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Let $V\subset \mathbb{A}^n$ be a variety defined by equations of degree $\leq D$, or, what is the same, an intersection of hypersurfaces of degree $\leq D$. Let $V^+_0$ be an irreducible component of $V$ of maximal dimension $d$. Is there a variety $W$ defined by equations of degree $O(n D)$ such that $V\cap W$ has $V^+_0$ as its only top-dimensional irreducible component?

(Additional questions/remarks: (a) Do Gröbner bases give an answer (presumably "yes")? (b) We know that asking for $W$ to be defined by equations of degree $\leq D$ is too much: see How does taking intersections and irreducible components affect the degree?)

UPDATE: $O(n D)$ does not always hold. Can we do $O(D^2)$? Or $O(n D^2)$?

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  • $\begingroup$ What happens if you don't take the intersection with $V$ i.e. ask for a variety $W$ defined by equations of degree $O(nD)$ such that $V_0^+$ is the only top-dimensional component? $\endgroup$
    – user178109
    Commented Apr 26, 2021 at 14:02
  • $\begingroup$ I'd also be happy, but it would be the same question! $\endgroup$
    – H A Helfgott
    Commented Apr 26, 2021 at 14:07
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    $\begingroup$ Daniele Dona has just shown me an example demonstrating that degree $\gg D^2$ can be needed. $\endgroup$
    – H A Helfgott
    Commented Apr 26, 2021 at 16:16
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    $\begingroup$ Here's the example needing $\gg D^{2}$: it's a generalization of Elkies's answer in here. Fix $k$, and take $x^{k-1}y-z$ and $y^{k}-z^{k-1}$: they define a variety $V$ in $\mathbb{A}^{3}$ that is the union of $L=\{y=z=0\}$ and of the curve $C$ parametrized as $(t,t^{(k-1)^2},t^{k(k-1)})$. Now let $P$ be a polynomial vanishing on $C$ and not on $L$: if $P$ doesn't vanish on $L$, it must have a monomial $x^r$ inside, and if $P=0$ on $C$ then $r\gg k^2$, because any other monomial with $y,z$ becomes $t^s$ with $s\geq\min\{(k-1)^2,k(k-1)\}\gg k^2$. $\endgroup$
    – D. Dona
    Commented Apr 27, 2021 at 6:42

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