By Brian Osserman

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10 Z has a unique irreducible component Z containing P , which has dimension equal to that of X. It then follows that X must also contain Z , and choosing U to be the complement of any other components of X or Z, we have X ∩ U = Z ∩ U = Z ∩ U , as desired. 13. When one studies this sort of question, there are two different strengths with which one can formulate the question. The weaker version is to look for f1 , . . , fc such that X = Z(f1 , . . , fc ). The stronger one is to actually ask that f1 , .

Suppose we have U1 = U2 = A1k , and set U = A1k (0). We consider two different possibilities for gluing U1 to U2 along U to obtain a prevariety. The first is to let X be the union of U1 and U2 glued along U , where we identify U ⊆ U1 and U ⊆ U2 simply by the identity map. In this case, X is almost the same as A1k , except that now it has two copies of the origin instead of one. In the usual real or complex topology, this satisfies the conditions to be a manifold except that it is not Hausdorff.

Now, V is defined by gluing two copies of Y = A1k (0) to each other via the map t → 1/t. Denote the two copies of Y by V1 and V2 . Recalling that Y is affine, isomorphic to Z(x, y) ⊆ A2k , we see that the Vi give an atlas for V as a prevariety. To construct an isomorphism to A1k , consider the regular function defined by 1/(t − 1) on V1 , and by 1/(1/t − 1) = t/(1 − t) on V2 . This defines a morphism to A1k , with inverse morphism defined by sending x to (x + 1)/x in V1 for x = 0, and to x/(x + 1) in V2 for x = −1.