Suppose $\varphi: B\rightarrow A$ is morphism of rings. This induces a morphism of ringed spaces as follows: We get a continuous map of topological spaces $\pi:\operatorname{Spec}A\rightarrow\operatorname{Spec}B$ given by $P\mapsto \varphi^{-1}(P)$. And to describe a morphism of sheaves $\mathcal{O}_{\operatorname{Spec}B}\rightarrow \pi_*\mathcal{O}_{\operatorname{Spec}A}$ on $\operatorname{Spec}B$ it suffices to describe a morphism of sheaves on the distinguished base. So on $D(g)\subset \operatorname{Spec}B$, we define $\mathcal{O}_{\operatorname{Spec}B}(D(g))\rightarrow \mathcal{O}_{\operatorname{Spec}A}(\pi^{-1}(D(g))=\mathcal{O}_{\operatorname{Spec}A}(D(\varphi(g)))$ by $B_g\rightarrow A_{\varphi(g)}$.
My question is why does this map actually give us a morphism of locally ringed spaces? That is, why does the induced map on stalks, $\pi^{\#}:\mathcal{O}_{{\operatorname{Spec}B},q}\rightarrow\mathcal{O}_{{\operatorname{Spec}A},p}$, map the maximal ideal of the first stalk to the maximal ideal of the second stalk?
My idea is that we start with an element of the first stalk, say $[(s,D(g))]$ and assume that $s$ vanishes at $q$. Now, since $s\in\mathcal{O}_{\operatorname{Spec}B}(D(g))$, we have that $s\in B_g$ and so $s=b/g^n$ forsome $b\in B$ and integer $n$. Then the induced stalk map maps $[(s,D(g))]=[(b/g^n,D(g))]$ to $[\varphi(b)/\varphi(g^n),D(\varphi(g))]$.
Now, it seems the result would follow if a) the reasoning above is correct, and b) $\varphi(b)/\varphi(g^n)$ vanishes at $p$.
Is this right?
Best Answer
Yes, this is one correct approach. Note that because you assumed that $\varphi^{-1}(p)=q$, this means that as $s\in q$, we have that $\varphi(s)\in p$ and thus $\varphi(s)/\varphi(g^n)\in p_p$.
It may also be helpful to note that the first part of this setup can be done directly from the fact that $\mathcal{O}_{\operatorname{Spec} A,p}=A_p$ without doing something like picking a representative $[(s,D(g))]$. The map $B\to A$, when composed with $A\to A_p$, gives an induced map $B_q\to A_p$ from the universal property of localization, and you can do the same game by writing your element as a fraction from here.