Suppose $R$ is a commutative Artinian local ring over an algebraically closed characteristic 0 field $k$. Suppose $f\in R$ is such that $df=0$ (in the sense that the element $df$ vanishes in the module of Kähler differentials). Is $f$ necessarily in $k$?
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3$\begingroup$ No. Take for $R$ is any finite extension of $k$; then $\Omega _{R/k}=0$. $\endgroup$– abxCommented Jan 5, 2021 at 8:30
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1$\begingroup$ Thanks, abx! Forgot to include that. $\endgroup$– jacobCommented Jan 5, 2021 at 8:36
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1$\begingroup$ @Zero, I don't think thats a counterexample, cause dx doesn't vanish. $\endgroup$– jacobCommented Jan 5, 2021 at 8:51
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1$\begingroup$ 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 $df=0$. $\endgroup$– MohanCommented Jan 5, 2021 at 17:46
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1$\begingroup$ @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. $\endgroup$– abxCommented Jan 6, 2021 at 15:24
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1 Answer
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I finally remembered the example (though not the reference). Take $f=x^2y^2+x^5+y^5\in R=\mathbb{C}[[x,y]]$. Then $f_x,f_y$ form a regular sequence in $R$ and thus $R/I$ where $I=(f_x,f_y)$ is an Artin local ring. One checks $f\not\in I$. Thus, $df=0\in\Omega^1_{R/I}$, but $f\neq 0$ in $R/I$.
