# $\mathbb A^n$ is not a complete quasiprojective variety

Suppose that $\mathbb K$ is an algebrically closed field. In the course of Algebraic Geometry I've attended we gave the following definition (in what follows every space is to be intended on the field $\mathbb K$):

Definition. A set $X \subset \mathbb P^N$ is said to be complete iff for every quasiprojective variety $Y$ the projection $\pi_2 \colon X \times Y \to Y$ is a closed map.

I would like to prove (as an exercise) that the affine space $\mathbb A^n$ is not complete.

If $n=1$ this is very easy: we take $Y=\mathbb A^1$ and we consider the closed subset $Z:=V(x_1x_2-1) \subset \mathbb A^{1}\times \mathbb A^{1}$; the projection onto the second factor $\pi_2(Z) \simeq \mathbb A^{1} \setminus \{0\}$ which is not Zariski closed.

Is there an easy trick to adapt this argument to higher dimensional case? How would you prove it?

Thanks.

• Hint: a closed subset of a complete algebraic set is again complete. Now use your example of A^1 above... – user64687 Jul 9 '13 at 16:15
• So if I'm not wrong you are saying: once you have prove that $\mathbb A^1$ is not complete you're done, since $\mathbb A^1$ can be seen as a closed subset of $\mathbb A^n$, $n \ge 2$ (e.g. $\mathbb A^1 \simeq V(x_n) \subset \mathbb A^n$). Am I right? Thank you very much. – Romeo Jul 9 '13 at 16:18
• Your example in $\Bbb{A}^1$ can be seen graphically: You're just projecting the hyperbola $xy = 1$ in the plane onto the $y$ - axis. Now how do you do this in higher dimensions? – user38268 Jul 9 '13 at 16:19
• Romeo: yes, that's exactly what I had in mind! Of course, there is a more direct argument, as @BenjaLim suggests. – user64687 Jul 9 '13 at 16:20
• @BenjaLim I think it is the projection onto the hyperplane $x_1=0$ of the hyperbola $x_1x_{n+1}=1$. But how can I prove that this projection is not closed? – Romeo Jul 9 '13 at 16:22

Here's a generalization of your idea in three dimensions. We let $X = \Bbb{A}^2$ and $Y = \Bbb{A}^1$. Consider $V(zy -1) \subseteq X \times Y = \Bbb{A}^3$ and think of $X$ as being the $y - z$ plane and $Y$ as the $x$ - axis. Then projection onto $Y$ is isomorphic to $\Bbb{A}^1 - \{0\}$ which is not Zariski closed (because $k$ - algebraically closed guarantees that $\Bbb{A}^1 - \{0\}$ is not a finite set of points). Now generalize to higher dimensions.
• Oh, I see. Thank you very much. In higher dimensions I think one can take $V(x_nx_{n+1}-1) \subset \mathbb A^{n} \times \mathbb A^{1} = \mathbb A^{n+1}$ and think of $X$ as the hyperplane $x_1=0$. The projection onto $X$ is still isomorphic to $\mathbb A^{1} \setminus \{0\}$ and we are set. Thank you very much. – Romeo Jul 9 '13 at 16:40