The Stacks project

Remark 10.127.12. Suppose that $R \to S$ is a local homomorphism of local rings, which is essentially of finite presentation. Take any system $(\Lambda , \leq )$, $R_\lambda \to S_\lambda $ with the properties listed in Lemma 10.127.10. What may happen is that this is the “wrong” system, namely, it may happen that property (4) of Lemma 10.127.11 is not satisfied. Here is an example. Let $k$ be a field. Consider the ring

\[ R = k[[z, y_1, y_2, \ldots ]]/(y_ i^2 - zy_{i + 1}). \]

Set $S = R/zR$. As system take $\Lambda = \mathbf{N}$ and $R_ n = k[[z, y_1, \ldots , y_ n]]/(\{ y_ i^2 - zy_{i + 1}\} _{i \leq n-1})$ and $S_ n = R_ n/(z, y_ n^2)$. All the maps $S_ n \otimes _{R_ n} R_{n + 1} \to S_{n + 1}$ are not localizations (i.e., isomorphisms in this case) since $1 \otimes y_{n + 1}^2$ maps to zero. If we take instead $S_ n' = R_ n/zR_ n$ then the maps $S'_ n \otimes _{R_ n} R_{n + 1} \to S'_{n + 1}$ are isomorphisms. The moral of this remark is that we do have to be a little careful in choosing the systems.


Comments (2)

Comment #9812 by Branislav Sobot on

I understand what you want to illustrate here, but your rings are not essentially of finite type over (unless maybe )?

Comment #9813 by Branislav Sobot on

Sorry, the first formula in the previous comment is , and the second one is


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Your email address will not be published.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 00QW. Beware of the difference between the letter 'O' and the digit '0'.