Lecture 4 discussed Spec A and its topology, and was mostly taken from [UCA], Section 5. The definition I gave said V(I) is closed. Xf is basis for topology. If X = union Xfi then 1 in (fi), so that we only need finitely many, and X is quasi-compact. (compact but of course not Hausdorf.) It is natural to associate A[1/f] to Xf and A_P to P in X. Lecture 4. The case that we understand fairly well: A = k[X]/J(X) is the affine coordinate ring of an algebraic set X in AA^n_k over an algebraically closed field k. The maximal spectrum m-Spec A corresponds to k-valued points of X. Every prime ideal is J(Y) for an irreducible subvariety Y in X; in particular every prime ideal P is the intersection of maximal ideals at points of Y, so that m-Spec A determines Spec A. X has a Zariski topology, with closed subsets the subvarieties Y in X. The sets Xf (the set of points with f <> 0) is the complement of the variety V(f). Every closed set is V(I) = intersect_{f in I} V(f) so that its complement if the union of V(f) taken over f in I. This means that the Xf form a basis for the Zariski topology of X. In fact Xf is itself isomorphic to an affine variety with coordinate ring k[Xf] = k[X][1/f]. A prime ideal P in A = k[X] corresponds to an irreducible closed subvariety Y = V(P) in X. This has its own Zariski topology (the subspace topology of Y in X or Y in AA^n). The localisation A_P consists of fractions f = g/h of k[X] having one possible denominator h that does not vanish on Y. This means that the fraction f restricts to a rational function on Y. We know A_P is a local ring with maximal ideal PA_P, so that the quotient ring A_P/PA_P is a field, the function field k(Y) of the irreducible variety Y. The fraction f = g/h has domain of definition dom f (in X) that intersections Y in a dense open set of Y, and restricts to a rational function f_|Y on Y whose domain includes (dom f) intersect Y. For a general ring A, we can't rely on any useful relation between prime ideals and maximal ideals, so m-Spec A is not especially useful. We work with Spec A. It has a Zariski topology with closed sets V(I) = { primes P containing I }. This means that f in I maps to 0 in the quotient ring A/P, so we can still view this as saying the f evaluates to 0 at P. The Zariski topology on X = Spec A has a basis by the open sets Xf = { P in X | f notin P }. This topology is far from Hausdorff or separated, but it is compact (without any finitess conditions on A). The ring of fractions A[1/f] has Spec A[1/f] = Xf. Each point P in X has the local rings A_P with maximal ideal PA_P and residue field A_P/PA_P. What is missing here is the idea that f is a function on Spec A in any conventional sense. Instead of evaluating f at a point P in Spec A, the operation that preserves all the information is mapping f to its class in A_P. The notion of variety is replaced by Grothendieck's definition of affine scheme, that puts together: X = Spec A, its Zariski topology, the partial localisation A[1/f] associated to Xf, the local rings A_P over P in Spec A. New chapter Chain conditions on modules and on rings: M Noetherian modules: every increasing chain terminates <=> every nonempty set of submodules has a maximal submodule <=> every submodule is f.g. You know lots of examples: If A is f.g. as a ring over k or over ZZ, or a localisation of such, a finite A-module is Noetherian. Another class of examples: if A is Noetherian, a formal power series ring A[[x1,x2,..xn]] is Noetherian. (later ex. -- these are of course not f.g. rings). M Artinian module: every decreasing chain of submodules terminates <=> every nonempty set of submodules has a minimal submodule This is not a frivolous issue. e.g. ZZ is not Artinian: take integer n not a unit (say n = 10), then the chain of ideals (n^i) is an infinite decreasing chain. Likewise k[x] is not Artinian: take nonunit polynomial f, then the chain of ideals (f^i) is infinite decreasing. What examples? Inverse polynomials: Let k[x,x^-1] be the ring of Laurent power series, and consider the quotient module M = k[x,x^-1] / k[x]. As a vector space, this is countable dimensional, with basis { x^-i }. As a k[x]-module, multiplication by x does x * x^-i |-> x^{-i-1}. Whereas in k[x] multiplication by x is injective, in M, multiplication by x is surjective. For any m in M, we can find a predecessor m' with x*m' = m. M is Artinian: every proper submodule N in M only involves finitely many x^-n, and is the k-vector space based by x^i for i in the range -n < i < -1. The longest chain of these is M subsetneq .. Nn subsetneq N{n-1} subsetneq .. N{-1} subsetneq N0 = 0. On the other hand, it is not finitely generated as module, and not Noetherian. As n gets bigger n, the modules (x^-n) get bigger, so infinite ascending sequences are the order of the day. You should think of M as a tempered dual of the polynomial ring k[x]. Since k[x] is infinite dimensional, the dual vector space k[x] = Hom(k[x], k) would be uncountable dimensional. Instead, think of k[x] as the union of the spaces k[x]_{<=n} of polynomials of degree <= n. Then M is the union of their duals Hom(k[x]_{<=n}, k). The duality between M and k[x] is the analog of Cauchy residue of a meromorphic function -- take inverse polynomial q and polynomial f into the residue of product q*f, that is, the coefficient of x^-1. M is called the module of Macaulay inverse systems. It contains a unique submodule N = k*x^-1 that is isomorphic to the residue field k[x]/(x), and M is injective as k[x] module in a sense that we will come to later. The same trick applies to the localisation ZZ_(p) of ZZ at a prime p. The module M = QQ/ZZ_(p) is generated by the negative powers of p. Any proper ZZ_(p)-submodule N in M has only finitely many p^-n, and the chains of modules have the same shape as above. The module QQ/ZZ is the sum of the QQ/ZZ_(p) taken over all p. It is Artinian but not Noetherian. Jordan-Hoelder filtration and modules of finite length Theorem Let M be an A-module. Then the following are equivalent: (1) M is both Artinian and Noetherian (2) There exists a finite filtration 0 = M0 inneq M1 inneq .. M_{n-1} inneq Mn = M. such that for each i, there is no A-module strictly intermediate between Mi and M{i+1}. The last condition is: if Mi inneq N inneq M_{i+1} then either Mi = N or N = M_{i+1}. You can also say that the module M_{i+1}/M_i is simple (has no nontrivial submodule). (1) => (2) is straightforward. M is Noetherian so (if <> 0) there is a maximal submodule M' in M. The inclusion M' in M has not no strictly intermediate property by construction. Apply the same to M' (assuming <>) and get a descending chain. Since M is Artinian, this must terminate, giving (2).