I have already proved that the projection map $p \colon \mathbb{A}^{n+1}_\mathbb{C}\setminus \{0\} \to \mathbb{P}^n_\mathbb{C}$ given by $(x_0,x_1,..x_n) \rightarrow [x_0,x_1,..,x_n]$ is a morphism. But a further question arises that whether this morphism is closed, i.e. it maps closed sets to closed sets (assuming equipped with Zariski topology).

The answer is negative, as Daniel Schepler has pointed out.

Moreover, this exercise is followed by another one, asking whether $\mathbb{C}^{n+1} \setminus \left\{ 0 \right\}$ is isomorphic to $\mathbb{P}^n \times \left( \mathbb{C} \setminus \left\{ 0 \right\} \right)$. Is there any correlation between these two problems? Any advice is welcomed and appreciated.

  • 7
    $\begingroup$ As a hint: if you take the closed subset $Z(xy-1)$ of $\mathbb{A}^2_{\mathbb{C}}$, what is its image in $\mathbb{P}^1_{\mathbb{C}}$? $\endgroup$ Nov 20 at 19:47
  • 1
    $\begingroup$ That description isn't independent of the choice of homogeneous coordinates $[x : y]$ for a point in $\mathbb{P}^1_{\mathbb{C}}$. So for example, $(i, -i)$ is in the given subset of $\mathbb{A}^2_{\mathbb{C}}$, and it has image $[i : -i] = [1 : -1]$. $\endgroup$ Nov 20 at 20:28
  • 2
    $\begingroup$ So, the image is the set of $[x : y]$ such that for some $(x', y') \in \mathbb{A}^2_{\mathbb{C}} \setminus \{0\}$, $x'y' - 1 = 0$ and $[x' : y'] = [x : y]$. You can show that's equivalent to $xy \ne 0$, and conclude that the image is $\mathbb{P}^1_{\mathbb{C}} \setminus \{ [0 : 1], [1 : 0] \}$. $\endgroup$ Nov 20 at 20:37
  • $\begingroup$ @DanielSchepler Thanks a lot for your elaboration. $\endgroup$ Nov 20 at 20:40


You must log in to answer this question.