TITLE: Maximum number of common zeros of n polynomials in n-1 variables
QUESTION [11 upvotes]: 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 $2^{n-1}-1$.

REPLY [14 votes]: 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.