1
$\begingroup$

Let $k$ be an algebraically closed field and let $\mathbb P^n_k=\text{Proj}(k[x_0,x_1,...,x_n])$ .

If $n\ge 2$, and $p\in \mathbb P^n_k$ is a closed point, then can $\mathbb P^n_k\setminus \{p\}$ be a Projective variety ?

Considering global section ring doesn't give any contradiction because since $\{p\}$ is a closed subset of codimension $\ge 2$, so $\mathcal O(\mathbb P^n_k\setminus \{p\})\cong \mathcal O(\mathbb P^n_k)\cong k$. Similarly, the Picard group or class group also doesn't give any contradiction.

Please help.

$\endgroup$

1 Answer 1

5
$\begingroup$

No. As any map $f:X\to Y$ with $X$ a projective variety is closed, if $\Bbb P^n_k\setminus\{p\}$ were projective, then it's image under the natural embedding in $\Bbb P^n_k$ would be closed. But it's not (if it were, $\Bbb P^n_k$ would be reducible, which is obviously nonsense).

$\endgroup$
1
  • 2
    $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$
    – KReiser
    Commented Jul 25, 2020 at 19:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .