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Let an irreducible, square-free complex polynomial $f\in \mathbb C[x,y]$ be given. It is well known that the curve $\mathcal C:=\{f=0\}$ is nonsingular if and only if $\mathbb C[x,y]=<f,\partial_xf,\partial_yf>$ as ideals. I'd like to know whether this criterion is effective, in the following sense.

Consider the linear mapping $$\phi : (A,B,C)\in\mathbb C[x,y]^3\longmapsto Af+B\partial_xf+C\partial_yf$$ The curve $\mathcal C$ is nonsingular if and only if $\phi$ is surjective. The question is the following:

Assume $\mathcal C$ nonsingular. Is there a computable integer $k\in\mathbb Z_{>0}$ such that the restricted operator $\phi|_{\mathbb C[x,y]_{\leq k}\times C[x,y]_{\leq k+1}^2}$ has maximal rank (i.e. $\dim_\mathbb C \mathbb C[x,y]_{\leq k+\deg f}$) ?

(Here $\mathbb C[x,y]_{\leq k}$ is the vector space of polynomials of degree $\leq k$.) I devised an argument based on Noetheriality that shows such a $𝑘$ (maybe not computable) always exists, but I'd like to know a bound on its value.

This question seems natural, hence it is very likely that it admits a (positive or negative) answer in the literature and I'm not aware of it (not being an algebraist myself). Any pointer would be much appreciated (I'm aware of Walcher's "Plane polynomial vector fields with prescribed invariant curves" but that doesn't quite answer the question).

The question popped up while looking for an effective way of computing a basis of (the projection on the last $2$ factors of) $\ker \phi$, which exists by a theorem of Saito (it is a free $\mathbb C[x,y]$-module of rank $2$). More precisely I'm looking for an apriori bound on the degree of such a basis (including the case where $\mathcal C$ can be singular, but the nonsingular case is a first step).

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  • $\begingroup$ Hmm. Are you sure about that? I think $C$ is nonsingular if and only if $\mathbb{C}[x,y] = \langle f, f_x,f_y \rangle$ - with $f$ included; not just $\langle f_x,f_y \rangle$. For example take $f = x^2+y^2-1$. It is nonsingular but $\langle f_x,f_y\rangle = \langle 2x,2y\rangle$, vanishing at the origin. $\endgroup$
    – Zach Teitler
    Commented Apr 15, 2023 at 7:44
  • $\begingroup$ @Zach: you're right, it was in the first version of the post, then I edited it and removed that part (don't know why really). Corrected $\endgroup$
    – Loïc Teyssier
    Commented Apr 15, 2023 at 12:07
  • $\begingroup$ Maybe change your notation. It's confusing to use $C$ for two different things, as the name of the curve, and in $(A,B,C)$, the variables in your function $\phi$. $\endgroup$
    – Joe Silverman
    Commented Apr 15, 2023 at 19:03
  • $\begingroup$ You might like to look into Castelnuovo-Mumford regularity; for example, section 1.8 of Lazarsfeld's book Positivity in Algebraic Geometry. If a sheaf $F$ on projective space $\mathbb{P}$ is $m$-regular, then the multiplication maps $H^0(F(m)) \otimes H^0(O_{\mathbb{P}}(k)) \to H^0(F(m+k))$ are surjective for all $k \geq 0$. I'm don't know if it helps; even if you homogenize to get a curve in $\mathbb{P}^2$ instead of $\mathbb{C}^2$, I'm not sure if this surjectivity corresponds to the one you're asking for. But perhaps it's worth a look. $\endgroup$
    – Zach Teitler
    Commented Apr 16, 2023 at 2:31

1 Answer 1

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I believe such bound exist. Perhaps you can deduce it from effective versions of Hilbert's Nullstellensatz. Take a look at Kollár's paper "Sharp Effective Nullstellensatz". The wikipedia page on Hilbert's Nullstellensatz has a quick discussion about it.

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  • $\begingroup$ Thanks a lot Jorge, I'll have a look at all this. $\endgroup$
    – Loïc Teyssier
    Commented Apr 15, 2023 at 20:52
  • $\begingroup$ You are welcome. Hope this help, even if I am not sure it is actually relevant for your question. At a first quick reading, I though that you were looking for bounds like the one in effective Nullstelensatz. Sorry about that. Anyway, you may also want to take a look in a recent preprint by Camacho and Movasati where they address the problem of generation of the module of vector fields tangent to a mildly singular curve algorithmically. $\endgroup$
    – Jorge Vitório Pereira
    Commented Apr 15, 2023 at 22:27
  • $\begingroup$ Your mention of the Nullstellensatz is spot-on, since in the case of a regular curve $f$ and its partial derivatives have no common zero, hence the Nullstellensatz provides polynomials $A,B,C$ such that $\phi(A,B,C)=1$, at which point $\phi$ has maximal rank (modulo a shift on the degree of its arguments, but that's probably ok for my purposes). The effective version you point to in your answer provides a sharp a priori bound on the degree of $A,B,C$. $\endgroup$
    – Loïc Teyssier
    Commented Apr 17, 2023 at 12:07
  • $\begingroup$ Let me reflect a little bit more about this, and I'll accept your answer. $\endgroup$
    – Loïc Teyssier
    Commented Apr 17, 2023 at 12:08

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