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Are there general results of the number/structure of critical points of a polynomial $p : \mathbb{C}^n \rightarrow \mathbb{C}$? To be precise, the set of $z\in\mathbb{C}^n$ such that $\nabla p(z) = 0$.

The example $p(z_1,z_2) = (z_1^2 + z_2^2 - 1)^2$ shows that the set may consist of both isolated and non-isolated points. If the set is discrete, does it have an upper bound on the number of elements? Do the non-discrete part consist of smooth manifolds? These are the kind of results I am looking for.

Jas Ter
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  • You mean apart from the Bezout bound $(\deg(p)-1)^n$ for the number of roots? – Lutz Lehmann Feb 19 '14 at 18:04
  • I have never heard of it, and Wikipedia's page doe sn't enlighten me about the content ... But clearly there are polynomials with critical points forming manifolds, say $(z_1^2 + z_2^2 - 1)^2$. For $z_i$ real, we have a circle plus the origin; for complex $z_i$ we have even more critical points. – Jas Ter Feb 19 '14 at 19:44
  • It depends on what you understand as critical point. In the first interpretation these are the points of zero gradient. Or did you mean critical points on the variety defined by $p$? The first thing to do is to care that $p$ is square free, factors of higher multiplicity naturally generate singular components. – Lutz Lehmann Feb 19 '14 at 19:49
  • Thank for the comment. I have updated the question, and I am thinking of points where the gradient vanishes. – Jas Ter Feb 19 '14 at 19:58
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    The gradient components form a system of polynomials of degree $\deg p-1$ or less. Any system of polynomials $f_1,...,f_n$ has at most $\deg f_1\cdot \deg f_2\cdots\deg f_n$ isolated solutions. I do see nothing that would restrict the dimension of components of non-isolated solutions. – Lutz Lehmann Feb 19 '14 at 20:17
  • Thanks, I think I see the bound on the isolated solutions. (However, one has not actually used that the system is a gradient.) It answers the question sufficiently for me. (Could you type the isolated points result as an answer?) – Jas Ter Feb 20 '14 at 07:44
  • Ah, that's Bézout's theorem! Thanks. – Jas Ter Feb 20 '14 at 07:47

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The gradient components $\partial_1p,...,\partial_np$ form a system of polynomials of degree $\deg(p)−1$ or less.

Bézout's theorem states that any system of polynomials $f_1,...,f_n$ has at most $deg f_1⋅deg f_2⋯deg f_n$ isolated solutions. If one also counts solutions at infinity, i.e., the solution set of the homogenized version of the system in projective space, and there are still only isolated solutions, the bound is exact.

So there will be at most $(\deg(p)−1)^n$ isolated critical points for $p$.

As an example that fills the bound consider polynomials of the form $p(x)=\sum_{k=1}^nq_k(x_k)$ with $\deg q_k=d$, $q_k$ "in general position". Then $\partial_kp=q_k'(x_k)$ depends only on one variable and has $d-1$ roots and the set of critical points is the cartesian product of $(d-1)^n$ combinations.

I do see nothing that would restrict the dimension of components of non-isolated solutions.

Lutz Lehmann
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