Timeline for $\mathbb{A}^1$ connectedness of open subsets of $\mathbb{A}^n$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 9, 2021 at 17:53 | comment | added | Evans Gambit | @dhy: I had noticed that $k$ infinite ensures that the complement of the union contains a $k$-rational point. I was wondering if there was a deeper reason to use the infiniteness of $k$. Thanks for your comment. | |
| May 9, 2021 at 17:45 | comment | added | dhy | @EvansGambit To conclude that there is a point in $\mathbb{A}^n$ that does not lie in a given proper subvariety (in this case the union of Yosemite Stan's comment) you need $k$ to be infinite. (Think about the case of the subvariety $x(x-1)\cdots(x-(p-1))=0$ in $\mathbb{A}^1_{\mathbb{F}_p}$.) | |
| May 8, 2021 at 9:41 | comment | added | Evans Gambit | @Yosemite Stan: Thanks for your comment. So for a point $r$ not in the union, the line joining $p$ and $r$ has to be disjoint from $Z$. Otherwise, if the line $l$ meets $Z$ at a point $s$, then $l=\vec{ps}$, so must be contained in the above union. This is much clearer. Thanks again! Slight tidbit: Though we know the claim in the post is false for finite fields, I am unclear as to where we are using $k$ is infinite in your argument. | |
| May 8, 2021 at 8:51 | comment | added | Yosemite Stan | Maybe try the following. The join of a point and a subvariety of dimension d (the closure of the union of all the lines containing both the subvariety and the point) is closed of dimension d+1. If Z is the complement of U and J(p,Z) and J(q,Z) are the joins of p to Z and q to Z then the union has dimension < n. Any point not in the union will work. | |
| May 8, 2021 at 7:49 | comment | added | Evans Gambit | @dhy: Thanks for your comment. Could you elaborate on why the line passing through $p$ and $r$ doesn't meet the complement of $U$? My reasoning was as follows: The line is defined over the function field $F=k(\mathbb{A}^n_k)$ of $\mathbb{A}^n_k$. After base changing to $F$, the line and $Z:=\mathbb{A}^n_F-U_F$ are both closed subsets of $\mathbb{A}^n_F$ of dimensions 1 and atmost $n-2$ respectively. So their intersection has to be empty. Is that correct? | |
| May 7, 2021 at 13:25 | comment | added | dhy | Let $p,q$ be points in $U$, and let $r$ be a generic point of $U$. Then the line through $p,r$ and the line through $q,r$ both lie entirely inside $U$ (this is where the codimension $\geq 2$ gets used.) | |
| May 7, 2021 at 13:21 | history | asked | Evans Gambit | CC BY-SA 4.0 |