Proof that the common zeros of multihomogeneous polynomials is a Zariski closed subset of a product of projective spaces Let $K$ be an algebraically closed field.
Let $n, m \ge 0$ be integers.
A polynomial $F \in K[x_0,\dots,x_n,y_0,\dots,y_m]$ is called bihomogeneous of bidegree $(p,q)$
if $F$ is a homogeneous polynomial of degree $p$(resp. $q$) when considered as a polynomial in $x_0\dots,x_n$(resp. $y_0\dots,y_m)$ with coefficients in $K[y_0\dots,y_m]$(resp. $K[x_0,\dots,x_m])$.
Let $P^n, P^m$ be projective spaces over $K$.
We consider $P^n$ and $P^m$ as topological spaces equipped with Zariski topology.
We consider $P^n\times P^m$ a topological space equipped with the product topology.
Let $(F_i)_{i\in I}$ be a family of bihomogeneous polynomials in $K[x_0,\dots,x_n,y_0,\dots,y_m]$.
Let $Z = \{(x, y) \in P^n\times P^m| F_i(x, y) = 0$ for all $i \in I\}$.
Then how do you prove that $Z$ is a closed subset of $P^n\times P^m$?
This is a related question

Use the open cover by the open sets $U_{i,j}$ where $x_i \neq 0$ and $y_j \neq 0$.  These open sets are isomorphic to affine space.  The set $Z$ is Zariski closed if and only if $Z \cap U_{i,j}$ is Zariski closed in $U_{i,j}$ for all $i$ and $j$.  To show that $Z \cap U_{i,j}$ is Zariski closed, simply dehomogenize the defining equations $F_i(x,y)$.
