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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 $P^n\times P^m$?

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Makoto Kato
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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 00:35
  • @MichaelJoyce I wonder why you think they are the same. – Makoto Kato Nov 22 '12 at 04:42
  • 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 05:16
  • @MichaelJoyce This is the converse of the other proposition. You know that the converse of a correct proposition is not necessarily correct. It is a mistake to think otherwise. – Makoto Kato Nov 22 '12 at 05:20
  • Yes, I am aware that the each statement is the converse of the other. In fact, the connection between the two involves an unpacking of the definitions, so they are equivalent. But since that is what you are asking about in the first place, it is understandable that you might perceive the two directions as being different. Still, if you are trying to understand a statement of the form $P \Leftrightarrow Q$, it is probably best not to divide into separate questions on why $P \Rightarrow Q$ and why $Q \Rightarrow P$. – Michael Joyce Nov 22 '12 at 06:23
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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-Álvarez Nov 22 '12 at 07:38
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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-Álvarez Nov 22 '12 at 07:43
  • @MarianoSuárez-Alvarez Please tell me what's wrong with separating questions which are related but different. – Makoto Kato Nov 22 '12 at 08:37
  • @MarianoSuárez-Alvarez I deleted the remark. – Makoto Kato Nov 22 '12 at 09:20

2 Answers2

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

Georges Elencwajg
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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 $F_i(x,y)$.

Michael Joyce
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