Compactification problem.

Question

How do I compactify the curve $Q(x,y)=0$ in $\mathbb{P}^1\times\mathbb{P}^1$ where $Q$ is a polynomial ?

Answer 1

If the two copies of $\mathbb P^1_k$ have homogeneous coordinates $x=(x_0:x_1),y=(y_0:y_1)$, then a non-constant polynomial $Q=Q(x_0,x_1;y_0,y_1)\in k[x_0,x_1;y_0,y_1]$ will define a curve in $\mathbb P^1_k\times\mathbb P^1_k$ if and only if it is homogeneous both in the variables $x_0,x_1$ and in the variables $y_0, y_1$.

For example the polynomial $Q=x_0^3y_0y_1-x_0x_1^2y_1^2$ defines a curve on $\mathbb P^1_k\times\mathbb P^1_k$, which is said to be of bidegree (3,2) because $Q$ is homogeneous of degree $3$ in $x$ and homogeneous of degree $2$ in $y$.
But the polynomial $R=x_0^3y_0-x_0x_1^2y_1^2$ does not define anything on $\mathbb P^1_k\times\mathbb P^1_k$ because it is not homogeneous in $y$: the first term has degree $2$ in $y$ but the second has degree $1$.

That said, if $Q$ is bihomogeneous it defines a closed sub-scheme $V_{proj}(Q)\subset \mathbb P^1_k\times\mathbb P^1_k$ (the locus of zeroes of $Q$), which is always complete.
If the base field is $k=\mathbb C$, that curve also inherits a transcendental (= classical) topology, strictly finer than its Zariski topology, and the curve is always compact in that topology.
This answers your question: you don't need to compactify it since it is already automatically compact!

Edit: why $Q$ has to be bihomogeneous
Suppose $R(x_0,x_1,y_0,y_1)=x_0^3y_o-x_0x_1^2y_1^2$.
Does this polynomial vanish on $P=((1:1),(1:1))\in \mathbb P^1_k\times\mathbb P^1_k$ ?
Apparently yes, because if you plug in $x_0=x_1=y_0=y_1$ into $Q$, you get $R(1,1,1,1)=0$.
However if you realize that you can also write $P$ as $P=((1:1),(2:2))$ you get $R(1,1,2,2)=-2$ so that now you would say that $R$ does not vanish at $P$.
This contradiction shows why it doesn't make sense to talk of the zero set of a polynomial $R$ if that polynomial is not bihomogeneous with respect to the two sets of variables $x=(x_o,x_1)$ and $y=(y_o,y_1)$.