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

Question

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 $R$?

Answer 1

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.