I am reading the proof that the canonical projection $\mathbb P ^n \times \mathbb P^m \to \mathbb P^m$ is closed from this lecture notes, page 57. Let $Z$ be a closed subset of $\mathbb P ^n \times \mathbb P^m $, then by serge embedding, we can write $Z = V(f_1, \ldots, f_r)$ for homogeneous polys $f_1, \ldots, f_r$ of the same degree in the Segre coordinates of $\mathbb P^n \times \mathbb P^m$, i.e. for bihomogeneous polynomials of degree $d$ in both the coordinates of $\mathbb P^n$ and $\mathbb P^m$.
I am not quite sure why that has to be the case. So $Z = V(f_1, \ldots, f_r)$ because a closed subset of $\mathbb P^n \times \mathbb P^m$(which can be realized as a closed subset by the segre embedding) is again closed in $\mathbb P^{(m + 1) * (n + 1)- 1}$ thus must be of the form $V(f_1, \ldots, f_r)$, and it is a fact that $f_1, \ldots, f_r$ can be chosen to be of the same degree. But I do not know why those generators have to be bihomogeneous.
Best Answer
Let $\{z_{ij}|i=0,\cdots, n, j=0,\cdots, m\}$. Then we have a map from $k[z_{00},\cdots, z_{ij},\cdots, z_{nm}]\rightarrow k[x_0,\cdots, x_n, y_0,\cdots, y_m]$ sending $z_{ij}$ to $x_iy_j$. Let $\mathfrak{a}$ be the kernel of this map, then we have the image of $\mathbb{P}^n\times \mathbb{P}^m\rightarrow \mathbb{P}^{n+m+mn}$ is exactly $Z(\mathfrak{a})$. In particular, your homogeneous polynomial in $\mathbb{P}^{n+m+nm}$ can be viewed as in $z_{ij}$ and we can map $z_{ij}$ to $x_iy_j$ to get the corresponding bihomogeneous function.