I.4).

Let $$X\subseteq k^n$$ be an irreducible algebraic set, $$R$$ its affine coordinate ring. Since $$X$$ is irreducible, $$I(X)$$ is prime and $$R$$ is an integral domain. Let $$K$$ be its field of fractions.

Let $$\underline{o}_x=\{f/g\mid f, g\in R, g(x)\neq 0\}\subseteq K$$. Now for $$U$$ open in $$X$$, let $$\underline{o}_X(U)=\bigcap_{x\in U}\underline{o}_x.$$

Let $$X_f=\{x\in X\mid f(x)\neq 0\}$$.

The author said: Since the sets $$X_f$$ are a basis of the Zaiski topology of $$X$$, we have $$\varinjlim_{x\in U}\underline{o}_X(U)=\varinjlim_{x\in X_f}\underline{o}_X(X_f).$$

My Question: Why? I think the first part I have to prove is that $$\bigcup_{x\in U}\underline{o}_X(U)=\bigcup_{x\in X_f}\underline{o}_X(X_f).$$ But I cannot prove it.

• If you know some complex analyis compare $\mathcal O_x$ with germs of meromorphic functions at a point. Might be insightful. – Ignorant Mathematician Apr 14 at 7:00
• @Ghosh Thx. But I am looking for a purely algebraic proof. I will come back to see your hint after learning complex analysis. – bfhaha Apr 14 at 7:06

The sets $$X_f$$ are cofinal in the poset of open sets of $$X$$ containing $$x$$ (this follows from the $$X_f$$ being a basis for the topology of $$X$$). Since stalks are calculated as colimits, it suffices to calculate along any cofinal collection - which is exactly the claim the author makes.
• I don't have a reference offhand - just look at the definition of cofinal: a subcollection $A$ of a poset $B$ is cofinal iff for any $b\in B$ there is an $a\in A$ so that $b\leq a$. In particular, this means that the morphisms from any $b\in B$ to the colimit factor through the cofinal collection, so the colimit of the original collection and the cofinal collection satisfy the same universal property and must be isomorphic. – KReiser Apr 14 at 20:34