If $X=\operatorname{Spec} A$, then $H^0(X,\mathcal O_X)=\mathcal O_X(X)=A$ is a ring.
(1) For a general scheme $X$, does the space $H^0(X,\mathcal O_X)$ of global sections of structure sheaf still form a ring? If so, what is $\operatorname{Spec} (H^0(X,\mathcal O_X))$, and how is it related to $X$? Any reference?
(2) Suppose we have several global sections $s_1,\dots, s_m\in H^0(X,\mathcal O_X)$, and consider the ideal $\mathcal I=(s_1,\dots, s_m)$ generated by them. Then, how do we understand $\operatorname{Spec} (H^0(X,\mathcal O_X) /\mathcal I)$? Is it just the base locus of these sections? (If $X=\operatorname{Spec} A$, this is clear. But what if $X$ is a general scheme.)
Best Answer
Yes, $\mathcal{O}_X(X)$ is a ring, because $\mathcal{O}_X$ is a sheaf of rings. The scheme $\operatorname{Spec} \mathcal{O}_X(X)$ is called the affinization of $X$, and it's the universal affine scheme to which $X$ maps (this follows from the adjunction between global sections and Spec - see here for more details).
It's the base locus of those sections on the affinization. The preimage of this under the affinization map is the base locus of the sections $s_i$ on $X$.