By Christian Peskine

ISBN-10: 0521108470

ISBN-13: 9780521108478

ISBN-10: 0521480728

ISBN-13: 9780521480727

Peskine does not supply loads of reasons (he manages to hide on 30 pages what frequently takes up part a booklet) and the workouts are difficult, however the e-book is however good written, which makes it lovely effortless to learn and comprehend. urged for everybody keen to paintings their means via his one-line proofs ("Obvious.")!

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

Example text

28, f is an isomor0 Proof Since f is injective, we have Supp(M/(QKi)) c lJi Supp(M/Ki). But Supp(M/Ki) c Supp(M/(niKi)) is obvious and we find Supp(M/(niKi)) = UiSupp(M/Ki). Consider P E Supp(M/(niKi)). There is a unique integer j such that P E Supp(M/Kj). 38 Let A be a Noetherian ring with only finitely many maximal ideals. If M is a finitely generated A-module such that M M is a free AMmodule of rank n for all maximal ideals M of A, then M is a free A-module of rank n. , r , be the maximal ideals of A.

This shows that the natural surjective homomorphism A/Z an isomorphism. 4. Dualizing module on an artinian ring 6. A first contact with homological algebra (v) for all injective homomorphisms N M of finitely generated A-modules, the natural homomorphism HomA(M, D ) + HomA(N, D) is surjective and for all maximal ideals M of A, one has A I M 11’ HomA(A/M, D ) . Pro0f (i) + (ii). Assume D is a dualizing A-module. 21 Let A be an artinian ring. A finitely generated A-module D is dualizing if the natural evaluation application shows that D is faithful.

X T / l E Mp generate this Ap-module. 9, that there exist prime ideals Pi and an increasing sequence (0) = MOc M1 c ... c Ml = M of submodules of M such that Mi/MiPl N Alpi. Clearly Pi E Supp(M). If Pi is maximal for all i , it is a composition series of M, which is therefore of finite length. U Our next result will be particularly useful when studying quasi-coherent sheaves. We believe nevertheless that this is the right time to state it and prove it. 37 Let M be an A-module. -0. We assumed (M/N)p = (0).

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


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