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 $I = {\langle \partial\_1(F), \dots, \partial\_n(F), \partial\_t(F) \rangle}$ ?



Deformation of isolated singularities and non zero divisors
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.


