The Stacks project

Example 29.50.4. Here are two examples of birational morphisms which are not isomorphisms over any open of the target.

First example. Let $k$ be an infinite field. Let $A = k[x]$. Let $B = k[x, \{ y_{\alpha }\} _{\alpha \in k}]/ ((x-\alpha )y_\alpha , y_\alpha y_\beta )$. There is an inclusion $A \subset B$ and a retraction $B \to A$ setting all $y_\alpha$ equal to zero. Both the morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$ and the morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ are birational but not an isomorphism over any open.

Second example. Let $A$ be a domain. Let $S \subset A$ be a multiplicative subset not containing $0$. With $B = S^{-1}A$ the morphism $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is birational. If there exists an open $U$ of $\mathop{\mathrm{Spec}}(A)$ such that $f^{-1}(U) \to U$ is an isomorphism, then there exists an $a \in A$ such that each every element of $S$ becomes invertible in the principal localization $A_ a$. Taking $A = \mathbf{Z}$ and $S$ the set of odd integers give a counter example.