Giving morphism of schemes $\pi$ : $\operatorname{Spec}A \rightarrow \operatorname{Spec}B$, by definition we have $\pi^{-1}\mathscr{O}_{\operatorname{Spec}B}$ is a sheaf of rings on $\operatorname{Spec}A$, i.e., $(\operatorname{Spec}A, \pi^{-1}\mathscr{O}_{\operatorname{Spec}B}$) is a ringed space. With $\pi$ being a mophism of schemes rather than just of ringed spaces, $(\operatorname{Spec}A, \pi^{-1}\mathscr{O}_{\operatorname{Spec}B}$) is also an affine scheme, by argument similar in For a morphism of affine schemes, the inverse of an open affine subscheme is affine. Then it shoud have form $(\operatorname{Spec}C, \mathscr{O}_{\operatorname{Spec}C}$)

Algebraically, which ring is the ring $C$ under the morphism $B \rightarrow A$ ? Considering an example like $\operatorname{Spec}k[x]/(x) \to \operatorname{Spec}k[x]$, I suspect that $C$ is the localization of $B$ with multiplicative set $\pi^{-1}(U(A))$, where $U(A)$ means the units of $A$. But I failed to give a proof of it.

And is there any geometric view point for it?

Thank you in advance.


1 Answer 1


Let $X$ be a scheme and $x \in X$ any point. Then we have a morphism of schemes $\pi:\mathrm{Spec}(\kappa(x)) \to X$, where $\kappa(x)$ is the residue field at $x$. Now consider the ringed space $(\mathrm{Spec}(\kappa(x)),\pi^{-1} \mathscr O_X)$. What is the structure sheaf? Our underlying space has only one element, so we just need to compute global sections. We get that $\pi^{-1}\mathscr O_X(\mathrm{Spec}(\kappa(x)))=\varinjlim\limits_{U \subset X, x \in U}\mathscr O_X(U)=\mathscr O_{X,x}$. If $\mathscr{O}_{X,x}$ has more than one prime ideal, this ringed space is not a scheme.

You're misunderstanding the statement of the linked question: for a morphism of affine schemes $f:X \to Y$ and an affine open $U \subset Y$, we get that $(f^{-1}(U),\mathscr O_X|_{f^{-1}(U)})$ is an affine scheme. This is what is meant by saying that "the inverse of an open affine subscheme is affine".


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