Systems of multivariate polynomials with less affine roots than roots at infinity. We are considering systems
of $n$ multivariate polynomials in $n$ variables with coefficients from $\mathbb{C}$ (or $\mathbb{R}$):
$p_1(x_1,\ldots,x_n)=0,~~\ldots~~,p_n(x_1,\ldots,x_n)=0$.
We are interested in the common roots of such systems and assume a zero-dimensional solution set (only isolated roots).
Let $d_i$ be the degree of $p_i$ (largest degree of monomials with nonzero coefficient). Then by Bezout's theorem there are $m:=\prod_{i=1}^nd_i$ solutions in the projective space $\mathbb{P}=\mathbb{C}^{n+1}\backslash\lbrace 0\rbrace$, counted with multiplicities.
These solutions split up into $m_a$ affine roots (points in $\mathbb{C}^n$) and $m_p$ projective solutions, which are the solutions of the homogenized system $p_i^h(x_0,x_1,\ldots,x_n)=0$ with $x_0$-component equal zero. Let's call those $m_p$ roots the roots at infinity (since they cannot be scaled back to affine space).
Now the question (sorry if the explanation was too extensive):
Can we find/construct systems of the above class that have strictly less affine roots than roots at infinity, $1<m_a<m_p$, and all roots are simple?
(of course, at least one $p_i$ should have degree$>1$).
We could show that this is not possible for simple cases (e.g., $n=2$, $d_1=d_2=2$) and suspect that this is not possible in general. We are no algebraic geometers and lack the tools to investigate this issue rigorously, so any help / tipps would be appreciated.
 A: I am not sure this is what you are looking for.
So, let $\deg p_i=d_i$ and we may assume $d_1\geq d_2\geq \cdots\geq d_n$. I will assume  that $d_n\geq 2$, linear equations create issues and am not sure how they can be handled.
We are assuming the homogenization of these have transversal intersections and thus there are $M=\prod d_i$ points in the intersection.
If $l_i$ are the leading forms of the $p_i$, then the points at infinity are the intersection of these $l_i$s. There are $n$ equations in $n-1$ projective space with finite intersection, so after suitable linear change (adding multiples of the last equation to the previous ones) we may assume $l_i , i<n$ intersect in finitely many points and $m_p$ is less than or equal to this number, since it is got by further intersecting with $l_n$. So, by Bezout, we get $m_p\leq \prod_{i=1}^{n-1} d_i=m$, while the total number of points are $M$. Thus, $m_a=M-m_p\geq m(d_n-1)\geq m\geq m_p$.
Here, I write down a linear case which is contrary to the above statements. Take $n=2, p_1=x_1, p_2=x_1+1$. Then $m_p=1, m_a=0$.
