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



Maximum number of common zeros of n polynomials in n-1 variables
There is a bound for the multiplicity of a (homogenous) almost complete intersection in Theorem 1 of this [paper](https://arxiv.org/pdf/0802.0469.pdf) 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](https://www.ams.org/journals/proc/2015-143-06/S0002-9939-2015-12612-3/home.html) and [this very recent work](https://arxiv.org/abs/2006.14717).


