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.

单词	ExampleOfQuasiaffineVarietyThatIsNotAffine
释义	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.
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