Affine schemes and rings: equivalence of categories

DEFINITION 10.1   Let $A,B$ be rings. Let $\varphi:A \to B$ be a ring homomorphism. We have already introduced $\operatorname{Spec}(\varphi)$ as a continuous map $Y=\operatorname{Spec}(B)\to \operatorname{Spec}(A)=X$. Now that the spaces $\operatorname{Spec}(A),\operatorname{Spec}(B)$ carry structures of locally ringed spaces, we (re)define $\operatorname{Spec}(\varphi)$ as a morphism of locally ringed spaces by defining $\operatorname{Spec}(\varphi)^\char93$ as in Example 9.11.

LEMMA 10.2   $\operatorname{Spec}(\varphi)$ is indeed a morphism of locally ringed space.

THEOREM 10.3   Let $(f,f^\char93 ):\operatorname{Spec}(B)\to \operatorname{Spec}(A)$ be a morphism of locally ringed space. Then there exists an unique ring homomorphism $\varphi:A \to B$ such that $f$ coincides with $\operatorname{Spec}(\varphi)$.

PROOF.. Let us put $Y=\operatorname{Spec}(B)$ and $X=\operatorname{Spec}(A)$. The data

$$f^\char93 : f^{-1}\mathcal{O}_X \to \mathcal{O}_Y$$

is equivalent to a data

$$f_\char93 : \mathcal{O}_X \to f_* \mathcal{O}_Y$$

which gives rise to a ring homomorphism

$$f_\char93 (X): A=\mathcal{O}_X(X) \to (f_* \mathcal{O}_Y)(X)=B.$$

Let us take this homomorphism as $\varphi$.

$$\begin{CD} A_{f(y)}=\mathcal{O}_{X,f(y)} @>{f^\char93 _{y}} >> \m... ...name{restr}_y} AA \\ \mathcal{O}_X(X) @>{\varphi} >> \mathcal{O}_Y(Y) \end{CD}$$

By the hypothesis of $f$ being a morphism of locally ringed spaces, $f^\char93 _{y}$ is local homomorphism. That means,

$$(f^{\char93 })^{-1} (y B_y)=f(y) A_{f(y)}.$$

$$\varphi^{-1}(y)=f(y).$$

From the definition of $\operatorname{Spec}(\varphi)$, we have

$$\operatorname{Spec}(\varphi)(y)=f(y).$$

We have thus proved that $\operatorname{Spec}(\varphi)$ is equal to $f$ as a map $Y\to X$.

$\qedsymbol$