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Read e-book online An Algebraic Introduction to Complex Projective Geometry: PDF

By Christian Peskine

ISBN-10: 0521108470

ISBN-13: 9780521108478

ISBN-10: 0521480728

ISBN-13: 9780521480727

Peskine does not provide loads of reasons (he manages to hide on 30 pages what frequently takes up part a e-book) and the routines are difficult, however the ebook is however good written, which makes it beautiful effortless to learn and comprehend. suggested for everybody prepared to paintings their manner via his one-line proofs ("Obvious.")!

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Extra info for An Algebraic Introduction to Complex Projective Geometry: Commutative Algebra

Example text

Fraction modules 7. 11 Let S be a multiplicatively closed part of the ring A and M an A-module. We denote by S-lM the quotient of M x S by the equivalence relation ( 2 ,s ) ( y ,t ) if there exists r E S such that r(xt - ys) = 0 and b y x / s E S-'M the class of ( x ,s ) . - Proof (i) and (ii) are obvious. We show (iii). If x / s E ker S-lq5, then ~ ( x ) /=s 0. Hence there exists t E S such that tq5(x) = 0 = q5(tx). Let y E M be such 0 that t x = +(y). We have x / s = $ ( y / s t ) . 12 (i) The operations x / s y / t = ( t x sy)/st (for x , y E M and s , t E S ) and (a/s)(x/t)= ab/st (for x E M , a E A and s , t E S ) are well defined.

27 Let M be an A-module. The following conditions are equivalent: (i) M 87 = 0; 0 (ii) Supp(M) = 0; (iii) Suppm(M) = 0. 30 If M as a finately generated A-module, then Supp(M) as a closed set of Spec(A) for the Zarzska topology. 7. 3. Support of a module Proof The set defined by the ideal ((0) : M) is closed. 31 Show that the support of the Z-module Q/Z is not a closed set of Spec@) for the Zariski topology. 32 A finitely generated A-module M such that Mp is a free Ap-module for all P E Spec(A) is called locally free.

Using (*) twice, we find + is different from zero. Note next that (iii) + (iv). By (*), lA(HomA(M,D ) ) 5 ~ A ( Mfor ) all finitely generated A-modules M. If M is finitely generated, there exists an integer n and an exact sequence 0 + K 4 nA 4 M -+0. But a non-zero homomorphism from A I M to A I M is obviously an isomormorhism. We have proved that eD,A/M is an isomorphism. Now, if 1 ~ ( h f= ) 1, there exists a maximal ideal M such that M N A I M . This shows that eD,M is an isomorphism. We can now prove, by induction on ~ A ( M )that , the evaluation homomorphism eD,M : M HomA (HOmA ( M ,D ), D ) This induces an exact sequence 0 ---f HomA(M, D ) + HomA(n-4, D ) 4 HomA(K, D ) -+ is an isomorphism for all finitely generated modules M .

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An Algebraic Introduction to Complex Projective Geometry: Commutative Algebra by Christian Peskine


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