Let $V$ be a closed subvariety of $\mathbb{A}^n$. Let $\pi:\mathbb{A}^n\to \mathbb{A}^1$ be the projection map that forgets the last $n-1$ coordinates (say). Assume that the Zariski closure of $\pi(V)$ is $\mathbb{A}^1$. Then $\pi(V)$ must be of the form $\mathbb{A}^1\setminus S$, where $S$ is a finite set of points in $\mathbb{A}^1$. What sort of useful bound can we give on the number of points $|S|$ in $S$?

I am looking for a bound of the form $|S|\leq (\deg V)^D$, where $D$ is a function of $n$. It would be very nice if $D$ were bounded by a polynomial on $n$. (I think I already know how to get a horrific $D$.)

  • $\begingroup$ Question arising from work with my former postdoc Jitendra Bajpai and my former student Daniele Dona. The formulation is Daniele's. $\endgroup$ Nov 26, 2020 at 14:22
  • $\begingroup$ Sure, that's what I meant. (Let's not get distracted, but there must be some difference here depending on your foundations; if you are just using classical foundations and define a variety as a closed algebraic subset of $\mathbb{A}^n$ or $\mathbb{P}^n$, there's of course no difference. You mean an embedding of an abstract variety in affine space?) $\endgroup$ Nov 26, 2020 at 14:43
  • $\begingroup$ Ah, right, good point. Changed the wording. $\endgroup$ Nov 26, 2020 at 14:49
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    $\begingroup$ If $V \subset \mathbb{A}^2$ I would say that the maximum number is $\deg V$. Compactifying, we get a projection $\pi \colon \bar{V} \to \mathbb{P}^1$. If there are missing points in the image of the original projection, they come from the points at infinity that I add to $V$. If these points are $x_1, \ldots, x_d$, where $d=\deg V$, then the maximum number of missing points occurs when $\pi$ is the projection from $x_1$ (say) and all the secant lines $x_1x_j$ are distinct (included the case $j=1$, giving a tangent line to $\bar{V}$). In this case we have $d$ missing points. $\endgroup$ Nov 26, 2020 at 14:53
  • $\begingroup$ uhm...not sure anymore of what I have said. If the line $x_1x_j$ intersects $\bar{V}$ in a further point $y \in V$, then there is no missing point corresponding to this line... $\endgroup$ Nov 26, 2020 at 15:06

1 Answer 1


It seems to me that one can bound $|S|\leq (n-1) \deg(V)$.

First, note that we can work projectively, that is, we will be able to work with the projective closure $\overline{V}\subset \mathbb{P}^n$. In the end, the points of $\overline{V}\setminus V$ will only contribute a point at infinity in $\mathbb{P}^1$, and we are not counting that point anyhow. We will write $V$ instead of $\overline{V}$ henceforth.

We can define a map $\pi_n:\mathbf{P}^n\setminus P_{0,n}\to \mathbf{P}^{n-1}$ by $\pi_n(x_0:x_1:\dotsc:x_n) = (x_0:\dotsc:x_{n-1})$, where $P_{0,n}\in \mathbf{P}^n$ is the point $(0:0:\dotsc:0:*)$. As pointed out in How many holes may a projection of an algebraic variety have?, either (a) $\dim(\overline{\pi_n(V)})=\dim(V)$ and $\pi_n(V)$ contains $\overline{\pi_n(V)}\setminus W$, where $W$ is a variety of dimension $\leq \dim(V)-1$ and degree $\leq \deg(V)$, or (b) $V$ is a cone whose vertex contains $P_{0,n}$, and so $\pi_n(V)$ is closed and of dimension $\dim(V)-1$. Clearly $\deg(\overline{\pi_n(V)})\leq \deg(V)$.

We iterate: we define $\pi_{n-1}:\mathbf{P}^{n-1}\setminus P_{0,n-1}\to\mathbf{P}^{n-2}$ just as above. If we are now in case (a), we have $\dim(\overline{\pi_{n-1}(\pi_n(V))})=\dim(\overline{\pi_n(V)})$, and $\pi_{n-1}(\pi_n(V))$ contains $\pi_n(\pi_{n-1}(V))\setminus (W' \cup \overline{\pi_{n-1}(W)})$, where $\deg(W')\leq \deg(V)$ and $\dim(W')\leq \dim(\overline{\pi_n(V)})-1$, and $W$ is as above (and is empty if we were in case (b) before). If we are in case (b), then we need not remove a new variety $W'$, and we also notice that what we must remove from $\pi_{n-1}(\overline{\pi_n(V)})$ is the variety consisting of the points of $\pi_{n-1}(W)$ whose preimage under $\pi_{n-1}$ is contained in $W$. That variety is either empty or of dimension $\leq \dim(W)-1$; its degree is presumably $\leq \deg(W)$.

We iterate further, and are done.

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    $\begingroup$ I think you even have $|S|\leq \deg(V)-1$. This follows from your argument and the observations that, assuming $V$ is not a cone over the point of projection, we have $\deg W + \deg \pi(V)=\deg V$. So the total amount of degree you pick up in the "variety of possibly missing points" by your sequence of projections is $\deg V -1$. (Thanks to Aaron Landesman for pointing out that the degree of $V$ should go down by projecting from a point on $V$!) $\endgroup$ Dec 4, 2020 at 1:20
  • $\begingroup$ Humor my slow brain: I take that $\deg(W) + \deg \pi_n(V) = \deg V$ follows from the observation that $\deg(W)$ equals the multiplicity of $V$ at $P_{0,n})$? (Thus, a space of dimension complementary to that of $V$ going through $P_{0,n}$ contains in fact $\deg(W)$ copies of $P_{0,n}$, and thus can have at most $\deg(V)-\deg(W)$ points of intersection with $V\setminus P_0$? $\endgroup$ Dec 4, 2020 at 22:43
  • $\begingroup$ Right, you need the relation between the multiplicity and the degree of $W$. A reference is Harris' "Algebraic geometry" p. 259, Ch. 20 in the part titled Multiplicity. The main point is that the class of the proper transform of $V$ is given by $[\widetilde{V}]=\deg V [L]-m[\Lambda]$, where $L$ is a linear subspace of $\mathbb{P}^n$ and $\Lambda$ a linear subspace of the exceptional divisor (both with the same dimension as $V$). $\endgroup$ Dec 5, 2020 at 2:20

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