Let $P_1,\dots,P_k \in \mathbb{C}[x_1,\dots,x_n,y_1,\dots,y_m]$ be a sequence of polynomials on $n+m$ variables, and let $Z\subset\mathbb{C}^{n+m}$ be the intersection of the zero sets (so $Z = \cap_{j=1}^k \{(\vec{x},\vec{y}) | P_j(\vec{x},\vec{y}) = 0\}$).

Let $\Pi:\mathbb{C}^{n+m}\to\mathbb{C}^n$ be the projection onto the first $n$ coordinates. When is it the case that $\Pi(Z)$ is also defined by a system of polynomials? I know that, for example, if $k=n=m=1$ and $P(x,y) = xy-1$, the projection onto the $x$-axis is everything except for the origin, so this does not always happen.

My understanding from the little I've been able to read and understand is falls under elimination theory, but my algebraic geometry is far from good enough to be able to understand the statements of the main results.

  • 1
    $\begingroup$ When this happens is too general a question. Only general theorem in this direction is given by Noether Normalization . $\endgroup$
    – Mohan
    Jul 2, 2020 at 15:43
  • $\begingroup$ Thanks, that is plenty helpful already, it is good to know what is possible and what is too general. $\endgroup$
    – felipeh
    Jul 2, 2020 at 16:09

1 Answer 1

  1. A short answer is:

Proposition 1: If no affine translate $a+V$, $a\in \mathbb C^{m+n}$ is asymptotic to the variety $Z$, then $\Pi(Z)$ is algebraic.

Here $V=\{x_1=\dots=x_n=0\}$ is the linear subspace of $\mathbb C^{n+m}$, which can be thought as "projection direction" of $\Pi$. When we say $a+V$ and $Z$ are asymptotic, it means there are sequences of points $p_i\in a+V$ and $q_i\in Z$, such that $||p_i-q_i||\to 0$. For example, the $y$-axis is asymptotic to $xy=1$.

  1. The following is an algebro-geometric criterion:

Let's consider the inclusion $\mathbb C^{n+m}\subseteq \mathbb P^{n+m}$ and take the closure $\bar{Z}$ of $Z$ inside the projective space. Set $D=\bar{Z}\setminus Z$, One can think of $D$ as the set of limiting points of $Z$ "at infinity".

Now, the projection $\Pi$ is the restriction to the $\{x_0=1\}$ chart of the linear projection

$$\bar{\Pi}:\mathbb P^{n+m}\setminus\mathbb P^{m-1}\to \mathbb P^n$$

$$[x_0,x_1,...,x_n,y_1,...,y_m]\mapsto [x_0,x_1,...,x_n]$$

where $\mathbb P^{m-1}$ is the projection center defined as the linear subspace consisting points whose the first $n+1$ coordinates are zero. By noting that the set $D\bigcap\mathbb P^{m-1}$ is exactly the limiting points of $Z$ at infinity asymptotically closed to $V$ direction, proposition 1 has the following equivalent form:

Proposition 2: If $D\bigcap \mathbb P^{m-1}=\emptyset$, then $\Pi(Z)$ is algebraic.

Proof: The condition implies that the restriction $$\bar{\Pi}|_{\bar{Z}}:\bar{Z}\to \mathbb P^n$$ is a regular map. Since regular map between projective varieties has closed image (Theorem 10.1 in this notes), $\bar{\Pi}(\bar{Z})$ is a projective variety, so $\Pi(Z)$ is the affine variety $\bar{\Pi}(\bar{Z})\bigcap\{x_0=1\}$.$\tag*{$\blacksquare$}$

  • $\begingroup$ Your first definition of "asymptotic" would also be trivially satisfied if $V+a$ and $Z$ intersect. I believe this wasn't intended? $\endgroup$
    – M. Winter
    Jul 16, 2023 at 19:45
  • $\begingroup$ You're right, one should require $q_i$ to go to infinity as well. Perhaps the algebro-geometric criterion is a better way to formulate. $\endgroup$
    – AG learner
    Jul 16, 2023 at 21:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .