Can we define $\operatorname{Spec}(H^0(X, \mathcal O_X))$ for a scheme $X$?

Question

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.)

Answer 1

1. 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).

2. 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 $X$.