# Defining the sheaf of the maximal spectrum $\mathcal{O}_{Spm(R)}$

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 $$R$$?

• read the linked post. If you have more questions - please post a question and I will explain the definition. Jun 15, 2021 at 11:49
• 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/… Jun 15, 2021 at 11:54
• @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 Jun 15, 2021 at 11:59
• 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. Jun 15, 2021 at 12:02
• 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. Jun 15, 2021 at 12:03

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.
• so you proved that we can define $\Gamma(D(J),\mathcal{O}_{Spec(R)})=\Gamma(D_m(J),\mathcal{O}_{MaxSpec(R)})$? Jun 18, 2021 at 9:04
• 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. Jun 18, 2021 at 9:10
• 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 $\mathcal{O}_{MaxSpec(R)}(D(J)_m)=\mathcal{O}_{Spec(R)}(D(J))$? Jun 18, 2021 at 9:38
• 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. Jun 18, 2021 at 9:48