3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

$\def\P{\mathbb P}$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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.