Anti-equivalence of affine schemes and rings

Remarks

Part 1 is a gluing argument: a global section is given locally by fractions $a_i / f_i^{n_i}$ on basic open sets $D(f_i)$ that cover $\operatorname{Spec} A$. Since the $f_i$ then generate the unit ideal, these fractions glue to a single element of $A$. For part 2, the stalk maps of a morphism must be local. This forces $f(\mathfrak{q})$ to be the preimage of $\mathfrak{q}$ under the map on global sections. Then the sheaf map is determined on each $D(g)$ by localizing. Example: for a field $k$, the morphisms $\operatorname{Spec} k \to \operatorname{Spec} k[x]$ over $k$ correspond to $k$-algebra maps $k[x] \to k$, that is, to the elements $a \in k$, the $k$-points of the line. The locality condition is needed. Let $R$ be a discrete valuation ring with fraction field $K$. Send the point of $\operatorname{Spec} K$ to the closed point of $\operatorname{Spec} R$, with the inclusion $R \to K$ on sections. This is a morphism of ringed spaces, but it is not the morphism induced by $R \to K$, which hits the generic point.

Going deeper

For a student

This is the reason the spectrum of a ring is the right local model for algebraic geometry. It extends to all schemes: for any scheme $X$, morphisms $X \to \operatorname{Spec} A$ correspond to ring maps $A \to \mathcal{O}_X(X)$. It is used constantly. For example, it shows that the fiber product of affine schemes is the spectrum of a tensor product. It also lets one describe a scheme through its functor of points on rings.