Proof that a Zariski closed subset of a product of projective space is the common zeros of multihomogeneous polynomials 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 $Z$ be a closed subset of $P^n\times P^m$.
Then how do you prove that $Z$ is the common zeros of bihomogeneous polynomials?
 A: $\def\P{\mathbb P}$Note that the closed sets in $X := \P^n \times \P^m$ are generated by sets of the form $X \setminus (U \times V)$ where $U \subseteq \P^n$ and $V \subseteq \P^m$ are open. 
So let $U \subseteq \P^n$ and $W \subseteq \P^m$ be open sets. Then there are homogeneous polynomials $f_i\in K[x_0, \ldots, x_n]$ and $g_j \in K[y_0, \ldots, y_m]$ such that $\P^n \setminus U = V(f_i \mid i \in I)$ and $\P^m \setminus W = V(g_j \mid j \in J)$. We may consider $f_i$ and $g_j$ as bihomogenous elements of $K[x_0,\ldots, x_n, y_0,\ldots, y_m]$, and compute their product $h_{ij} := f_i\cdot g_j$. For $(p,q) \in \P^n \times \P^m$ we have
\begin{align*}
  (p,q) \in U \times W &\iff p \in U \land q \in W\\
                       &\iff \exists i: f_i(p) \ne 0\land \exists j: g_j(q) \ne 0\\
                       &\iff \exists i,j : f_i(p)g_j(q) \ne 0\\
                       &\iff \exists (i,j) : h_{ij}(p,q) \ne 0
\end{align*}
So $X \setminus (U \times W) = V(h_{ij}\mid (i,j) \in I \times J)$. 
If $F \subseteq X$ is any closed set, write $F = \bigcap_k X \setminus (U_k \times W_k)$ find sets $H_k \subseteq K[x_0,\ldots, x_n, y_0, \ldots, y_m]_{\text{bihom}}$ with $X \setminus U_k \times W_k = V(H_k)$ as above and let $H = \bigcup_k H_k$. Then $F = V(H)$ is as wished.
