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Let $X$ be a smooth scheme of finite type over an algebraically closed field of characteristic zero and with a trivial class group $Cl(X)=0$. Let $Y$ be a dense open subscheme of $X$ such that:

1) $Y$ is a quasi-affine scheme s.t. $\Gamma(\mathcal{O}_Y,Y)$ is of finite type; and

2) $X \setminus Y$ is irreducible of codimension at least two.

Does these conditions imply that $X$ is itself a quasi-affine scheme? If not, can anyone provide a counter-example please?

Thank you in advance!

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  • $\begingroup$ Do you have an example of a finite type scheme as in your question with class group zero, but not quasi affine? $\endgroup$
    – Mohan
    Commented Apr 22, 2016 at 0:40
  • $\begingroup$ Actually no. I'm aware that there are complete toric varieties with trivial Picard group (Eickelberg "Picard groups of compact toric varieties..." 1993). But they don't have a trivial class group. $\endgroup$
    – sabrebooth
    Commented Apr 22, 2016 at 6:19
  • $\begingroup$ For a smooth variety, isn't Picard and class groups the same? $\endgroup$
    – Mohan
    Commented Apr 22, 2016 at 12:32
  • $\begingroup$ Yes but the complete toric varieties that I refereed to are not smooth. $\endgroup$
    – sabrebooth
    Commented Apr 22, 2016 at 12:57

1 Answer 1

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Here is a counterexample. Unfortunately it is not separated, so I don't know how interesting it is to you.

Consider $X = \mathbf{A}^2 \cup \mathbf{A}^2$, glued along $\mathbf{A}^2 \setminus \{(0,0)\}$, and let $Y$ be one copy of $\mathbf{A}^2$. $X$ is not quasi-affine since it is not separated. $\operatorname{Cl}(X) \cong \operatorname{Cl}(Y) = 0$ since the complement of $Y$ in $X$ is a point, which has codimension 2. Also, $Y$ is affine, and $\Gamma(Y,\mathcal{O}_Y) = k[x,y]$ is of finite type.

Finally, some remarks which I commented earlier:

  1. Schröer has an example of a complete normal variety with $\operatorname{Pic}X = 0$, but I didn't compute its class group since you wanted a smooth example.
  2. Hamm and Lê show that a complex algebraic variety with $H^1 = H^2 = 0$ (the actual condition is weaker) would have trivial class group, so this might be a place where you could find a counterexample that is also a variety.
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    $\begingroup$ Thank you for the counter-example, this is enlightening. However, it turns out that if you also ask the scheme $X$ to be separated, then it has to be quasi-affine. Actually, any normal variety with a trivial class group is a quasi-affine variety. This can be seen for instance by using the notion of ample family of line bundles. A normal variety with trivial class group always has such an ample family which turns out to be a singleton (since the Picard group is trivial), and thus the structure sheaf itself is ample, which is equivalent to being quasi-affine. $\endgroup$
    – sabrebooth
    Commented Apr 30, 2016 at 7:56

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