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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 $I = {\langle \partial_1(F), \dots, \partial_n(F), \partial_t(F) \rangle}$ ?

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    $\begingroup$ What does the notation $\mathbb C\{x_1,\ldots,x_n\}$ mean? Formal power series which has a positive convergence radius? And what is "a deformation"? Does that mean that $F\vert_{t=0}=f$? $\endgroup$
    – Z. M
    Commented Apr 27, 2022 at 16:20
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    $\begingroup$ @Z.M $\mathbb{C}\{x_1, \dots, x_n\}$ is the ring of germs (at the origin) of analytic maps in $n-$variables. $F$ is a deformation of $f$ if $F\_{t=0}=f$ (as you said). $\endgroup$
    – Serge the Toaster
    Commented Apr 28, 2022 at 6:08
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    $\begingroup$ In that case, what about $F=3(x_1^5+x_2^5)-5t^2x_1^3x_2^3$? Modulo $J=\langle \partial_1(F),\partial_2(F) \rangle$, we have $x_1^4\equiv t^2x_1^2x_2^3$, so that $x_1^8\equiv t^4x_1^4x_2^6$. We also have $x_2^4 \equiv t^2 x_1^3x_2^2$, so that $x_2^6\equiv t^2x_1^3x_2^4 \equiv t^4x_1^6x_2^2$. Thus, $x_1^8\equiv t^4 x_1^4 x_2^6 \equiv t^8 x_1^{10} x_2^2$, i.e., $x_1^8(1-t^8x_1^2x_2^2) \equiv 0$. Thus, $x_1^8\equiv 0$. Similarly, $x_2^8\equiv 0$. So $\partial_t(F) = -10tx_1^3x_2^3$ is nilpotent modulo $J$. Yet $x_1^3x_2^3$ is not zero modulo $I$. So $t$ is a zero divisor modulo $I$. $\endgroup$
    – Jason Starr
    Commented Apr 28, 2022 at 10:33

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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.

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  • $\begingroup$ Good example! But what about the case where where the degree is preserved? That is, if $f$ is homogeneous of degree $p$ and for every $t_0$ we have that $F_{t_0}$ is homogeneous of degree $p$ as well? $\endgroup$
    – Serge the Toaster
    Commented Apr 28, 2022 at 13:11
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    $\begingroup$ I do not understand this comment. I thought that you clarified that you were working with germs of holomorphic functions, not with polynomials. Obviously if you change the statement of a problem after it has been answered, then you can "invalidate" the answer. If you want to ask a second question, I suggest that you make that a second post. $\endgroup$
    – Jason Starr
    Commented Apr 28, 2022 at 13:16

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