Lecture 6 The previous lecture discussed a.c.c. and d.c.c. and the related notion of Noetherian and Artinian modules. This lecture has 2 aims: (I) to treat modules that satisfy both chain conditions, and characterise them as modules of finite length. (II) To discuss Artinian rings, which (as opposed to Artinian modules) are necessarily also Noetherian, and are basically analogous to k-algebras that are finite dimensional as vector spaces over a field k. Definition: Simple module. M is a simple A-module if its only submodules are 0,M. Obviously, if M is a simple module and N any module, any homomorphism M -> N is 0 or injective any homomorphism N -> M is 0 or surjective If M, N are both simple, any M -> N is 0 or an isomorphism. Definition: A a ring, and M an A-module. A Jordan-Hoelder filtration is a chain of submodules 0 = M0 in M1 in .. M_{n-1} in M_n = M (*) with no strictly intermediate modules between M_{i-1} and M_i. The latter condition holds if and only if M_i/M_{i-1} is simple. Theorem Equivalent conditions on M: (1) M is Artinian and Noetherian. (2) M has a Jordan-Hoelder filtration. Moreover, if they hold, the set of simple quotient modules Mi/M_{i-1} up to isomorphism in any JH filtration is unique up to permutation. If the conditions hold we say M has finite length, or has length n as an A-module where n is in (*). (This depends on A -- if we view M as a module over a smaller ring, the length will increase.) (1) => (2) is the obvious follow-your-nose argument: First, define Si = set of all nonzero submodules. If M <> 0 this is nonempty, so by the Artinian condition it has a minimal element M1. Next, set Si = set of all submodules strictly bigger that M1. If this is empty, M1 = M and we are finished. If nonempty, it has a minimal element M2. Continue by induction: we construct 0 = M0 < M1 < .. Mi <= M where each step cannot be refined. At any stage, either we have reached M or we can take another step. So far, this has only used Artinian. Now since M is also Noetherian the increasing chain must terminate, so at some point Mn = M. (You can also do this from the other end, working down from M and taking maximal nontrivial M' < M.) The proof of (2) => (1) and the "moreover" final clause involve standard arguments using the isomorphism theorems. Given one JH filtration (*), and any chain of submodules {Ni}, if none of the Ni contain M1, they are part of a chain for M/M1, which has a JH chain of length n-1. In the contrary case, there is some i s.t. N_{i-1} does not contain M1 but Ni does, so necessarily Ni/N{i-1} iso M1. etc. (Clean up the proof for yourself.) ====== Artinian rings ===== For modules, the a.c.c. and d.c.c. conditions are in a sense logically dual, and there are (vague kinds of) duality relations between the two classes of Noetherian and Artinian modules. At least the modules and the classes of modules are "the same size". For rings, however, Artinian is a much more restrictive condition: an Artinian ring A is also Noetherian, and has finite length as A-module. A particular case is that A is an algebra over a field k. Then A is finite dimensional as k-vector space, so k in A is similar to the finite field extensions k in K you studied in Galois theory. As a preview, consider the example A = k[x]/(F) with k a field. Suppose F = prod (fi)^ni with fi irreducible and coprime. Then the maximal ideals of A are (fi), and A = directsum (k[x]/(fi^ni)) each of the summands has a JH filtration with ni steps and each quotient isomorphic the field k[x]/(fi), which is a finite extension field of k and a simple A-module. Proposition (I) If an integral domain A is Artinian, it is a field. (II) If A is an Artinian ring, every prime ideal is maximal (that is, A has Krull dimension = 0). (I) Pick x in A, and consider the descending chain of ideals (x) > (x^2) > .. > (x^n) .. By the d.c.c., eventually (x^n) = (x^{n+1}), so that x^n is a multiple of x^{n+1}, say x^n = y*x^{n+1}. Now either x = 0 or x*y = 1. This condition characterises a field. (II) follows by passing to the quotient A/P, which is still Artinian. Theorem. An Artinian ring A is Noetherian Step 1 A has only finitely many maximal ideals m_i. Proof (An exercise. See for example [A&M, Lemma 1.11]). Step 2 Write J = product mi = intersection mi. Since every prime ideal is maximal, J is also the intersection of all prime ideals of A, which is its nilradical. So every x in J is nilpotent. Claim. J^n = 0. J = intersection mi = intersect of all primes, so every x in J is nilpotent. Consider the descending chain J > J^2 > .. > J^n. This must terminate, say in N = J^n with J*N = N^2 = N. We prove that N = 0 by contradiction. If N <> 0 the set Si of ideals b such that b*N <> 0 is nonempty, so has a minimal element c. Now some x in c has x*N <> 0, and c = (x), else (x) would be smaller than minimal. Now (x*N)*N = x*N^2 = x*N <> 0. So x*N is an ideal contained in (x), and again by minimality, x*N = (x). It follows that x = x*y for some y in N, and therefore x = x*y = x*y^2 = .. x*y^n. But y is an element of N, and every element of J is nilpotent, so x = x*y^n = 0. This is a contradiction, hence N = 0. Step 3. An Artinian ring A has a JH filtration, and so is Noetherian as A-module. In fact we proved that the finitely many prime ideals m_i have prod m_i^n = 0. So taking prod m_i^{ni} with one of the exponents increasing by 1 at a time, we get a decreasing sequence with each quotient a copy of the A-module A/m_i, which is a field. Q.E.D. The above argument divides into a number of fairly tricky steps. [A&M] and [Matsumura] give essentially the same proof, and I guess there is no short-cut. This result in the converse direction is much easier: Theorem: Noetherian and dim A = 0 => Artinian. As in the preceding argument, all primes are maximal, so the intersection J of all maximal ideals is the nilradical, the set of all nilpotent elements. However, Noetherian gives that J is finitely generated, so J nilpotent is clear. Now A/J is a product of fields, and Noetherian implies there are only finitely many of them. etc.