A question on infinitely many closed points on a smooth projective variety and their behavior under embeddings

Question

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.

Answer 1

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 $CH_0(X)$.

Answer 2

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!