Can the complement of a projective variety be isomorphic to an affine variety?

Question

Let's say I have a Projective variety and take its complement. This should be open, since the variety is closed. Now is it possible for this to be isomorphic to an affine variety? Since affine varieties are closed, but the complement described above is open, and morphisms are continuous, shouldn't closed sets only map to closed sets in an isomorphism?

Answer 1

An isomorphism between two varieties is just an isomorphism between those varieties, not between their ambient spaces. So if you have an open subvariety $U\subseteq\mathbb{P}^n$ and a closed subvariety $C\subseteq\mathbb{A}^n$, an isomorphism between $U$ and $C$ is just a morphism $U\to C$, not a morphism $\mathbb{P}^n\to\mathbb{A}^n$. So, there is no reason that such an isomorphism would force $C$ to be open in $\mathbb{A}^n$ or $U$ to be closed in $\mathbb{P}^n$.

For example, note that for any $n$, $\mathbb{A}^n$ is the complement of the hyperplane $H$ at infinity in $\mathbb{P}^n$. So, $\mathbb{P}^n\setminus H$ is isomorphic to $\mathbb{A}^n$ even though $\mathbb{A}^n$ is closed in $\mathbb{A}^n$ but $\mathbb{P}^n\setminus H$ is not closed in $\mathbb{P}^n$.

Answer 2

Let $X\subset\mathbb P^n$ be a closed, irreducible projective subvariety and $U=\mathbb P^n\setminus X$ its open complement. Then:

$U$ is isomorphic to an affine variety $\iff$ $\dim X=n-1$