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.

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01RQ. Beware of the difference between the letter 'O' and the digit '0'.