By Akhil Mathew
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It will be convenient to prove this more generally for ringed spaces. Recall that Sh(X) consists of the sheaves of modules over the constant sheaf of rings Z on X. 2. Let (X, OX ) be a ringed space. Then the abelian category Mod(OX ) has enough injectives. Proof. It is a well-known fact of algebra that the category of A-modules for any ring A has enough injectives. Let F ∈ Mod(OX ). Then we can imbed each Fx into an injective Ox -module Mx . 3. Let Mx be an injective Ox -module and G(x) the skyscraper sheaf of Mx at x (equivalently, the push-forward).
Then M → N is an isomorphism. 6. LOCALLY FREE SHEAVES 37 Proof. Indeed, M → N is surjective by Nakayama. But M and N are isomorphic. It is a theorem in commutative algebra (cf. ) that a map of a finitely generated module into itself which is surjective is an isomorphism. We now prove the proposition. Proof. We have Lx ⊗Ox Nx = Ox at each x. So Lx is free of rank one for each x ∈ X by the first lemma. Fix now x ∈ X and a generator r ∈ Lx . This means that r extends to a generator of L in a neighborhood of x.
Then we define Spec(A) as the set of prime ideals of A. Recall that the points of an algebraic variety are in bijection with the maximal ideals of the coordinate ring. This is thus a more general definition. For p ∈ SpecA, we write Ap for the localization and κ(p) = Ap /pp for the residue field. 1. The topology on SpecA. Fix a commutative ring A. We now define a topology on SpecA. 2. For a ⊂ A an ideal, we define V (a) as the set of primes containing a. We could also define V (S) for S ⊂ A a set in the same way; it is clear that V (S) is the samet thing as V of the ideal generated by S.