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Let $f\in \mathbb{C}[x_1, \dots,x_n]$ be a homogeneous polynomial of degree $p$ and let $F \in \mathbb{C}[x_1, \dots,x_n,t]$ be a polynomial such that $F(x_1, \dots, x_n,0)=f$, for every $t_0$ we have that $F(x_1, \dots,x_n,t_0)$ is a homogeneous polynomial of degree $p$, and $\partial_t(F) \in \sqrt{\langle \partial_1(F), \dots, \partial_n(F) \rangle} \subset \mathbb{C}[x_1, \dots,x_n,t]$.

Can we conclude that $t$ is a non zero divisor of the quotient ring $\mathbb{C}[x_1,\dots,x_n,t]/I$, where $I = {\langle \partial_1(F), \dots, \partial_n(F), \partial_t(F) \rangle}$ ?

Note that this question is related to the question posed in Deformation of isolated singularities and non zero divisors .

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  • $\begingroup$ If $f$ has an isolated singularity at the origin, then this corresponds to a smooth hypersurface in projective space. Then your hypothesis involving $m$ implies that the corresponding one-parameter family of hypersurfaces is constant in moduli. This then implies that, up to a change of homogeneous coordinates to $y_i=g_i(x_1,\dots,x_n,t)$, the polynomial $F$ is $H(y_1,\dots,y_n,t) = h(y_1,\dots ,y_n)$. Then the quotient ring is $\mathbb{C}[t]\otimes_{\mathbb{C}}(\mathbb{C}[y_1,\dots,y_n]/\text{Jacobian}(h))$. In this ring, the image of $t$ is a nonzerodivisor. $\endgroup$
    – Jason Starr
    Commented Apr 28, 2022 at 13:55
  • $\begingroup$ @JasonStarr why does this condition imply that the corresponding one-parameter family of hypersurfaces is constant in moduli? $\endgroup$
    – Serge the Toaster
    Commented Apr 28, 2022 at 17:02
  • $\begingroup$ I recommend that you look this up in a book on parameter spaces and / or deformation theory, e.g., p. 13 of "Lectures on singularities" by Michael Artin: math.tifr.res.in/~publ/ln/tifr54.pdf $\endgroup$
    – Jason Starr
    Commented Apr 28, 2022 at 22:32
  • $\begingroup$ @JasonStarr Many thanks for your recommendation. Yet, I think there might be a problem, since if $p \geq 4$, then for a generic $g$ of homogeneous degree $p$, the deformation $f+tg$ will not be constant in moduli, and the moduli can be changed by changing $g$ accordingly. $\endgroup$
    – Serge the Toaster
    Commented May 7, 2022 at 12:27
  • $\begingroup$ The choice of $g$ is not generic. If you choose $g$ so that the polynomial $F=f+tg$ satisfies your condition that $\partial_t F$ is contained in the ideal generated by the partials $\partial_i F$, then this imposes strong constraints on $g$. $\endgroup$
    – Jason Starr
    Commented May 7, 2022 at 12:59

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