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Graded Rings and Dimension ALGBOOK GRADING 1.1 29. april 2002 Graded rings and dimension 1. Gradings and filtrations. 1 n (1.1) Definition. A ring A is graded if it is the direct sum !!©n=0An of subgroups An such that AmAn ⊆ Am+n for all natural numbers m and n. We say that an 1 n A-module M is graded if it is the direct sum !©n=0Mn! of subgroups Mn such that AmMn ⊆ Mm+n for all natural numbers m and n. The elements in An and Mn are called homogeneous of degree n. When x 2 M Pm and x = n=1 xn with xn 2 Mn we call the elements x1; x2; : : : ; xm the homogeneous components of x. For all negative integers n we let An = 0 and Mn = 0. 1 An ideal a of A is homogeneous if a = ©n=0an with an ⊆ An. (1.2) Remark. It follows from the definitions that A0 is a ring, and that Mn is an A0-module for each n. We have that A is an A0-algebra via the inclusion of A0 in A. An ideal is homogeneous if and only if it is generated by homogeneous elements. (1.3) Example. Let A be a ring and A[t®]®2I be the polynomial ring in the variables ¹ Q ¹(®) (I) Pft®g®2I for some index set I. Then the elements t = ®2I t® with ¹ 2 N and ®2I ¹(®) = n generate an A-module (A[t®]®2I )n, and A[t®]®2I is a graded ring with homogeneous elements (A[t®]®2I )n of degree n. Pm Let a ⊆ A[t®]®2I be an ideal such that for each element f = i=1 fi in a with 1 fi 2 Ai we have that fi 2 a. Then a = ©n=0an, with an = a \ An, is a graded A[t®]®2I -module. That is, the ideal a is homogeneous. Moreover we have that 1 A=a = ©n=0Ai=ai is a graded ring. n (1.4) Example. Let A be a ring and a an ideal in A. Then the direct sum !!Ra(A) = 1 n n 0 ©n=0a of the ideals a , with a = A, is a graded ring where the multiplication of elements in am with elements in an is given by the multiplication in A. The ring Ra(A) is called the Rees-ring of a, or the Rees-algebra of a when it is considered as an A0-algebra. 1 n Let M be an A-module. We have that the direct sum Ra(M) = ©n=0a M is an m Ra(A)-module, where the multiplication of the elements of a with the elements in anM is defined by the operation of A on M. n (1.5) Example. Let A be a ring and a an ideal in A. The direct sum !!Ga(A) = 1 n n+1 n n+1 0 ©n=0a =a of the A=a-modules a =a , with a =a = A=a, is a graded ring. To m m+1 n n+1 define the multiplication we let gm 2 a =a and gn 2 a =a be the classes of grading1 ALGBOOK GRADING 1.2 29. april 2002 m n m+n m+n+1 fm 2 a , respectively fn 2 a . We define gmgn 2 a =a as the class of the m+n element fmfn 2 a . It is clear that the definition is independent of the choice of the representatives fm and fn of the classes gm, respectively gn. We call the ring Ga(A) the graded ring of a, or the graded algebra of a when we consider Ga(A) as an A=a-algebra. 1 n n+1 Let M be an A-module, and let Ga(M) = ©n=1a M=a M. Then Ga(M) is an m m+1 Ga(A)-module. To define the operation of Ga(A) on Ga(M) we let g 2 a =a and y 2 anM=an+1M be the classes of f 2 am respectively of x 2 anM. Then we define the product gy 2 am+nM=am+n+1M as the class of fx 2 am+n. It is clear that the product is independent of the choice of the representatives f and x for the classes in am=am+1 respectively anM=an+1M. 1 (1.6) Proposition. Let A = ©n=0An be a graded ring. The following two condition are equivalent: (1) The ring A is noetherian. (2) The ring A0 is noetherian and A is a finitely generated A0-algebra. Proof. (2) ) (1) It follows immediately from The Hilbert Basis Theorem (CHAINS ! 2.10) that, when condition (2) is fulfilled, then condition (1) is fulfilled. (1) ) (2) We have that the ring A0 is isomorphic to the residue ring of A modulo 1 ! the ideal A+ = ©n=1An. Hence it follows from Corollary (CHAINS 2.6) that A0 is noetherian. Since A is noetherian we have that A+ is a finitely generated A-module. If we, if necessary, take all the homogeneous components of a finite set of generators for A+, we obtain homogeneous generators f1; f2; : : : ; fm of the A-module A+. Let fi 2 Api and write B = A0[f1; f2; : : : ; fm]. We shall show that A = B. To show that A = B it suffices to show that An ⊆ B for all n 2 N. This is proved by induction on n. For n = 0 it is clear. Assume that An¡1 ⊆ B. For every element g 2 An we have Pm that g = i=1 gifi with gi 2 An¡pi . Since pi > 0 for i = 1; 2; : : : ; m it follows from the induction assumption that gi 2 B for i = 1; 2; : : : ; m. Hence we have that g 2 B and we have proved that An ⊆ B. Hence we have that A = B, and A is a finitely generated A0-algebra. (1.7) Definition. Let A be a ring and a an ideal of A. Moreover let M be an nn A-module. A filtration !! fMngn2N of M is a sequence of submodules !!M = M0 ¶ M1 ¶ M2 ¶ ¢ ¢ ¢ of M. The filtration is an a-filtration if aMn ⊆ Mn+1 for all n. We say that an a-filtration is a-stable if aMn = Mn+1 for all sufficiently large n. For all sufficiently large n means that there is an integer m such that the property holds for n ¸ m. (1.8) Remark. Let M be an A-module and let fMngn2N be an a-stable filtration, and m an integer such that aMn = Mn+1 for n ¸ m. Then we have that Mm+n = n a Mm for n = 0; 1;::: . ALGBOOK GRADING 1.3 29. april 2002 (1.9) Example. Let A be a ring and a an ideal in A. Moreover let M be an a-module. Then we have a filtration M = a0M ¶ aM = a1M ¶ a2M ¶ ¢ ¢ ¢ of M. This filtration is a-stable. (1.10) Example. Let A be a ring and let a be an ideal in A. Moreover let M be an 1 A-module and let fMngn2N be an a-filtration. Then the direct sum ©n=0Mn of the 1 n m A-modules Mn is a graded Ra(A) = ©n=1a -module. The product of f 2 a with m x 2 Mn is fx 2 a Mn ⊆ Mm+n. (1.11) Example. Let A be a ring and a an ideal of A. Moreover let M be an 1 A-module, and let fMngn2N be an a-filtration. Then the direct sum ©n=0Mn=Mn+1 1 n n+1 of the A=a-modules Mn=Mn+1 is a graded Ga(A) = ©n=0a =a -module. The m m+1 m product of the class in a =a of f 2 a and the class in Mn=Mn+1 of x 2 Mn is the class in Mm+n=Mm+n+1 of fx 2 Mm+n. Again the definition of the product is independent of the representatives of the classes of f and x. n 1 n n+1 When Mn = a M for n = 0; 1;::: we obtain that ©n=0a M=a M is the Ga(A)- ! module Ga(M) defined in Example (1.5). (1.12) Lemma. Let A be a ring and a an ideal in A. Moreover let fMngn2N and 0 fMngn2N be a-stable filtrations of an A-module M. Then there is a positive integer m such that 0 0 Mm+n ⊆ Mn and Mm+n ⊆ Mn for n = 0; 1;:::: 0 0 n n Proof. Since aMn ⊆ Mn+1 and aMn ⊆ Mn+1 we have that a M ⊆ Mn and a M ⊆ 0 0 Mn for n = 0; 1;::: . The filtrations fMngn2N and fMngn2N are a-stable. Hence we 0 n 0 n 0 can find natural numbers p; p in N such that Mn+p = a Mp and Mn+p0 = a Mp0 0 0 n+p0 n 0 for n ¸ p + p . Let m = p + p . Then we have that Mm+n = a Mp ⊆ a M ⊆ Mn 0 n+p 0 n and Mm+n = a Mp0 ⊆ a M ⊆ Mn for n = 0; 1; 2;::: . (1.13) Lemma. Let A be a noetherian ring and let a be an ideal in A. Moreover let M be a finitely generated A-module and let fMngn2N be an a-filtration. The following conditions are equivalent: 1 (1) The graded Ra(A)-module ©n=0Mn is finitely generated, where Ra(A) = 1 n ©n=0a .
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