INTRODUCTION TO ALGEBRAIC GEOMETRY-page64

INTRODUCTION TO - I is finite This is enough for our applications since we can always choose a finite covering subfamily The proof of injectivity

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60 LECTURE 7. MORPHISMS OF PROJECTIVE ALGEBRAIC VARIETIES Let us now state and prove the lemma. Recall first that for any ring A a local line M P n ( A ) defines a collection { M a i } i I of lines in A n +1 a i for some covering family { a i } i I of elements in A . Let us see how to reconstruct M from { M a i } i I . We know that for any i, j I the images m i of m M in M a i satisfy the following condition of compatibility: ρ ij ( m i ) = ρ ji ( m j ) where ρ ij : M a i M a i f j is the canonical homomorphism m/a r i mf r j / ( a i f j ) r . For any family { M i } i I of A a i -modules let limind i I M i = { ( m i ) I Y i I M i : ρ ij ( m i ) = ρ ji ( m j ) for any i, j I } . This can be naturally considered as a submodule of the direct product Q i I M i of A -modules. There is a canonical homomorphism α : M limind i I M a i defined by m ( m i = m ) i I . Lemma 7.1. The homomorphism α : M limind i I M a i is an isomorphism. Proof. We assume that the set of indices
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Unformatted text preview: I is finite. This is enough for our applications since we can always choose a finite covering subfamily. The proof of injectivity is similar to the proof of Lemma 5.11 and is left to the reader. Let us show the surjectivity. Let ( m i a n i i ) i ∈ I ∈ limind i ∈ I M a i for some m i ∈ M and n i ≥ . Again we may assume that all n i are equal to some n . Since for any i, j ∈ I ρ ij ( m i a n i ) = ρ ji ( m j a n j ) , we have ( a i a j ) r ( a n j m i-a n i m j ) = 0 for some r ≥ . Let p i = m i a r i , k = r + n . Then m i ra n i = p i a k i , f k j p i = a k i p j ....
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This note was uploaded on 01/08/2012 for the course MATH 299 taught by Professor Wei during the Spring '09 term at SUNY Stony Brook.

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