Timeline for Kähler differentials on an Artinian local ring
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2021 at 4:35 | vote | accept | jacob | ||
| Jan 19, 2021 at 2:17 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Jan 18, 2021 at 23:32 | answer | added | Mohan | timeline score: 2 | |
| Jan 8, 2021 at 2:50 | comment | added | jacob | @Mohan Thanks! That makes a lot of sense. If you wanna post as an answer I'd gladly accept it :) | |
| Jan 6, 2021 at 16:12 | comment | added | Mohan | @abx No, it is not necessary , just mentioned it . | |
| Jan 6, 2021 at 15:24 | comment | added | abx | @Mohan: Thank you! By the way, I don't see why you need this. It seems to me that any $f\notin(f'_x,f'_y)$ will do the job. | |
| Jan 6, 2021 at 15:02 | comment | added | Mohan | @abx It is called Briancon-Skoda theorem and it is more general. For $n$ variable power series $f^n$ belongs to the ideal of partial derivatives . | |
| Jan 6, 2021 at 7:43 | comment | added | abx | @Mohan: Do you have a reference for the "well known theorem"? That doesn't seems obvious to me. | |
| Jan 5, 2021 at 17:46 | comment | added | Mohan | These examples exist, though I have forgotten the author of the first one. If $f\in k[[x,y]]$ in the maximal ideal, it is a well known theorem that $f^2\in (f_x,f_y)$. An example can be constructed such that $f$ itself is not in this ideal. Further, we can find an $f$ such that $f_x,f_y$ form a regular sequence . Then take $A=k[[x,y]]/(f_x,f_y)$. So, $f\neq 0\in A$, but $df=0$. | |
| Jan 5, 2021 at 12:06 | comment | added | user130903 | Something like this could work: 1. If $df=0$ in $\Omega_{K/k}$ for a field extension $K$, then $f\in k$, as you have a transcendence basis. 2. Using the exact sequence of $k\to R\to R/M$, where $M$ is the maximal idea, you see that if $df=0$, then $f\in k +M$. Subtracting the $k$-part, you might assume $f\in M$. 3. Replace $R$ by $k+M$, then the differential module is isomorphic to $M/M^2$, by which you conclude that $f\in M^2$. Then your replace $R$ my $k+M^2$ and you iterate this to find $f\in M^k$ for every $k$. For $k$ large, you get $f=0$. | |
| Jan 5, 2021 at 8:51 | comment | added | jacob | @Zero, I don't think thats a counterexample, cause dx doesn't vanish. | |
| Jan 5, 2021 at 8:50 | comment | added | user130903 | Let $k=\mathbb C$ and $R=k[X]/X^2$. It becomes true, if you insist that $R$ is an integral domain. | |
| Jan 5, 2021 at 8:36 | comment | added | jacob | Thanks, abx! Forgot to include that. | |
| Jan 5, 2021 at 8:36 | history | edited | jacob | CC BY-SA 4.0 |
added 22 characters in body
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| Jan 5, 2021 at 8:30 | comment | added | abx | No. Take for $R$ is any finite extension of $k$; then $\Omega _{R/k}=0$. | |
| Jan 5, 2021 at 8:07 | history | edited | jacob | CC BY-SA 4.0 |
edited body; edited title
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| Jan 5, 2021 at 7:56 | history | asked | jacob | CC BY-SA 4.0 |