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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

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How is this question different than the original question you posted? It is bad etiquette to post the same question multiple times. –  Michael Joyce Nov 22 '12 at 0:35
    
@MichaelJoyce I wonder why you think they are the same. –  Makoto Kato Nov 22 '12 at 4:42
    
Because the first paragraph and a half are copied verbatim and the last few sentences are the two different halves of a single biconditional? –  Michael Joyce Nov 22 '12 at 5:16
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I think it would be a good idea, next time you find yourself copying large portions of a question into another, to not do it and instead refer to it: that is the whole point of being able to make links... (and, if I understand correctly, a big part of your motivation to use this site to build a database of sorts; you are well aware of the linking capabilities, in any case!) Also, it would probably have been less disruptive to simply add the converse question to the old one. –  Mariano Suárez-Alvarez Nov 22 '12 at 7:38
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Condescending remarks like the one in your comment above are quite out of place, by the way. You are surely capable of avoiding them; if not, try harder. –  Mariano Suárez-Alvarez Nov 22 '12 at 7:43

2 Answers 2

up vote 5 down vote accepted

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 $\mathbb P^n\times \mathbb P^m \hookrightarrow \mathbb P^{(n+1)(m+1)-1}$.

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Oops. I forgot that fact. Thanks for reminding me. –  Makoto Kato Nov 22 '12 at 8:00
    
Dear Makoto, yes I was sure you knew that and you had just been absent-minded! –  Georges Elencwajg Nov 22 '12 at 8:04

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 $F_i(x,y)$.

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