Lectures 9-10 I described completion in general terms last time, and gave a major consequence of it (Hensel's lemma). Now I treat it more formally. A directed set La is a p.o. set so that any two la, mu in La have a bound nu in La, with la, mu <= nu. The case La = NN would be perfectly adequate and in practice is the main one. Let A be a ring and M an A-module. The starting point is a set {M_la}_{la in La} of submodules of M indexed by a directed set La, with M_mu < M_la for every mu > la. (Finer and finer as mu gets bigger.) Lemma (1) There is a topology on M ("linear topology") determined by (a) the {M_la} form a basis for the neighbourhoods of 0, and (b) the module operations are continuous. (2) If we give the quotients M/M_la the discrete topology, the quotient maps M -> M/M_la are continuous. (3) The topology is separated (Hausdorff) iff the intersection of all M_la is zero. Pf. Requiring addition by x in M and by -x in M to be continuous determines that every x in M has a basis of neighbourhoods given by the cosets {x+M_la}. The "directed" property of La gives that the intersection of M_la, M_mu contain M_nu, so is still a neighbourhood of 0. (2) Under M -> M/M_la, the inverse image of any subset of the quotient is a union of cosets x+M_la, so open. (3) The topology separates two distinct elements x, y in M if and only if there exists M_la not containing x-y. qed ==== Construction of completion The {M_la} correspond to the inverse system M/M_mu -> M/M_la taking x mod M_mu to x mod M_la The _completion_ of M w.r.t. the topology {M_la} is defined as the inverse limit Mhat = M^ = lim_{<-} M/M_la. This consists of compatible sets of elements { x_la in M/M_la }_{la in La} such that x_mu -> x_la for every mu > la. There is a homomorphism M -> M^ that takes x in M to x mod M_la for all la. This has kernel the intersection (bigcap M_la). In any argument, either you assume (bigcap M_la), and can work with M subset M^ or you have to divide M by this kernel to get its image in M^. By construction, M^ has surjective homomorphisms to each M/M_la. Its kernel is (M_la)^, the completion of the submodule M_la in M w.r.t. to the subspace topology. These in turn define a topology on M^ with M^/(M_la)^ = M/M_la, and M^ is complete w.r.t. this. The particular case M = A gives filtrations of A by ideals I_la and completion A^ = lim_{<-} A/I_la, which is a ring having surjective maps A^ -> A/I_la to each A/I_la. ==== Philosophy. This type of completion appears in all areas of math. For example, consider all the rational roots of unity in CC^x. This is the union (= direct limit) of mu_n with inclusions mu_n into mu_{m*n}: the roots of z^{m*n} = 1 include the roots of z^n = 1 as a subgroup. Since the mu_n form a direct system, their character groups ZZ/n = Hom(mu_n, CC^x) form an inverse system lim_{<-} ZZ/n, that has limit ZZ^ the _profinite completion_ of ZZ. This is an uncountable group, equal to the direct product over all p of the p-adic integers ZZ_p. You know that the real line RR is the universal cover of the unit circle, with RR -> S^1 in CC^x given by exp(2*pi*i*theta), with kernel ZZ^+. The exponential function is not algebraic. But in algebra I can define the usual n-fold cover z |-> z^n from CC^x -> CC^x or from S^1 -> S^1, with the advantage that these are algebraic varieties and morphisms, and correspond to the inverse system CC^x/mu_{mn} -> CC^x/mu_n for all m. This idea replaces the exponential cover CC^+ -> CC^x or RR^+ -> S^1 in CC^x familiar in analysis or topology by the algebraic inverse limit lim_{<-} CC^x/mu_n which is "much bigger". For example, the inverse image of 0 is uncountable: it contains the profinite completion of the mu_n, a group that is isomorphic to ZZ^x (the argument depend the axiom of choice), but with a nontrivial structure of Galois module ("Tate module"). As you know, a finite Galois field extension K in L has a finite Galois group Gal(L/K). Now an infinite Galois extension K in L is the union (= direct limit) of normal finite subfields L_i: in fact each individual element x in L is algebraic, so belongs to a finite extension, and the corresponding normal subfield (the splitting fielf of x in L). The Galois group Gal(L/K) takes L_i to L_i, so has a surjective map Gal(L/K) -> Gal(L_i/K) to the finite group, and this makes Gal(L/K) = lim_{<-} Gal(L_i/K), which is therefore a profinite group. (Everything to do with the group is determined by its finite quotients, but these get bigger and bigger, and there are infinitely many of them.) The group Gal(QQbar/QQ) is a central object of study in algebraic number theory. For example, Wiles' 1994 proof of Fermat's Last Theorem depended on work on the representation theory of Gal(QQbar/QQ), in particular Serre's conjecture that its algebraic representations are "modular". (The progress since Wiles' work has only solved a small fraction of this conjecture.) ==== Exactness properties of completion The next issue is the following question on exactness: suppose 0 -> N \into M ->> M/N -> 0 is an exact sequence of modules. This means N subset M with quotient module M/N. Suppose we take the completion of N,M,M/N (with respect to some topology specified later). Under what circumstances can we prove that 0 -> N^ -> M^ ->> (M/n)^ -> 0 is again an exact sequence? Let me give a formal argument first, and understand what exactly it proves later. We know the Snake Lemma: given 0 -> P -> Q -> R -> 0 | | | arrows down c_P, c_Q, c_R 0 -> P' -> Q' -> R' -> 0, we obtain a long exact sequence 0 -> ker c_P -> ker c_Q -> ker c_R -> -> coker c_P -> coker c_Q -> coker c_R -> 0. where you have to think how the boundary map de: ker c_R -> coker c_P is defined. (Lift an element of R to Q anyhow, map it down by c_Q to an element of Q' that goes to 0 in R', so belongs to P', then check the result is independent of the choice, and that the resulting sequence is exact.) The argument of [A&M] applies this to an exact sequence of inverse systems. Define an inverse system to be a system of A-modules Pi with homomorphisms pi_{i+1}: P_{i+1} -> Pi, initially with no further assumptions. Its inverse limit P^ = lim_{<-} Pi is defined as the set of sequences {xi in Pi} with pi_{i+1}(x_{i+1}) = xi for every i. ==== Fact. By definition, the inverse limit P^ = lim_{<-} Pi is the set of compatible sequences of elements of Pi, which is the same thing as the kernel of c_P: prod_{all i} Pi -> prod_{all i} Pi where c_P takes sequence {x_i} |-> new sequence {pi_{i+1}(x_{i+1}) - xi}. To unwrap this, at the end of the sequence, the image of {..,x2,x1} is {.., pi(x3)-x2, pi(x2)-x1}. Taking ker c_P means exactly that the sequence is compatible. Note this refers to the direct _product_ of the Pi: any elements xi are allowed at each i (including infinitely many different choices), as opposed to the usual direct _sum_ of algebra, that assumes only finitely many xi are nonzero. ==== A homomorphism between inverse systems P and Q is a system of homomorphism fi: Pi -> Qi for each i forming commutative squares with the down maps P_{i+1} -> Q_{i+1} | | Pi -> Qi It is clear that this induces a homomorphism P^ -> Q^ of the respective limits. A short exact sequence of inverse systems 0 -> P -> Q -> R -> 0 is a pair of homomorphisms f: P -> Q and g: Q -> R of inverse systems such that for each i the homomorphisms fi and gi give short exact sequences 0 -> Pi -> Qi -> Ri -> 0. (This means of course simply that fi: Pi into Qi is injective, and gi is the quotient morphism gi: Qi ->> Ri = Qi/fi(Pi).) The fact just discussed, together with the snake lemma implies the following result: ==== Proposition [Exact] (1) A s.e.s. 0 -> P -> Q -> R -> 0 of inverse systems induces an exact sequence between their completions: 0 -> P^ -> Q^ -> R^ (2) If moreover the morphism al_{i+1}: P_{i+1} -> P_i are all surjective then 0 -> P^ -> Q^ -> R^ -> 0 is again a short exact sequence. (1) comes directly from the snake lemma. For (2), we just need to deduce that c_P is surjective from the given assumption that all pi_{i+1}: P_{i+1} are surjective. That is, given a sequence {ai in Pi}, required to find a sequence {xi in Pi} with c_P({xi}) = ai. Choose x1 = 0, then x2 in P2 with al_2(x2) = a1. At each successive step, we have the target ai in Pi and our current choice of xi (to cover a_{i-1}) so choose x_{i+1} such that al_{i+1}(x_{i+1}) = ai+xi. Then of course c_P applied to the sequence .., x_{i+1}, xi, .. x1 has the ith entry al_{i+1}(x_{i+1})-xi. Thus we can construct by induction a sequence {xi in Pi} such that c_P({xi}) = ai. QED. ==== The Artin--Rees lemma [Matsumura, p. 59]. There is still a gap in applying the [Exact] proposition to I-adic completions: the assumptions of the proposition is that we have 3 inverse systems P,Q,R with s.e. sequences 0 -> Pi -> Qi -> Ri -> 0 for each i. However, we start from a submodule, N in M and the quotient M/N, take the I-adic filtrations of the three modules I^n*N, I^n*M and I^n*(M/N), and the inverse systems corresponding to the quotients. It is not true that these quotient form short exact sequences at each step. The next result fills this gap: under the standard finiteness assumptions of commutative algebra, it gives a compatibility between the I-adic filtration {I^n*N} of the submodule and the restriction to N of I-adic filtration {I^n*M} of the module. Theorem [Artin-Rees lemma] Assume A is Noetherian and I an ideal of A. Let M be a finite module and N subset M a submodule. Then exists c > 0 such that I^n*M intersect N = I^{n-c}*(I^c*M intersect N) for every n > c. Proof. The right-hand side is obviously contained in the left. Let (a1,.. ar) generate I. Define the graded polynomial ring B = A[x1,.. xr] with independent indeterminates xi corresponding to the generators ai of I. Write (m1,.. ms) for generators of the finite A-module M. Then any element of I^n*M can be written as sum_i fi(a)*mi where fi = fi(x1,.. xr) in B are homogeneous polynomials of degree n. That sum involves coefficients (f1,.. fs) in B^s. To work in the framework of Noetherian assumptions, define for each n the submodule Jn = { (f1,.. fs) in B^s | the fi are homog of deg n and sum fi(a)*mi in N }. We see that by construction, the sum sum fi(a)*mi is in N and also in I^n*M. Take U = Union_n Jn over all n and the ideal C generated by U. [The key point:] Since B^s is a Noetherian B-module, the ideal C is finitely generated: there are finitely many elements u1,.. ut in U such that C = B*u1 +.. B*ut each uj is an s-tuple of homogeneous elements of B of some degree nj, say uj = (u_{j1},.. u_{js}) in J_{nj}. Setting c = max nj gives us the c in the statement. It only remains to wrap up the conclusion. Suppose given y in I^n*M intersect N. We can certainly write y = sum fi(a)*mi with fi in B homogeneous of degree n, and then the s-tuple (f1,.. fs) in Jn (by definition of Jn). But Jn is in the B-module C, so is a B-linear combination of the generators uj. That is (f1,.. fs) = sum pj(x)*uj with polynomials pj in B = A[x1..xr]. Now if we replace each pj by its homogeneous part of degree n - nj the equality still holds. (Because the fi are all homog of deg n and the uj are homog of deg nj -- once we've matched the terms of deg n, all the other cancel, so we can throw them away.) Now complicated formula gives y in I^{n-c}*(I^c*M intersect N). Complicated formula is [Matsumura, p. 59, line -8]: y = sum_i fi(a)*mi = sum_j pj(a) * sum_i u_{ji}(a)*mi where the first sum consists of elements of I^{n-nj} and the second sum of elements of I^nj intersect N. QED Corollary Under the same assumptions (A Noetherian and M finite), take the I-adic topology on M and the subspace topology it induces on N in M. Then this coincides with the I-adic topology of N. ==== Exactness, I-adic completion is an exact functor, the I-adic completion A^ of A is a flat A-algebra M_n = M/(I^n*M). In particular, working with I-adic completions, we know that if L in M is a submodule then L^ in M^ is a submodule, and (L^)/(M^) = (L/M)^. Let A be a ring and I an ideal of A. We have just seen that I-adic completion gives an exact functor on A-module. At the same time, it is clear that the I-adic completion M^ is a module over A^, and is the same thing as M tensor A^. The exactness result just proved for I-adic localisations implies that A^ is a flat A-algebra. ==== Comparison with exactness statements for S^-1 and flatness of S^-1A. For A a ring and S a multiplicative sequence, we know how to construct the partial ring of fractions S^-1A. We can make essentially the same construction for an A-module M, obtaining an A-module S^-1M. It consists of expressions {m/s} modulo the same kind of equivalence relation, and the construction gives that S^-1M is an A-module on which every s in S acts bijectively. This means that S^-1M is also an S^-1A-module, and in fact one sees that S^-1M = S^-1A tensor_A M. Proposition. Let S be a multiplicative set in A and suppose that morphisms al: L -> M and be: M -> N give a sequence L -> M -> N that is exact (only in the middle, im(al) = ker(be)). Then al, be induce an exact sequence on the localised modules S^-1L -> S^-1M -> S^-1N (with morphisms al' and be'). In particular, working with localisation, we know that if L in M is a submodule then S^-1L in S^-1M is a submodule, and (S^-1L)/(S^-1M) = S^-1(L/M). Proof from [UCA], 6.6. Suppose m/s in S^-1M. Then be'(m/s) = 0 <=> exists u in S such that u*be(m) = 0 <=> exists u in S such that be(u*m) = 0. Now since im(al) = ker(be) in L -> M -> N, this happens <=> exists u in S and exists n in L s.t. u*m = al(n) <=> m/s = al'(n/u*s). Q.E.D. Localisation S^-1 applied to M can be thought of as S^-1M = S^-1A tensor M, and the exactness statement just proved can be stated as S^-1A is a flat A-algebra.