By Christian Peskine
Peskine does not supply loads of motives (he manages to hide on 30 pages what frequently takes up part a e-book) and the routines are tricky, however the publication is however good written, which makes it beautiful effortless to learn and comprehend. urged for everybody keen to paintings their approach via his one-line proofs ("Obvious.")!
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Extra info for An Algebraic Introduction to Complex Projective Geometry: Commutative Algebra
Since alA = aA, this implies alA = v ( M ) . From this we deduce a2 = v(a2e2) E v ( M ) = alA and the theorem is proved. 3. 28 If A is a domain and N an A-module, then the set T ( N ) of all x E N such that (0) : x # ( 0 ) is a submodule of N . If ax = 0 and by = 0, then ab(cx + d y ) = 0 for all c, d E A. 29 The module T ( N ) is the torsion submodule of N Consider now, as in the theorem, two decompositions of M : M N m m’ 1 1 ($ A/aiA) $r A N ($ A/a:A)$r’A. They induce isomorphisms m T ( M )N @ A / Q A 1 m’ $A/u:A, 1 and M / T ( M )cv rA N r’A This last relation shows r = r’.
2. Prove 1A(coker ( U ) ) = lA(A/det(u)). ZkTkNmTm = Nm+kTm+',in other words Z k ( F M n N) = P f m Mn N , and the lemma is proved. 0 56 5. 35 (Krull’s theorem) Let A be a Noetherian ring and M afinitely generated A-module. 5. Exercises 57 6. Let R be a principal ideal ring and M a finitely generated R-module. If M’ is a submodule of M , show ~ R ( T ( M5 ) ) ~R(T(M’))~R(T(M/M’)). + Show that if equality holds, then M / T ( M ) N M’/T(M’) @ ( M / M ’ ) / T ( M / M ’ ) Proof Put E = n,Z”M. We have ZnM n E = E for all n 2 0.
5 A function X defined on the categoy of A-modules (resp. finite length A-modules, finitely generated A-modules if A is Noetherian), with value in a n abelian group G, is additive ih for all exact sequences 0 -+ M’ M -% M” + 0, we have X ( M ) = X(M’) + X(Mll). It may be surprising for the reader, but the next result is truly important! The fundamental piece of information is that the commutative diagram presented induces a “canonical map” ker g“ + coker g’, a “connection homomorphism”. That this homomorphism fits nicely in a long exact sequence, makes it even more useful.