Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 :
$$Z=Y \cap \mathcal H\subset \mathbb P^n\times \mathbb P^{n\ast} $$

share|cite|improve this answer
Thank you for your answer, Georges. In I) you get a non algebraic $Z$ because you start with non algebraic $Y_p$. But the $Y_p$'s are assumed to be varieties (it is their "collection" $Y$ which might fail to be); however I see that I should assume that $Y$ is a subvariety of $\textbf P^n\times\textbf P^{n\ast}$. I will do so. Thank you! – Brenin Apr 16 '12 at 8:05
Dear atricolf, you are perfectly right. I have modified my previous counter-example in order that each $Y_p$ be a variety. (This is for the benefit of other users, since you write that you are now convinced that $Y$ must be assumed to be a subvariety.) – Georges Elencwajg Apr 16 '12 at 9:53
Now it is very clear! Thank you. – Brenin Apr 17 '12 at 22:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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