Timeline for Is the following local map unramified?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2021 at 2:11 | comment | added | user237522 | With the help of the following resources, things are more clear for me now, so there is no need to further discuss my above question, unless you wish to add something, then you are welcome; thank you for your help! stacks.math.columbia.edu/tag/024L jstor.org/stable/2372926 | |
| Apr 22, 2021 at 18:16 | comment | added | user237522 | Anyway, I like your last idea, that each of $R$ and $S$ is essentially of finite type over $k$ (our perfect base field) which means that each of $R$ and $S$ is a localization at a prime of an affine $k$-algebra. Now, assuming that $R$ and $S$ are essentially of finite type over $k$, please, could you prove that $R \subseteq S$ is unramified? This sounds a nice result (without assumung that $S$ is a finitely generated $R$-module!). | |
| Apr 22, 2021 at 18:09 | comment | added | user237522 | Condition (3) says that $S$ is the localization at a prime of a finitely generated $R$-algebra, and I had this condition in mind, but have not written it explicitly, so I apologize for not being clear. Summarizing, it remains to show that the residue field extension is finite dimensional. Now, if I am not wrong, finite dimensionality of that field extension will follow if we further assume that $S$ is finitely generated as an $R$-module. | |
| Apr 22, 2021 at 18:09 | comment | added | user237522 | @R.vanDobbendeBruyn, thank you very much for your comment. Please, let me explain: I guess we are all talking about the following definition: stacks.math.columbia.edu/tag/024L which includes 3 conditions: Condition (1) is assumed to hold. Condition (2) says that the residue field extension is finite and separable. In the edit I have assumed that the base field is perfect, hence it remains to prove that the field extension is finite. | |
| Apr 22, 2021 at 15:18 | comment | added | R. van Dobben de Bruyn | I gave you a counterexample to the claim in the second paragraph. Note that the linked question was talking about characteristic $0$, where all field extensions are separable. If $R$ and $S$ are essentially of finite type over a perfect field, there is still a chance, but you did not put this assumption anywhere. | |
| Apr 22, 2021 at 6:13 | comment | added | user237522 | @R.vanDobbendeBruyn, please, if the fields $R$ and $S$ are perfect and $R \subseteq S$ is algebraic, then the extension is separable (hence, since $0S=0$, the extension is unramified) or am I still missing something? Also, is the original claim (= first two lines in the question) true? It is assumed that $S$ is finitely generated as an $R$-algebra. (Is this enough or should $S$ be finitely generated as an $R$-module?). | |
| Apr 22, 2021 at 5:32 | comment | added | user237522 | Thank you very much, R. van Dobben de Bruyn and David Benjamin. | |
| Apr 22, 2021 at 4:22 | review | Low quality posts | |||
| Apr 22, 2021 at 5:07 | |||||
| Apr 22, 2021 at 3:07 | comment | added | David Benjamin Lim | Take an imperfect field $k$, $a \in k - k^p$. Then $k \to k(a^{1/p})$ satisfies all the assumptions in your question but is not separable. | |
| Apr 22, 2021 at 1:54 | comment | added | R. van Dobben de Bruyn | This is not quite right; think about the case where $R$ and $S$ are fields. | |
| Apr 21, 2021 at 23:52 | history | edited | user237522 | CC BY-SA 4.0 |
added 12 characters in body
|
| Apr 21, 2021 at 23:39 | history | edited | user237522 | CC BY-SA 4.0 |
added 12 characters in body
|
| Apr 21, 2021 at 23:13 | history | edited | user237522 | CC BY-SA 4.0 |
added 215 characters in body
|
| Apr 21, 2021 at 22:49 | history | edited | user237522 | CC BY-SA 4.0 |
added 219 characters in body
|
| Apr 21, 2021 at 21:45 | history | edited | user237522 | CC BY-SA 4.0 |
added 10 characters in body
|
| Apr 21, 2021 at 21:39 | history | asked | user237522 | CC BY-SA 4.0 |