By Brian Osserman

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Proof: Given an open subset U ⊂ X we define OX (U ) as the k– algebra of all the functions f : U → k such that for all i ∈ I, f |U ∩Ui ∈ OUi (U ∩ Ui ). It is clear that OX is a sheaf, and it follows from the very definition that OX (Ui ) = OUi (Ui ). The uniqueness is also clear and the assertion about the stalks follows from the fact that locally we are dealing with affine varieties whose stalks are local rings. 29. (1) If there is no danger of confusion we omit the reference to the base field k, and refer to affine atlas and algebraic varieties instead of affine k–atlas and algebraic k–varieties.

See Exercise 18. 26. Since the ideals of k[X1 , . . , Xn ]/I correspond to the ideals of k[X1 , . . , Xn ] that contain I, the closed subsets of X in the Zariski topology correspond to the ideals in k[X]. In particular, the points in X correspond to the maximal ideals of k[X]. 11 is Xf : f ∈ k[X] . 27. In the case that X and Y are abstract sets and F : X → Y is a function, define a k–algebra homomorphism F # : kY → kX as F # (f ) = f ◦F . The following definition of morphism between algebraic sets generalizes and is motivated by the construction of k[X].

Xn ] form a basis for the Zariski topology of X. Proof: The proof of this result is left as an exercise (see Exercise 7). 10, the Zariski topology in general is not Hausdorff. In fact, an algebraic set is Hausdorff if and only if it is a finite collection of points (see Exercise 8). We leave as an exercise the proof that algebraic sets are quasi–compact (see Exercise 9). 13. The Zariski topology when restricted to an arbitrary algebraic set of an affine space is noetherian. Proof: Clearly it is enough to prove this result for An .