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Given a system of complex polynomial equations in $n$ variables, giving rise to an affine variety $V \subseteq \mathbb C^n$, is there a bound $b \in \mathbb R$ such that if $V(\mathbb C) \neq \emptyset$ then there is a point $a \in V(\mathbb C)$ such that $||a|| < b$?

I am looking for some bound $b$ in terms of $n$, the degrees of the polynomials and the absolute values of the coefficients. Or indeed is anything known in this sort of direction?

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  • $\begingroup$ A bound for any hypersurface follows from a bound for a one variable polynomial, which I think is not difficult. $\endgroup$
    – Ben McKay
    Commented May 31, 2023 at 12:08
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    $\begingroup$ Consider, for example, two almost parallel lines in $\mathbb{C}^2$. Clearly, it's impossible to bound the distance to the point of intersection in terms of absolute values of coefficients. $\endgroup$
    – Oleg Eroshkin
    Commented May 31, 2023 at 13:18
  • $\begingroup$ @OlegEroshkin In that case, the corresponding algebraic variety (in complex affine space) is empty. The OP does stipulate that the algebraic variety is nonempty. $\endgroup$
    – Jason Starr
    Commented May 31, 2023 at 13:24
  • $\begingroup$ @JasonStarr Wait, I wrote "almost parallel", like $az+bw=c$ and $az+(b+\epsilon)w=d$. The distance to the intersection point depends on how small is $\epsilon$ $\endgroup$
    – Oleg Eroshkin
    Commented May 31, 2023 at 13:44
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    $\begingroup$ @JasonStarr Already for several linear equations one needs to bound from below the lowest singular value of the matrix of coefficients. I don't know the analog for non-linear algebraic equations. $\endgroup$
    – Oleg Eroshkin
    Commented May 31, 2023 at 13:55

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No (for one interpretation of the question). There does not exist a continuous function $b $ from the set of systems of polynomial equations to $\mathbb R^{>0} \cup \{\infty\}$ that takes a finite value on all systems with a solution, and such that if there is a solution there exists one of norm at most the value of $b$.

One considers the vanishing locus of the polynomials $xy, y(1-y), ty+tx+y-1$. This has the solution $(1/t,0)$ for $t\neq 0$ and $(0,1)$ for $t=0$, and these are the only solutions.

So $b$ would have to go to $\infty$ as $t\to 0$ but then it would have to take the value $\infty$ at $0$ which contradicts the assumption.

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  • $\begingroup$ Of course there is a locally closed stratification of the Chow variety of projective space that "flattens" the intersection of the universal cycle with the hyperplane at infinity. Over each locally closed stratum of this flattening stratification, there is, indeed, a continuous bound $b$. $\endgroup$
    – Jason Starr
    Commented May 31, 2023 at 15:32
  • $\begingroup$ Thank you. My question is evidently too naive and general to be true. In the counterexample, you have a fixed variety $V$ given by the first two equations $xy=0$ and $y(1-y) = 0$, intersecting with a family of curves dependent on the parameter t. $V$ is not irreducible, and the discontinuity comes from that. If we replace $V$ by its component $y=0$, we get the solution $(1/t,0)$ and so taking $b=1/t$ works. I am actually interested in this situation where $V$ is fixed and positive dimensional, intersected with a varying family of varieties (but with unbounded degree, not an algebraic family). $\endgroup$
    – Jonathan Kirby
    Commented Jun 2, 2023 at 7:28
  • $\begingroup$ @JonathanKirby I don't think your distinction is meaningful. We can take $V$ in $\mathbb A^4$ defined by the equations $xy=z, y(1-y)=w$ and then intersect with the family with equations $ty+tx+y-1 = w=z=0$. Then $V$ is irreducible, smooth, and otherwise nice and the varying family is a smooth family of affine subspaces. $\endgroup$
    – Will Sawin
    Commented Jun 2, 2023 at 10:13
  • $\begingroup$ @WillSawin Thank you, that's a good point. Although in the subject of atypical intersections in diophantine geometry (which is related to the motivation for my question), one does sometimes have to first project then take components in order to see the geometry of the situation, and in this example that would mean projecting back to the xyt coordinates. But it makes it difficult to come up with precise questions. $\endgroup$
    – Jonathan Kirby
    Commented Jun 5, 2023 at 6:48

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