Deformation of isolated singularities and non zero divisors Consider $f \in \mathbb{C}\{x_1,\dots,x_n\}$ such that $(V(f),0)$ has an isolated singularity.
Let $F \in  \mathbb{C}\{x_1,\dots,x_n,t\}$ be a deformation of $f$ such that there exists some integer $m$ such that $(\partial_t(F))^m \in \langle \partial_1(F), \dots, \partial_n(F) \rangle$.
Can we conclude that $t$ is a non zero divisor of the quotient ring $\mathbb{C}\{x_1,\dots,x_n,t\}/I$, where $I = {\langle \partial_1(F), \dots, \partial_n(F), \partial_t(F) \rangle}$ ?
 A: I am just posting my comment as one answer.  No, you cannot conclude that the image of $t$ is a nonzerodivisor modulo the ideal $I$.
Let $\ell$ and $k$ be positive integers $\geq 2$ with $\ell \geq 2k$.  Consider the following polynomial,
$$ F=\ell(x_1^{\ell+k}+x_2^{\ell+k})-(\ell+k)t^2x_1^\ell x_2^\ell. $$  The partial derivatives are, $$\partial_1 F = \ell(\ell+k)(x_1^{\ell+k-1}-t^2x_1^{\ell-1}x_2^\ell), \ \ \partial_2 F = \ell(\ell+k)(x_2^{\ell+k-1}-t^2x_1^{\ell}x_2^{\ell-1}),$$ $$\partial_t F = -2(\ell+k)tx_1^\ell x_2^\ell.$$  Thus, modulo the ideal $J=\langle \partial_1 F, \partial_2 F \rangle$, we have the following congruences, $$x_1^{\ell+k-1} \equiv t^2x_1^{\ell-1}x_2^{\ell}, \ \ x_1^{2(\ell+k-1)} \equiv t^4 x_1^{2(\ell-1)}x_2^{2\ell}, \ \ x_2^{\ell + k-1} \equiv t^2 x_1^{\ell}x_2^{\ell-1}, \ \ x_2^{2\ell} \equiv t^2x_1^{\ell} x_2^{2\ell-k}.$$  Taken together, this gives  $$ x_1^{2\ell+2k-2} \equiv t^6 x_1^{3\ell-2}x_2^{2\ell-k}, \ \text{i.e.,}\ \ x_1^{2\ell+2k-2}(1-t^6x_1^{\ell-2k}x_2^{2\ell-k})\in J.$$  Since the second factor is invertible in $\mathbb{C}\{x_1,x_2,t\}$, this gives that $x_1^{2(\ell+k-1)}\in J$.  By symmetry, also $x_2^{2(\ell+k-1)}\in J$.  Therefore, also $(\partial_t F)^3$ is in $J$.
However, $x_1^\ell x_2^\ell$ is not congruent to $0$ modulo $I$.  Indeed, the quotient of $I$ by the ideal generated by the image of $t$ equals $$\mathbb{C}\{x_1,x_2,t\}/\langle t,x_1^{\ell+k-1},x_2^{\ell+k-1} \rangle.$$  Since $k\geq 2$, also $k-1\geq 1$, so that $x_1^\ell x_2^\ell$ is nonzero in this quotient ring.  Since $tx_1^\ell x_2^\ell$ is in $I$, the image of $t$ modulo $I$ is a zerodivisor.
