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Given $n$ quadratic polynomials in $n-1$ variables over the complex field, what is the maximum number of common zeros? Can we have $2^{n-1}-1$ common zeros? I assume that a linear combination of the polynomials is always different from zero and the number of zeros is finite.

With $4$ polynomials, the maximum is not smaller than $6$. Using a projective space $(x_1,x_2,x_3,x_4)$, an example with $6$ roots is given by the polynomials.
$P_1=x_1 x_2$,
$P_2=x_1 x_3$,
$P_3=L_1 x_2+L_2 x_3$,
$P_4=(\text{a general quadratic polynomial})$,
$L_k$ being general linear polynomials. Indeed, if $x_1=0$, then the first two polynomials are equal to zero and the remaining two polynomials in $x_2, x_3, x_4$ give four roots, which are distinct for a general choice of $L_k$ and $P_4$. If $x_2=x_3=0$, then the first three polynomials are equal to zero and the fourth polynomial in $x_1,x_4$ gives 2 additional roots.

This construction has a natural generalisation to $n$ polynomials, giving $2^{n-2}+2^{n-3}$ roots, which is about $3/4$ of the desired bound $2^{n-1}-1$.

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    $\begingroup$ By "polynomials over an $n-1$-dimensional space," do you mean polynomials in $n-1$ variables? Are we working over the reals, the complex numbers, some other field? What's an "intersection point" of two or more polynomials? Is it a common zero? $\endgroup$
    – Gerry Myerson
    Commented Sep 26, 2020 at 1:22
  • $\begingroup$ Thanks for the comment. I modified the question accordingly. $\endgroup$
    – Alm
    Commented Sep 26, 2020 at 7:07
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    $\begingroup$ I think this is an interesting question, let me rewrite it more geometrically: can we find $n+1$ quadrics in $\mathbb{P}^n_{\Bbb{C}}$ which intersect exactly in $2^n-1$ distinct points? For example, 4 quadrics in $\mathbb{P}^3$ intersecting in 7 points? $\endgroup$
    – abx
    Commented Sep 26, 2020 at 7:17
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    $\begingroup$ You need to make some 'general position' assumption to get a good answer. For example, when $n=5$, there are 6 linearly independent quadratic polynomials in 4 variables that vanish identically on the rational curve $(t,t^2,t^3,t^4)$, so the number of common zeros of $5$ linearly independent quadratic polynomials in 4 variables can be infinite. $\endgroup$
    – Robert Bryant
    Commented Sep 27, 2020 at 10:37
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    $\begingroup$ You are right. I was assuming implicitly that the number of zeros is finite. Let me state explicitly in the question $\endgroup$
    – Alm
    Commented Sep 27, 2020 at 15:57

1 Answer 1

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There is a bound for the multiplicity of a (homogenous) almost complete intersection in Theorem 1 of this paper by Engheta. In case of $n$ quadrics in $n-1$ variables, that bound is $2^{n-1}-(n-2)$. So for $n\geq 4$, you can not get $2^{n-1}-1$.

(In Theorem 1 there was a condition that the first $n-1$ generators form a complete intersection, but I don't think that this is serious in your case, one could always make a linear change to turn the first $n-1$ generators into a regular sequence, as the whole ideal has maximal height)

There are further improvements and examples of Engheta's bound in this work and this very recent work.

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  • $\begingroup$ Is the bound sharp? I looked through the papers and did not see examples in the form here. $\endgroup$
    – user44143
    Commented Sep 28, 2020 at 21:21
  • $\begingroup$ @MattF.: I don't think it is sharp for large $n$, but I am not an expert on this topic. $\endgroup$
    – Hailong Dao
    Commented Sep 28, 2020 at 23:09
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    $\begingroup$ Btw, I noted that bound given in the last paper cited by @HailongDao gives 3,6,12,24,51 common zeros for 3,4,5,6,7 polynomials. The example in my question gives 3,6,12,24,48 common zeros. Thus, the example is optimal for up to 6 polynomials. If the next improvement of the bound will give 48 common zeros for n=7, I will start thinking that my example is optimal. $\endgroup$
    – Alm
    Commented Oct 19, 2020 at 13:37

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