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Let $R$ be a finitely generated algebra over an algebraically closed field $k$ and consider $\operatorname{Spm}(R)$ the maximal spectrum of $R$ equipped with the Zariski topology. For an ideal $J$ of $R$ we denote $V(J)$ the closed set of $\operatorname{Spec}(R)$ and $D(J)$ the open set. Similarly we denote $V(J)_m,D(J)_m$ the closed and open sets of $\operatorname{Spm}(R)$.

Question:

i) Prove that $V(J)_m=V(J)\cap \operatorname{Spm}(R)$ and $D(J)_m=D(J)\cap \operatorname{Spm}(R)$.

ii) Show that the sheaf of commutative rings $\mathcal{O}_{Spm(R)}$ can be defined as

$$\Gamma(D_m(J),\mathcal{O}_{\operatorname{Spm}(R)})=\Gamma(D(J),\mathcal{O}_{\operatorname{Spm}(R)})$$

I can do i) because $V_m(I)=\{\mathfrak{p}\in \operatorname{Spm}(R):\mathfrak{p}\supset I\}=\{\mathfrak{p}\in \operatorname{Spec}(R)\cap \operatorname{Spm}(R):\mathfrak{p}\supset I\}=V(I)\cap \operatorname{Spm}(R)$ and similarly for $D_m(J)$ but I am not sure how to proceed for ii). Why would this equality of ringed spaces define the sheaf $\mathcal{O}_{\operatorname{Spm}(R)}$? How would I use the finitely generated property of $R$?

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  • $\begingroup$ read the linked post. If you have more questions - please post a question and I will explain the definition. $\endgroup$
    – hm2020
    Jun 15, 2021 at 11:49
  • $\begingroup$ Let me remark that equation $SH$ in your post is not a definition - you cannot use $\mathcal{O}_{Spm(R)}$ to define $\mathcal{O}_{Spm(R)}$. Included is a link to a post at MO where I give a correct definition of the structure sheaf for the max-spectrum: mathoverflow.net/questions/377922/… $\endgroup$
    – hm2020
    Jun 15, 2021 at 11:54
  • $\begingroup$ @hm2020 Yes that was exactly my problem, perhaps there is a typo and it meant $\Gamma(D_m(J),\mathcal{O}_{Spec(R)})=\Gamma(D_m(J),\mathcal{O}_{Spm(R)})$? That morphism would come from the inclusion map $\endgroup$
    – idocomb
    Jun 15, 2021 at 11:59
  • $\begingroup$ If you read the attached link you will find that I use the topological inverse image $i^{-1}(\mathcal{O}_{Spec(R)})$ to define a structure sheaf on $mSpec(R)$ - I explained this in the deleted post - Ask the moderators to undelete the post. $\endgroup$
    – hm2020
    Jun 15, 2021 at 12:02
  • $\begingroup$ I recieved the following message: "This post is hidden. It was deleted 1 hour ago by José Carlos Santos, Harish Chandra Rajpoot, amWhy." I do not understand why it was deleted. $\endgroup$
    – hm2020
    Jun 15, 2021 at 12:03

1 Answer 1

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The key idea is that the map $D(J)\mapsto D(J)_m$ from open subsets of $\operatorname{Spec} R$ to open subsets of $\operatorname{MaxSpec} R$ is a bijection between the lattices of open sets of $\operatorname{Spec} R$ and $\operatorname{MaxSpec} R$. Once you know this, it is easy to conclude that the assignment $D(J)_m \mapsto \Gamma(\mathcal{O}_{\operatorname{Spec} R},D(J))$ satisfies the conditions to be a sheaf because $\mathcal{O}_{\operatorname{Spec} R}$ does.

First, $\operatorname{MaxSpec} R$ embeds topologically as a subspace of $\operatorname{Spec} R$: this is essentially what (a) is asking you to show. This means that every open subset of $\operatorname{MaxSpec} R$ is obtained by intersecting an open subset of $\operatorname{Spec} R$ with $\operatorname{MaxSpec} R$, so the map $D(J)\mapsto D(J)_m$ is surjective.

Next, suppose $U_1$ and $U_2$ are two open subsets of $\operatorname{Spec} R$ with the same closed points. By replacing $U_1$ with $U_1\cap U_2$, we may assume $U_1\subset U_2$ and therefore that $U_2\setminus U_1=V(I)$ is a closed subset of $\operatorname{Spec} R$ with no closed points. But $V(I)$ (at least set-theoretically) is the spectrum of $R/I$, which is a finitely generated $k$-algebra, so by the Nullstellensatz if it is not the empty scheme it must have a closed point. Therefore $V(I)$ is empty and $U_1=U_2$, proving that $D(J)\mapsto D(J)_m$ is injective and thus a bijection.

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  • $\begingroup$ so you proved that we can define $\Gamma(D(J),\mathcal{O}_{Spec(R)})=\Gamma(D_m(J),\mathcal{O}_{MaxSpec(R)})$? $\endgroup$
    – idocomb
    Jun 18, 2021 at 9:04
  • $\begingroup$ I proved that $D(J)\mapsto D(J)_m$ is a bijection between the open subsets of the spectrum and the maximal spectrum. This implies that the assignment of $D(J)_m\mapsto\mathcal{O}_{\operatorname{Spec} R}(D(J))$ gives a sheaf. $\endgroup$
    – KReiser
    Jun 18, 2021 at 9:10
  • $\begingroup$ Apologies, most concepts are still new to me. How does this assignment give a sheaf? Isn't your assignment meant to be defining an element $\mathcal{O}_{MaxSpec(R)}(D(J)_m)=\mathcal{O}_{Spec(R)}(D(J))$? $\endgroup$
    – idocomb
    Jun 18, 2021 at 9:38
  • $\begingroup$ I'm guessing a little bit as to what you're saying there (you write "an element" but then follow it up with something not an element) but I am indeed defining $\mathcal{O}_{\operatorname{MaxSpec} R}(D(J)_m)$ as $\mathcal{O}_{\operatorname{Spec} R}(D(J))$. A sheaf is an assignment of an object to each open and a restriction map to each inclusion of opens subject to conditions, and frequently they're just specified in terms of the assignment of objects - this is what I was saying in the previous comment. $\endgroup$
    – KReiser
    Jun 18, 2021 at 9:48

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