Suppose $p(x_1,...,x_n)$ is a polynomial in $n$ real variables whose Hessian is not identically zero. Can we say that there is a set $Z \subset {\mathbb R}^n$ of measure zero and a constant $M$ such that for all real numbers $c_1,...,c_n$ the cardinality of the set of $(x_1,...,x_n) \in {\mathbb R}^n - Z$ for which ${\partial p \over \partial x_i}(x_1,...,x_n) = c_i$ for each $i$ is at most $M$? It's possible I might be able to exclude some $(c_1,...,c_n)$ but I'd have be careful what I was excluding.

I've looked around online at resources on systems of polynomial equations and nothing really applied. Also, the above might generalize to systems $p_i(x_1,...,x_n) = c_i$ for $(p_1,...,p_n)$ whose Jacobian is not identically zero, but I only need it in the above situation.

Basically, if this follows from something well-known, or someone has pointers on where to look I'd be appreciative.

  • $\begingroup$ What role does $Z$ play (it is mentioned only once)? $\endgroup$ Commented Jun 15, 2022 at 15:31
  • $\begingroup$ I think they mean "for all $(c_i) \notin Z$" $\endgroup$ Commented Jun 15, 2022 at 15:45
  • $\begingroup$ I mean the set of $(x_1,...,x_n)$ outside of $Z$, and I've edited the question. I'd prefer it to hold for each $(c_1,...,c_n) \in {\mathbb R}^n$, but if this isn't possible a carefully chosen excluded set of $(c_1,...,c_n)$ might also serve my purposes. $\endgroup$
    – Zarrax
    Commented Jun 15, 2022 at 15:54

1 Answer 1


This follows from basic facts about polynomials, to your generalization of a polynomial map $F: \mathbb R^n \to \mathbb R^n$. Since the Jacobian determinant of $F$ does not vanish identically, its zero set $Z$ (the critical points of $F$) has measure 0, since $F$ is a polynomial. Away from critical values, the inverse image is a dimension 0 manifold, i.e. a finite set. Finally, the size of this finite set is bounded by the product of the degrees of the $F_i$'s by Bezout, since it is the intersection of the $F_i^{-1}(c_i)$, each of which is a hypersurface of degree $\deg F_i$

  • $\begingroup$ Can you provide a reference for saying the size of the finite set is uniformly bounded by the degree of $F$? Also, would this extend to real analytic functions? $\endgroup$
    – Zarrax
    Commented Jun 15, 2022 at 16:32
  • $\begingroup$ Sorry, actually the simplest thing I can think of bounds it by the product of the degrees, which in your case would be $(n-1)^n$. I can try to think if there's a better bound in your case. It definitely does not extend to real analytic, both bc you need to switch Z to the codomain bc of bump functions, and bc there could be infinitely many preimages (consider sine) $\endgroup$ Commented Jun 15, 2022 at 17:18
  • $\begingroup$ Ok thanks. For the real analytic case it would have to be a local theorem. Is this possible? $\endgroup$
    – Zarrax
    Commented Jun 15, 2022 at 17:51
  • $\begingroup$ By the way, the reason I was hesitant to use Bezout's theorem is the assumption in that theorem that the zero sets have no common component, which can happen in some cases here. Are you saying there's a way around this? $\endgroup$
    – Zarrax
    Commented Jun 15, 2022 at 18:04
  • 1
    $\begingroup$ Re: "the assumption" in Bezout's theorem, see the text of this question: mathoverflow.net/q/417812/1508 $\endgroup$
    – pinaki
    Commented Jun 15, 2022 at 19:47

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