# Steps in defining $\mathscr O_{Spec R}$

I'm trying to understand the definition of $$\mathscr O_{Spec R}$$ from a larger perspective. Here's what steps I was told about (I might be misinterpreting something). I'll write $$\mathscr F$$ for $$\mathscr O_{Spec R}$$.

(1) Define the presheaf on the base by $$F(D(f))=R_f$$

(2) Prove that this is a sheaf

(3) Define the stalks by $$F_P=R_P$$

(4) For an open set $$U$$, define $$\mathscr F(U)=\{(s_P\in R_P)_{P\in U}: s_P \text{ are compatible}\}$$ (in our case, compatible means $$\forall P\in U\ \exists f\in R\text{ s.t. } P\in D(f)\subset U \ \text{ and } \exists\ a/f^n\in \mathscr F(D(f))=R_f \text{ with } s_Q=(a/f^n)_Q \text{ for all } Q\in D(f)$$)

(5) Prove that $$\mathscr F(U)$$ is isomorphic to $$F(U)$$

Here are my questions:

• Regarding (3), Vakil's Theorem 2.5.1 says that $$F_P$$ should be $$colim F(D(f))$$, if I understand correctly. So is it the case that $$colim F(D(f))=R_P$$? If so, what is the definition of the colimit map $$R_f\to R_P$$? If $$P\in D(f)$$, then $$f\notin P$$ and I guess this map can be defined as $$a/f^n\mapsto a/f^n$$, but do we know that $$P\in D(f)$$? This confusion may be coming from not fully understanding the notation $$colim F(D(f))$$ -- it is the colimit of which diagram $$E:I\to \text{Ring}$$? What is $$I$$? Update: Is it true that $$I$$ depends on $$P$$, and $$I=I_P$$ contains (as objects) elements $$f\in R$$ such that $$f\notin P$$? And the coprojection maps are, I suppose the maps described here?

• (Continuation of the above question.) Suppose $$D(f_j)\subset D(f_i)$$. What is the map $$R_{f_i}=F(D(f_i))\to F(D(f_j))=R_{f_j}$$? My naive guess would be $$a/(f_i)^n\mapsto a/(f_j)^n$$, but how is the fact $$D(f_j)\subset D(f_i)$$ used? Vakil says that $$F(D(f_j))$$ is a further localization of $$F(D(f_i))$$, but I'm not sure if I understand this -- these rings, by definition are $$R_{f_j}$$ and $$R_{f_i}$$, and we only know that if $$f_j\notin P$$, then $$f_i\notin P$$, I don't see how this makes $$F(D(f_j))$$ a further localization of $$F(D(f_i))$$ (there's no reference to $$P$$ in, say, $$F(D(f_j))=R_{f_j}$$). This question is answered here (although I still don't understand why it is a "further" localization).

• Is (5) just a particular case of Vakil's theorem 2.5.1? For any sheaf on the base $$F$$, $$\mathscr F(U)$$ is isomorphic to $$F(U)$$. (Also I suppose the fact that our $$\mathscr F$$ is a sheaf is a consequence of the same theorem.)

For your first question, I think it is easier to understand it using the concrete description of filtered colimits given in Vakil's exercise 1.4.C.

The diagram does depend on $$P$$. To make the connection, we treat the presheaf on a base $$F$$ as a contravariant functor $$D \to \operatorname{Ring}$$, where $$D$$ is the category of distinguished open sets containing $$P$$ (with morphisms given by inclusion). The induced functor $$D^{\text{Op}} \to \text{Ring}$$ will be our diagram in Ring.

We note that $$D^{\text{Op}}$$ is filtered. After all, if $$U_1$$ and $$U_2$$ are open sets containing $$P$$, then $$U_1 \cap U_2 \subseteq U_1$$ and $$U_1 \cap U_2 \subseteq U_2$$ so that there are 'morphisms' $$U_1 \to U_1 \cap U_2$$ and $$U_2 \to U_1 \cap U_2$$. The second condition follows similarly.

As such, we can consider $$F_P$$ as ordered pairs $$[D(f), s]$$ for distinguished neighborhood $$P \in D(f)$$ and $$s \in F(D(f)).= R_f$$ under the equivalence relation $$[D(f), s] \sim [D(g), t]$$ if there is some neighborhood $$D(h) \subseteq D(f) \cap D(g)$$ of $$P$$ where $$s|_{D(h)} = t|_{D(h)}$$.

With this description, we define a surjective homomorphism $$F_P \to R_P$$ given by $$[D(f), s] \mapsto s$$. This makes sense since $$f \notin P$$ by the definition of $$D(f)$$ and the fact that $$s \in R_f$$.

To show that this is an isomorphism, suppose $$[D(f), s] \mapsto 0$$. Then $$s = t/f^n \in R_f$$ is $$0$$ in $$R_P$$. Since $$t/f^n = 0/1$$ in $$R_P$$, there is a $$g \notin P$$ so that $$g(f^n * 0 - t * 1) = - gt = 0.$$ We then claim that $$s|_{D(fg)} = \frac{tg^n}{(fg)^n} = 0$$ in $$F(D(fg)) = R_{fg}$$. This follows from the above equation by multiplying each side by $$g^{n - 1}$$ to obtain $$(fg)^n * 0 - (tg^n) * 1 = - g^{n - 1}(gt) = 0.$$

This proves that $$F_P \cong R_P$$.

For your last question, I think the rest does indeed follow from the theorem 2.5.1 cited, and the fact that $$\mathscr{F}_P \cong F_P$$ follows quickly from the definition of a base.