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Let $\mathbb{C}[x_1,x_2,\cdots,x_n]_{= d}$ denote the set of polynomials in $\mathbb{C}[x_1,x_2,\cdots,x_n]$ of total degree $d$. Is there exists a sequence of polynomials $f_1,f_2,f_3,\cdots$ in $\mathbb{C}[x_1,x_2,\cdots,x_n]_{=d}$, such that for every $i_1<i_2<\cdots<i_n$, the affine variety $V(f_{i_1},f_{i_2},\cdots,f_{i_n})$ has size $\leq d^n$?

For $d=1$, we can let $f_i=x_1+ix_2+i^2x_3+i^3x_4+\cdots+i^{n-1}x_n$.

For $n=2$, we can let $f_i(x,y)=x+g_i(y)$, where $g_i(y)\in \mathbb{C}[y]_{=d}$, and $g_i\neq g_j$ for $i\neq j$.

How to construct $\{f_i\}$ when $n>2$ and $d>1$?

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    $\begingroup$ Do you want to avoid trivial examples where $f_i$ all differ by constants, making the varieties empty? $\endgroup$
    – Robert Israel
    Mar 28, 2023 at 1:44
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    $\begingroup$ You just want any $n$ of the polynomials to have zero loci which intersect transversely, since then by Bezout's theorem they will intersect in at most $d^n$ points. If you go by induction, you just need to construct $f_{k+1}$ in such a way that its zero locus intersects all the curves defined by any $n-1$ of the previous polynomials transversely. By Bertini's theorem, any general $f_{k+1}$ will do. So basically, any "random" sequence will work. $\endgroup$
    – Jack Huizenga
    Mar 28, 2023 at 5:19
  • $\begingroup$ @ Robert Israel, thanks for the comment, yes, you are right! $\endgroup$
    – Yuting Li
    Apr 4, 2023 at 11:54
  • $\begingroup$ @Jack Huizenga, thank you for pointing out Bezout's theorem and Bertini's theorem! Yes, I think random sequence will work, but I am looking for explicit constructions. $\endgroup$
    – Yuting Li
    Apr 4, 2023 at 11:57

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