COMPLETION AND RELATED CONSTRUCTIONS BRETT NASSERDEN Filtrations,the Artin Reese Lemma and Krull's Theorem Abstract: Pete Clark says I very much appreciate that nding the right bit of structure can make the solutions to your problem self evident. In this paper we will look at some bits of structure, that can at least help make some solutions to certain problems evident. The over arching idea, as I see it, is the construction of new rings and modules from known rings and modules that behave nicely can help us learn about the original ring in suitable circumstances. Specically, we will look at constructions that arise from chains of ideals and or submodules. We start by considering the associated graded rings that arise give descending sequences of submodules and various constructions and prove the Artin Reese Lemma and Krull's Theorem. We then turn to completion, and dene it in two dierent ways. Namely via Cauchy sequences and the inverse limit. After showing the denitions are equivalent we prove some things about completion. By and large we will follow Eisenbud's Commutative Algebra through chapters 5 and 7 and Atiyah McDondald's Introduction to Commutative Algebra with more of the details lled in. Denition. Let fIjgj be a descending sequence of ideals. We call such a sequence a ltration if IiIj ⊆ Ii+j. The n obvious example is the I − adic ltration In = I . Some examples: Take in for concreteness. Then n n . So n . On the other hand, when I = (x) C[x] I = (x ) \nI = (0) is the intersection non-empty? Take R = Z2 Then take I =< (1; 0) >. So (1; 0)n = (1; 0) for all n and so (1; 0) 2 In for all n. This means In = I. We can also make things strictly decreasing. Take, (1; 0) and (0; 2) in R as above. Put I = (1; 0)Z2 and J = (0; 2)Z2. Now dene T = I + J. Note that if i 2 I and j 2 J then ij = 0. Furthermore, In = I n n n but J is strictly decreasing. So \n≥1T 6= (0) and the T are strictly decreasing. Denition. Let M be an R module and suppose that fIng is a descending ltration.A descending sequence fMng of submodules is called an I ltration if IMn ⊆ Mn+1 for all n. We say an I ltration is I stable if for all but nitely many n we have that IMn = Mn+1. n Example. Let M be any R module, and I any ideal. Set Mn = I M. Then Mn is I stable. Denition. Let be an ideal of a ring . We dene 1 n n+1. (Where 0 We make into I R grI R := ⊕n=0I =I I := R) grI R a ring as follows. If a 2 In and b 2 Im then we set (a + In+1) · (b + Im+1) = ab + In+m+1. More generally, let be a ltration, we dene 1 . We make into a graded F := M = M0 ⊇ M1 ⊇ :::: I grF M := ⊕n=0Mn=Mn+1 grF M grI R module in the obvious way. Namely, if in + In+1 2 In=In+1 and mk 2 Mk=Mk+1 then (in + In+1) · (mk + Mk+1) = inmk + Mn+k+1. This is well dened because F is an I ltration. Proposition. Let I be an ideal of a ring R and F and I stable ltration of a R module M. If the sub modules of the ltration are nitely generated then so is the associated graded module over grI R. Proof: There is an n such that for i ≥ n we have IMi = Mi+1. In this case, note that if mi+1 2 Mi+1 then there 2 is some t 2 I and mi 2 Mi with tmi = mi+1. Then working in the graded structure gives (t + I )(mi + Mi+1) = 2 tmi + Mi+2 = mi+1 + Mi+2. That is, (I=I )(Mi=Mi+1) = Mi+1=Mi+2. So let G be the generators of the rst n sub 2 i i+1 modules. So we can write, (I=I )(Mn=Mn+1) = Mn+1=Mn+2. Continuing inductively gives (I =I )(Mn=Mn+1) = Mi+n=Mn+i+1. Since elements of grF M are nite sums, the generators of Mn=Mn+1 and the generators of the rst n sub modules are enough to give everything. Remark. The intuition is that stable ltration's are kind of like nitely generated modules. The above translates the intuition into concrete facts about the graded constructions. Denition. Let M be an R module with a ltration F. Given f 2 M we set s : M ! N[f1g be the map that sends to the largest integer such that and to innity if 1 . Now set f m f 2 Mm f f 2 \n=1Mn in(f) = f + Ms(f)+1 ⊆ grF M if s(f) is nite, and in(f) = 0 otherwise. Given an ideal I of R and a I ltration of M with N ≤ M we dene in(N) to the module generated by the initial forms of elements of N. I.E in(N) =< in(f): f 2 N > in grF M. Example. The in operator can behave in not so nice ways. Consider J =< xy + y3; x2 >⊆ k[x; y] and let I = (x; y). Put the I-adic ltration on k[x; y]. in(J) =< in(f): f 2 J >. Notice that in(x2) = x2 + (x; y)3k[x; y] and in(xy + y3) = xy + y3 + (x; y)2k[x; y] = xy + (x; y)2k[x; y]. So the question is whether, in(J) is generated by the initial forms of the generators of J. But this is not the case as we can look at what the degrees of elements spanned by xy + (x; y)2k[x; y] and x2 + (x; y)3k[x; y]. Now, y5 2 J so y5 + I6k[x; y] 2 in(J) but we cannot generate y5 + I6k[x; y] with our given generators. 1 COMPLETION AND RELATED CONSTRUCTIONS 2 Denition. Let be a ring and an ideal. We set 1 n 2 ∼ . If is a ltration R I BI R := ⊕n=0I = R ⊕ I ⊕ I ⊕ :::: = R[tI] F I then the blow up will be 1 . BF M = ⊕i=0Mi ∼ Lemma. BI R=IBI R := grI R i i i+1 P Proof: Let πi : I ! I =I be the canonical projection. Consider π : πi ! grI R. π is clearly a surjection. Now, suppose that Pn Pn . Then we must have that for each and so k+1. π( k=0 ik) = k=0 πk(ik) = 0 πk(ik) = 0 k ik 2 I k+1 Now, ik 2 I means that ik 2 IBI R and so the sum of the ik are in IBI R. So ker π ⊆ IBI R . On the other hand, any element in IBI R is in the kernel so we have the result. Remark. So we can regard as .(Where Pn k Now, given what is grI R R[tI]=IR[tI] R[tI] = f k=0 ikx : ik 2 Ikg) f 2 R in(f) intuitively? Well, if in(f) 6= 0 let in(f) = f + In+1. On the otherhand, in BI R we have that f can be thought of as the element, n. So in this is the element Pn j . Notice that Pn j . f + fx + :::: + fx grI R j=0 fx + IR[tI] j=0 fx 2= IR[tI] So in this sense, in(f) = xnf + IR[tI] and the initial term picks o the highest order the element f in the blow up algebra, and then projects this down into the graded ring. So in this sense, the in operator behaves somewhat like picking o the highest order term in a monomial ordering. ( If the reader is not familiar with monomial orders and wishes to learn then Eisenbud's chapter 15 treats them. For an easier read, there is Ideals,Varieties, and Algorithms by David A. Cox, John B. Little and Don O'Shea.) Note. The blow up algebras have geometric interpretations and roles. For example, see Eisenbud 5.2 for a brief discussion. The proposition below is interesting, it takes a question about the stability of ltrations of nitely generated modules, and then relates this to a module theoretic question about the blow up ring and blow up algebra. Since we have a large body of theory to draw upon when dealing with modules this can be useful. Proposition. Let R be a ring and I an ideal. Let M be a f.g. R module with I ltration F by nitely generated R modules. F is stable if and only if BF Mis a nitely generated BI R module. Proof: Suppose that BF M is nitely generated. We can take a nite set of homogenous elements of the Mi for i ≤ n (for some n) with the homogenous elements also generating each Mi. Now, let m 2 Mn+i for some i > 0. Then P there are some generators mj;k with mj;k 2 Mj for j ≤ n. Then, if rj;kmj;k = m we know, because of the graded n+i−j structure, that rj;kmj;k 2 Mn+i. That is, rj;k 2 I . But then, using the grading we can look at this as a sum of elements in multiplied by elements in i. So we have that, as a module, generates 1 or that Mn I BI R Mn ⊕i=0Mn+i i Mn+i = I Mn so that the ltration is I stable.