While working on a research problem (algebraic cycles), I bumped into a question that I want to prove, though I couldn't yet prove. After several days of attempts, I realized that if the following statement on projective geometry holds, then my original question is most likely answered affirmatively.

Statement: Let $k$ be an infinite perfect field (or even suppose algebraically closed, if you wish). Let $X$ be a smooth projective variety over $k$ of dimension $d$, and let $x_0, x_1, x_2, \cdots$ be an infinite sequence of distinct closed points of $X$. Then, there exist an embedding $X\hookrightarrow \mathbb{P}^N$ and a sequence of hyperplanes $H_1, H_2, \cdots, H_d$, each containing $x_0$, such that $X \cap H_1 \cap \cdots \cap H_d$ is a finite set contained in the given infinite countable set $\{ x_0, x_1, x_2, \cdots \}$.

I think that the statement is true when $\dim X = 1$, simply by taking sufficiently many points $x_0, \cdots, x_{2g+3}$, say, to form a very ample divisor $D = \sum_{i=0} ^{2g+3} x_i$. But, I do not know what to do if $\dim X >1$. Can someone help me in figuring out if this statement is true? I hope I could get some good ideas form mathoverflow. Thank you.


2 Answers 2


Let me explain why this statement cannot be true in general. I will give a counterexample where $X$ is a complex K3 surface.

By a result of [Beauville and Voisin, On the Chow ring of a K3 surface], it is possible to find a point $y\in X$ such that whenever $C_1$, $C_2$ are curves on $X$, $C_1\cap C_2$ is proportional to $y$ in $CH_0(X)$.

Now, I will construct $x_0, x_1,\dots...$ inductively. Suppose that $x_0,\dots,x_i$ have been constructed. Then choose $x_{i+1}$ such that it is independent of $y,x_0,\dots,x_i$ in $CH_0(X)$. To prove that such a point exists necessarily, let me suppose for contradiction that it does not exist : for every $z\in X$, a nonzero multiple of $z$ is in the subgroup of $CH_0(X)$ generated by $y,x_0,\dots,x_i$. Choose a curve $C$ through $y,x_0,\dots,x_i$. Our hypothesis is that every $z\in X$ has a nonzero multiple in the image of $Pic(C)\to CH_0(X)$. Since $Pic^0(C)$ is divisible and $CH_0^0(X)$ has no torsion by a theorem of Roitman [The torsion of the group of 0-cycles modulo rational equivalence], this implies that $Pic^0(C)\to CH_0^0(X)$ is surjective. This contradicts a theorem of Mumford [Rational equivalence of $0$-cycles on surfaces] stating that $CH_0^0(X)$ is infinite-dimensional.

Now, if $C_1, C_2$ are two curves on $X$, their intersection is a multiple of $y$ in $CH_0(X)$, and cannot be supported on the $x_i$ because $y$ and the $x_i$ have been chosen independent in $CH_0(X)$.

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    $\begingroup$ Thank you very much. It is unfortunate for me that the proposed statement is not true, but it is also fortunate for me to know that very deep results of Beauville, Voisin, Roitman and Mumford give a counterexample to this. It was not easy to forsee for me that there is a counterexample..... I guess I should change my approach, then. Thank you very much again. $\endgroup$ Sep 4, 2013 at 7:47

I think this should be true: This is a sort of Bertini-like argument. Since $X$ is smooth projective (say over $\mathbb{C}$) Bertini says that the set of hyperplanes not containing $X$ and with smooth intersection with it is even dense in $|H|$. Cutting out by generic hyperplane lowers the degree by one and since the dimension of $dimX=d$, the intersection $X\cap H_{1} \cap...\cap H_{d}$ is zero dimensional for generic choices of hyperplanes. We can arrange the Hyperplanes so that the intersection contains the points you said. Just, I am afraid that my argument depends on the position of these points. But I think it might be true in general, you can probably fill in the details!

  • $\begingroup$ Thank you for your answer, but I am a bit unsure of how to fill in the details. For instance, to use Bertini arguments, we should a priori have an embedding. But, even for the case of curve, I am not sure if I can achieve what I want to do simply by starting with some embedding, without having an option to reimbedd into a bigger projective space. $\endgroup$ Sep 4, 2013 at 6:16
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    $\begingroup$ I don't see why you should need to reimbed anything. Take an embedding $X\rightarrow \mathbb{P}^{N}$ for some $N$. Then you have an infinite set of points on the image too. In the linear system of hyperplanes $|H|$ there is a dense open subset (and since $X$ is projective, the intersections are even irreducible). Now, in this open set, the set of hyperplanes containing $x_{0}$ and such that and such that $X\cap H_{1}\cap...\cap H_{d}$ is a given finite set is a closed non-empty subset, because this intersection is zero-dimensional for general hyperplanes. $\endgroup$ Sep 4, 2013 at 6:30
  • $\begingroup$ Yes, I understand what you are trying to say. In such case, $X \cap H_1 \cap \cdots \cap H_d$ in general is indeed zero dimensional. But, my concern is that, the number of points in the intersection could be too large, to be suitably "transported" into $\{x_0, x_1, \cdots \}$ by a suitable linear automorphism of $\mathbb{P}^N$. If I increase the dimension of $\mathbb{P}^N$, then so does the degree of $X$ in the embedding. This was the main source of my confusion. $\endgroup$ Sep 4, 2013 at 6:36
  • $\begingroup$ @JinhyunPark, As I said, the intersections are even irreducible for $X$ projective and $dim(X) \geq 2$. This means you can even choose the Hyperplanes so that $X\cap H_{1}\cap ...\cap H_{d} =\{x_{0}\}$. Is this enough for what you need? $\endgroup$ Sep 4, 2013 at 6:42
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    $\begingroup$ I am glad that you understand what I am saying. I also believe that Bertini should help, but as I said at the beginning, to apply it we should have an embedding first, and for this problem, a "good enough" embedding has to be chosen, for otherwise it may not work. My puzzlement lies exactly where the relationship between the degree of X and the dimension of the ambient space of X both increase together... $\endgroup$ Sep 4, 2013 at 7:17

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