# 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 |
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Canonical name | ExampleOfQuasiaffineVarietyThatIsNotAffine |

Date of creation | 2013-03-22 14:16:39 |

Last modified on | 2013-03-22 14:16:39 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 7 |

Author | Mathprof (13753) |

Entry type | Example |

Classification | msc 14-00 |