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We work over an algebraically closed field $k$. Let $X=V_+(f_1,\dots,f_r)$ be a (smooth, if you want) projective subvariety of $\textbf P^n$, so $f_i(x_0,\dots,x_n)$ are homogeneous polynomials. I write $\textbf P^{n\ast}$ for the parameter space of hyperplanes $H\subset\textbf P^n$. Let us assume that for any point $p\in X$ we are given a subvariety $Y_p\subset\textbf P^{n\ast}$. The collection $(Y_p)_{p\in X}$ is not assumed to form a family over $X$ (that is, a subvariety of $X\times_k\textbf P^{n\ast}$), but if you prefer you may assume it. Finally, we can define a subset

\begin{equation} Z=\{(p,H)\,| \,\,p \textrm{ is a point of }X \textrm{ and }H\in Y_p \}\subset X\times_k\textbf P^{n\ast}. \end{equation}

My questions are:

How to see that $Z$ is a subvariety of $X\times_k\textbf P^{n\ast}$? Is it possible to give the polynomials defining it, perhaps in terms of the $f_i$'s?

I tried to introduce homogeneous coordinates $a_0,\dots,a_n$ on the dual projective space and to plug them somehow into the $f_i$'s but I was not able to conclude. My idea was to follow the strategy that one uses when looking for the polynomials defining same universal locus, like the universal hyperplane, the universal conic... For example the universal hyperplane $\mathcal H\subset \textbf P^n\times_k\textbf P^{n\ast}$ is given by the polynomial \begin{equation} h(x_0,\dots,x_n;a_0,\dots,a_n)=\sum_{i=0}^na_ix_i. \end{equation}

But this trick does not seem to work in this case.

Thanks in advance.

1 Answers 1

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I) I think you should suppose that the subset $Y=\bigcup_{p\in X} \lbrace p\rbrace \times Y_p \subset \mathbb P^n\times \mathbb P^{n\ast}$ is algebraic (i.e. is what you call a family).
Else you get easy counterexamples:

For instance take $n=2$ and choose for $X$ a line $X=L \subset \mathbb P^2$ .
Partition $L$ into two non algebraic subsets $L=L_1\sqcup L_2$.
Now for $p\in L_1$ choose $Y_p=$ all lines through $p$ and for $p\notin L_1$ take $Y_p=\emptyset $
It is then clear that $Z\subset L\times\textbf P^{2\ast}$ is not algebraic since its image under the first projection of $L\times\textbf P^{2\ast}$ onto $L$ is the non algebraic $L_1\subset L$ .

II) If $Y$ is assumed algebraic, then indeed $Z$ is algebraic because it is the intersection of two algebraic subvarieties :
$Z=Y \cap \mathcal H\subset \mathbb P^n\times \mathbb P^{n\ast} $

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    Now it is very clear! Thank you.2012-04-17