Question about the proof that $(D(f),\mathcal{O}_{\operatorname{Spec}A}|_{D(f)})\cong (\operatorname{Spec}A_f,\mathcal{O}_{\operatorname{Spec}A_f})$

I am trying to prove/understand why $$(D(f),\mathcal{O}_{\operatorname{Spec}A}|_{D(f)})\cong (\operatorname{Spec}A_f,\mathcal{O}_{\operatorname{Spec}A_f})$$. This problem appears in Vakil's algebraic geometry notes as problem 4.3.B.

I know that since $$D(f)=\{P\in\operatorname{Spec}A\mid f\not\in P\}$$, we can identify $$D(f)$$ and $$\operatorname{Spec}A_f$$. So let $$\pi:D(f)\rightarrow \operatorname{Spec}A$$ be the natural map.

I'd now like to show that $$\mathcal{O}_{\operatorname{Spec}A_f}\rightarrow \pi^*\mathcal{O}_{\operatorname{Spec}A}|_{D(f)}$$ is an isomorphism of sheaves. The hint given is to notice that distinguished open sets of $$\operatorname{Spec}A_f$$ are already distinguished open sets in $$\operatorname{Spec}A$$.

If we consider $$D(g/f^n)=\{P\in\operatorname{Spec}A_f\mid g/f^n\not\in P\}$$, then how can we think of this as a distinguished open set in $$\operatorname{Spec}A$$? It doesn't make sense to ask if $$g/f^n$$ is not in a prime ideal of $$A$$. Is really saying that the corresponding prime ideal of $$A$$ doesn't contain $$g$$?

Further, I know that $$\mathcal{O}_{\operatorname{Spec}A_f}(D(g/1))$$ is the localization of $$A_f$$ is the localization of $$A_f$$ at all elements that do not vanish outside of $$V(g/1)$$. That is, the localization of $$A_f$$ at $$\{a/f^n\in A_f\mid D(g/1)\subset D(g/f^n)\}$$.

And how do we describe $$\mathcal{O}_{\operatorname{Spec}A}|_{D(f)}(D(g))$$?

How can I go about finish this problem/seeing the isomorphism?

• Hint: clear denominators. Multiplying by an invertible element does not change the vanishing set. Aug 22, 2020 at 22:17

There's a lot of confusion in your post: as is, your proposed morphism of sheaves does not make any sense. The map you consider should not be the embedding $$\pi \colon D(f) \to \mathrm{Spec}(A)$$, but rather the embedding $$\mathrm{Spec}(\alpha) \colon \mathrm{Spec}(A_{f}) \to \mathrm{Spec}(A)$$ induced by the canonical localization map $$\alpha \colon A \to A_{f}$$. As you note, $$\pi := \mathrm{Spec}(\alpha)$$ is an open embedding whose image is $$D(f)$$, so we may view it as an isomorphism of topological spaces $$\mathrm{Spec}(A_{f}) \to D(f)$$.

Moving to sheaves, let me recall what the sheaf $$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)}$$ is. For any open set $$U \subset D(f)$$, $$U$$ is likewise an open set of $$\mathrm{Spec}(A)$$, and by definition we have $$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)}(U) = \mathcal{O}_{\mathrm{Spec}(A)}(U)$$. The key, then, is understanding which distinguished opens of $$\mathrm{Spec}(A)$$ are contained in $$D(f)$$ - more on this shortly. Moreover, the map $$\pi$$ comes with an associated morphism of sheaves $$\mathcal{O}_{\mathrm{Spec}(A)} \to \pi_{\ast}\mathcal{O}_{\mathrm{Spec}(A_{f})}$$, which on global sections is $$\alpha$$ and on distiniguished opens is the (induced) localization map. The corresponding morphism of sheaves $$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)} \to \pi_{\ast}\mathcal{O}_{\mathrm{Spec}(A_{f})}$$ is induced by $$\mathcal{O}_{\mathrm{Spec}(A)} \to \pi_{\ast}\mathcal{O}_{\mathrm{Spec}(A_{f})}$$ in the obvious way; on global sections, it is the identity map $$A_{f} \to A_{f}$$, since $$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)}(D(f)) = \mathcal{O}_{\mathrm{Spec}(A)}(D(f)) = A_{f}$$, and $$\pi^{-1}(D(f)) = \mathrm{Spec}(A_{f})$$.

All that remains is to understand why $$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)} \to \pi_{\ast}\mathcal{O}_{\mathrm{Spec}(A_{f})}$$ is an isomorphism of sheaves on $$D(f)$$. It suffices to check this on a basis for the topology on $$D(f)$$, which is given by the distinguished opens of $$\mathrm{Spec}(A)$$ contained in $$D(f)$$. With the above details settled, here is a guide to the approach, which I leave to you.

(1) First, show that we have a containment of distinguished opens $$D(g) \subset D(f)$$ if and only if $$f$$ is a unit of $$A_{g}$$. (This is exercise 3.5F of Vakil - very much worth doing, if you haven't done so yet.)

(2) Next, show that $$\pi^{-1}(D(g)) = D(\alpha(g)) = D(g/1)$$ for any $$g \in A$$. (There is nothing special about $$\pi$$ here, to be clear: for any morphism of rings $$u \colon A \to B$$ and any $$g \in A$$, one has $$\mathrm{Spec}(u)^{-1}(D(g)) = D(u(g))$$.)

(3) Finally, we put things together. Let $$D(g)$$ be a distinguished open of $$\mathrm{Spec}(A)$$ which is contained in $$D(f)$$, which by (1) ensures that $$f$$ is a unit in $$A_{g}$$. We have

$$\pi_{\ast}\mathcal{O}_{\mathrm{Spec}(A_{f})}(D(g)) = \mathcal{O}_{\mathrm{Spec}(A_{f})}(D(\pi(g))) = (A_{f})_{g/1}$$

and

$$\mathcal{O}_{\mathrm{Spec}(A)}|_{D(f)}(D(g)) = \mathcal{O}_{\mathrm{Spec}(A)}(D(g)) = A_{g}$$.

The map $$A_{g} \to (A_{f})_{g/1}$$ is the universal map induced by $$\alpha \colon A \to A_{f}$$. Your task is to show that this map $$A_{g} \to (A_{f})_{g/1}$$ is an isomorphism, which I leave to you. (I would use the universal property of localization to get a map $$(A_{f})_{g/1} \to A_{g}$$. You will use that $$f$$ is invertible in $$A_{g}$$ to get a map $$A_{f} \to A_{g}$$ first.)

• For the isomorphism in part (3), isn't it enough to say that since $f$ is invertible in $A_g$, and we know what invertible elements of $A_g$ look like, we have $f=g^m$ for some $m$ and so $(A_f)_{g/1} = (A_{g^m})_{g/1}=A_g$? Aug 25, 2020 at 15:11
• @ponchan: careful. Invertible elements of $A_{g}$ are a bit more complicated than what you describe - again, see Vakil exercise 3.5F. I really recommend using the universal property here: with some practice, it is the most natural and effective way to work with localizations. Aug 25, 2020 at 20:25
• A downvote (and unupvote) on this answer so long after it was first posted is perplexing to me. If the downvoter would comment, I would happily consider their suggestions. Jan 31, 2021 at 20:18