# Maximum number of common zeros of n polynomials in n-1 variables

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 $$2^{n-1}-1$$.

• 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? – Gerry Myerson Sep 26 '20 at 1:22
• Thanks for the comment. I modified the question accordingly. – Alberto Montina Sep 26 '20 at 7:07
• 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? – abx Sep 26 '20 at 7:17
• 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. – Robert Bryant Sep 27 '20 at 10:37
• You are right. I was assuming implicitly that the number of zeros is finite. Let me state explicitly in the question – Alberto Montina Sep 27 '20 at 15:57

## 1 Answer

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.

• Is the bound sharp? I looked through the papers and did not see examples in the form here. – Matt F. Sep 28 '20 at 21:21
• @MattF.: I don't think it is sharp for large $n$, but I am not an expert on this topic. – Hailong Dao Sep 28 '20 at 23:09
• 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. – Alberto Montina Oct 19 '20 at 13:37