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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?

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    $\begingroup$ Hint: clear denominators. Multiplying by an invertible element does not change the vanishing set. $\endgroup$
    – Zhen Lin
    Commented Aug 22, 2020 at 22:17

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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.)

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  • $\begingroup$ 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 $(A_f)_{g/1} = (A_{g^m})_{g/1}=A_g$? $\endgroup$
    – ponchan
    Commented Aug 25, 2020 at 15:11
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    $\begingroup$ @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. $\endgroup$ Commented Aug 25, 2020 at 20:25
  • $\begingroup$ 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. $\endgroup$ Commented Jan 31, 2021 at 20:18

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