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Fix an integer $n \ge 5$. Let $\mathcal{V}$ be a countable collection of closed subvarieties of $\mathbb{P}^n_{\mathbb{C}}$ of codimension at least $2$. Choose a point $p \in \mathbb{P}^n$. Does there exist a curve $C$ (affine or projective) containing the point $p$ and not intersecting any subvariety $V \in \mathcal{V}$ away from $p$ i.e., for any $V \in \mathcal{V}$, $C \cap V$ is either $p$ or $\emptyset$?

EDIT The underlying field is $\mathbb{C}$.

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    $\begingroup$ Are you working over $\mathbb C$? This is obviously false over countable fields. $\endgroup$
    – Wojowu
    Commented May 3, 2022 at 14:05
  • $\begingroup$ @Wojowu Yes, the underlying field is $\mathbb{C}$. I have made it explicit in the question. $\endgroup$
    – user45397
    Commented May 3, 2022 at 16:40
  • $\begingroup$ In the parameter space of all lines passing through $p$ (this is a projective space), interesecting a codimension two subvariety away from $p$ is a proper closed condition. Hence we are asking whether this projective space can be covered by countably many proper closed subvarieties, which is not the case. Thus there exsits a line with the required properties. $\endgroup$ Commented May 4, 2022 at 7:39
  • $\begingroup$ @EvgenyShinder Thanks for the answer. $\endgroup$
    – user45397
    Commented May 4, 2022 at 11:57

2 Answers 2

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Consider the case when $C$ is a line through $p$. Lines through $p$ correspond to points of $\mathbb P^{n-1}$, and this gives a projection map $\mathbb P^n \setminus p \to \mathbb P^{n-1}$ Then $C$ intersects $V$ if and only if the image of $V \setminus p$ under the projection map doesn't contain the point corresponding to $C$. Since $V$ has codimension at least $2$, the image under the projection map has codimension at least $1$.

The countable union of codimension $1$ subvarieties cannot contain all the points for the usual reasons (you can prove this with measure theory, Baire category theory, or purely algebraically by induction). So there must exist such a line $C$.

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  • $\begingroup$ Thanks for your answer. $\endgroup$
    – user45397
    Commented May 3, 2022 at 21:14
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$\DeclareMathOperator{\scrV}{\mathcal{V}}$Yes. As a test case consider the case that $n = 2$ and $\scrV$ is a countable collection of points. Then for any $d \geq 1$, the space $P_d$ of homogeneous polynomials of degree $d$ vanishing at $p$ is a positive dimensional vector space, and there is polynomial $F \in P_d$ which does not vanish at any other point of $\scrV$, since a vector space over an uncountable field can not be the union of countably many proper subspaces.

The general case follows in the same way. We can of course assume each $V$ is irreducible (which you probably mean anyway). For each $V \in \scrV$, choose a point $p_{1, V} \in V \setminus \{p\}$. By the same argument as above, there is a homogeneous polynomial $F_1$ which vanishes at $p$, but does not vanish on any $p_{1, V}$. Let $\scrV_2$ be the countable collection of irreducible components of $V(F_1) \cap V$ for all $V \in \scrV$. Choose $p_{2,V} \in V \setminus \{p\}$ for all $V \in \scrV_2$, and continue as above.

After step $m+1$, where $m$ is the largest dimension of all varieties in $\scrV$, you will have that $Z := V(F_1) \cap \cdots \cap V(F_{m+1})$ satisfies the property you seek. Note that $\dim(Z) \geq n - m - 1 \geq 1$.

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  • $\begingroup$ Thanks, I think I understand. You first prove in the case n=2, then reduce the general case to the n=2 case. Right? $\endgroup$
    – user45397
    Commented May 3, 2022 at 21:00
  • $\begingroup$ It reduces to the $n=2$ case if you choose all $F_i$ from the set of linear polynomials. $\endgroup$
    – pinaki
    Commented May 3, 2022 at 21:51

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