An Algebraic Introduction to Complex Projective Geometry pdf

[...]... functor F from an abelian category C to another abelian category, C’, transforms an object M of C into an object F(M) of C’ and an arrow of C into an arrow of C’ The functor is called covariant if it preserves the directions, in other words if the transform of an arrow f : M + N is an arrow F(f) : F(M) -+ F(N) It is called contravariant if it inverts these directions, in other words if the transform... exists b E A such that b is not nilpotent and that ab = 0 Show that A is Artinian 4 Let M be an Artinian module Show that an injective endomorphism of M is an automorphism of M ; 44 4 Artinian rings and modules 5 Let A be an Artinian ring, M a finitely generated A-module and u an endomorphism of M Show that the ring A[u] is Artinian 6 Let A be a local Artinian ring Show that if a homomorphism u : A"... and (iii) =+ (i) is an immediate consequence of Theorem 5.18 0 Corollary 5.22 Let A be an Artinian ring If an A-algebra B is finitely generated us an A-module, a is an Artinian ring t Proof Let r be the rank of E as a K-vector space There exists an isomorphism of K-vector spaces E N K‘ Since this is an isomorphism of k-vector spaces, we are done 0 Exercise 5.25 Let k be a field and A a k-algebra such... Fs 1 be an increasing sequence of submodules of MIN Put M, = Ni for i 5 r and, for i = 1, , s, let MT+ibe the submodule of M , containing N , and such that M,+i/N = Fi We have 1~(Mi) 3 If A is an Artinian ring and M a free A-module of rank n, then ~ A ( M ) = nlA ( A ) Proposition 5.24 Let K be a field and k c K a subfield such that K is a finite rank k-vector space If E is a finite rank K-vector space,... 4 Artinian rings and modules 41 4.1 Artinian rings Proof Since A is Artinian, there are positive integers ni such that MY 3 The ring @[X, Y]/(X2,Y2,X Y ) is Artinian, but it has infinitely many ideals = MY+' Let us show (0) = M y ' M y M F To understand the last example, note first that @[X,Y]/(X2,Y2,XY) is a @-vector space of rank 3 and that (l,cl(X),cl(Y))is a basis of this vector space Check next... is a domain, x is not a zero divisor and 1 = ax T My'M? MF MY, 1 by Lemma 1.61, and we are done 0 We can now prove the main result of this chapter Corollary 4.6 In an Artinian ring all prime ideals are maximal Theorem 4.9 A ring is Artinian if and only if it is Noetherian and all its prime ideals are maximal Proof Let P be a prime ideal The ring A l p is an Artinian domain, hence a n field In other... M,/MylMY MF-l, for 1 I i 5 r if : 42 4.2 Artinian modules 4 Artinian rings and modules 4.2 nT > 1 and for 1 5 i < r if n, = 1 Furthermore, there is, in A' a relation (0) = MY'MP M;?' If A is Artinian or Noetherian, so is A' Therefore, by the induction hypothesis, A' is both Artinian and Noetherian Consider now Z an increasing sequence of ideals of A It induces an in, creasing sequence Z,A' of ideals of... of rank n - 1 if and only if M has an invertible l-minor U 7 Let U be an endomorphism of a finitely generated free A-module If N = coker U , show that there exists a principal ideal having the same radical as the annihilator (0) : N of N (I know that I have already asked this question, but this is important and I want to be sure that you remember!) 8 Let L be a finitely generated free A-module and... n 2 m 3.2 Noetherian UFDs Theorem 3.8 A Noetherian domain A is a UFD if and only if any irreducible element of A generates a prime ideal Proof It suffices to show that in a Noetherian domain A any non-zero element is a product of units and irreducible elements First note that if aA = a'A, then a is a product of units and irreducible elements if and only if a' is such a product We can therefore consider... exists a prime ideal P' with a E P' and P'JaA E Ass(A/aA) and such that P c P' Assume that b E A is such that cl(b) E A/aA is not a zero divisor and prove that b is not a zero divisor 6 Let A be a Noetherian ring and a E A Assume (an+ 1): an large enough Show that a is not a zero divisor = ( a ) for n 38 3 Noetherian rings and modules 7 Let A be a Noetherian ring and a E A Show that the subring A[aT] . Grothendieck and Dieudonnk. With this book, I want to prepare systematically the ground for an algebraic introduction to complex projective geometry. . use in geometry. As indicated in the title, I maintain throughout the text a view towards complex projective geometry. In many recent algebraic geometry