# example of quasi-affine variety that is not affine

Let $k$ be an algebraically closed field. Then the affine plane $\mathbb{A}^{2}$ is certainly affine. If we remove the point $(0,0)$, then we obtain a quasi-affine variety $A$.

The ring of regular functions of $A$ is the same as the ring of regular functions of $\mathbb{A}^{2}$. To see this, first observe that the two varieties are clearly birational, so they have the same function field. Clearly also any function regular on $\mathbb{A}^{2}$ is regular on $A$. So let $f$ be regular on $A$. Then it is a rational function on $\mathbb{A}^{2}$, and its poles (if any) have codimension one, which means they will have support on $A$. Thus it must have no poles, and therefore it is regular on $\mathbb{A}^{2}$.

We know that the morphisms $A\to\mathbb{A}^{2}$ are in natural bijection with the morphisms from the coordinate ring of $\mathbb{A}^{2}$ to the coordinate ring of $A$; so isomorphisms would have to correspond to automorphisms of $k[X,Y]$, but this is just the set of invertible linear transformations of $X$ and $Y$; none of these yield an isomorphism $A\to\mathbb{A}^{2}$.

Alternatively, one can use Čech cohomology to show that $H^{1}(A,\mathcal{O}_{A})$ is nonzero (in fact, it is infinite-dimensional), while every affine variety has zero higher cohomology groups.

For further information on this sort of subject, see Chapter I of Hartshorne’s (which lists this as exercise I.3.6). See the bibliography for algebraic geometry for this and other books.

Title example of quasi-affine variety that is not affine ExampleOfQuasiaffineVarietyThatIsNotAffine 2013-03-22 14:16:39 2013-03-22 14:16:39 Mathprof (13753) Mathprof (13753) 7 Mathprof (13753) Example msc 14-00