Assume $f\in k[x_1,\ldots, x_n]$ is irreducible. Let for $g\in k[x_1,\ldots, x_n]$, $\partial(g)$ is divisible by $f$ for each derivation $\partial$ with $f\in\ker\partial$. Is it true that $g-t$ is divisible by $f$ for some $t\in k$?
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$\begingroup$ Your definition of slice does not refer to the derivation $D$. $\endgroup$– LSpiceCommented Aug 16, 2020 at 19:05
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$\begingroup$ Why? Slice is $f$ with $D(f) = 1$. $\endgroup$– A.SkutinCommented Aug 16, 2020 at 19:14
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$\begingroup$ OK; I don't know the definition of a slice, so it is what you say it is. You said "being a slice is equivalent to $(f'_{x_1}, \dotsc, f'_{x_n}) = (1)$", where I guess $f'_x$ means $\partial_x f$, and that doesn't mention $D$. The conclusions of your Question and Question' also don't seem to depend on $D$, but maybe that is intentional. $\endgroup$– LSpiceCommented Aug 16, 2020 at 20:22
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$\begingroup$ By a slice I mean $f$ for which exists $D$. And if the ideal is $1$ then such $D$ can be constructed. $\endgroup$– A.SkutinCommented Aug 17, 2020 at 5:08
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Consider the map $g:X\to\mathbb{A}^1$, where $X$ is defined by $f=0$ in $\mathbb{A}^n$. Your condition implies $dg=0$ and thus this map must be constant. This is what you wanted to prove.
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$\begingroup$ Is there a reference of the proof that if $dg = 0$ on the variety $X$ then $g$ is constant on $X$? $\endgroup$– A.SkutinCommented Aug 17, 2020 at 5:21
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$\begingroup$ Did you use that $(f'_{x_1},\ldots, f'_{x_n}) = (1)$ in the proof? $\endgroup$– A.SkutinCommented Aug 17, 2020 at 6:16
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$\begingroup$ @Algorithm Actually $dg=0$ implies $g$ is locally constant (just use the definition). So, my proof gives what you want only if $X$ is irreducible. I don't know a single slice which is not irreducible . $\endgroup$– MohanCommented Aug 17, 2020 at 12:41
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$\begingroup$ $X$ is irreducible because $f$ is irreducible. Are you sure that $dg = 0$ is equivalent to $f|D(g)$ for each $D$ with $0=D(f)$, or it is just equivalent to $f|D(g)$ for each derivation $D$? $\endgroup$– A.SkutinCommented Aug 17, 2020 at 12:45
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2$\begingroup$ @Algorithm I do not understand what you mean. For a smooth irreducible variety over complex numbers, the kernel of the map $d:O_X\to \Omega^1_X$ is just the constants. $\endgroup$– MohanCommented Aug 17, 2020 at 13:44