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
 A: The result you ask about is not true.
For example take $n=m=1$ and the family consisting of the sole polynomial $F=x_0y_1-x_1y_0$, which is bihomogeneous  of bidegree $(1,1)$.
Its zero locus is the diagonal $\Delta =\lbrace (a,a)\mid a\in \mathbb P^1\rbrace \subset \mathbb P^1\times \mathbb P^1$ which is not closed in the product topology of  $\mathbb P^1\times \mathbb P^1$, since the  only closed sets of that topology are unions of points, vertical lines and horizontal lines.
The result  holds however if you endow $\mathbb P^n\times \mathbb P^m$ with the topology induced by the Segre embedding $\mathbb P^n\times \mathbb P^m \hookrightarrow \mathbb P^{(n+1)(m+1)-1}$.
A: 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 $F_i(x,y)$.
