This is the final preprint version of a paper which appeared atJournal of Algebra and its Applications, 11 (2012) No.5, 1250090The published version is accessible to subscribers athttp://dx.doi.org/10.1142/S0219498812500909 .BILINEAR MAPS ON ARTINIAN MODULESGEORGE M. BERGMANAbstract. It is shown that if a bilinear map f : A × B → C of modules over a commutative ring k isnondegenerate (i.e., if no nonzero element of A annihilates all of B, and vice versa), and A and B areArtinian, then A and B are of finite length.Some consequences are noted. Counterexamples are given to some attempts to generalize the above resultto balanced bilinear maps of bimodules over noncommutative rings, while the question is raised whetherother such generalizations are true.Below, rings and algebras are associative and unital, except where the contrary is stated.Examples of modules over a commutative ring k that are Artinian but not Noetherian are well known; forexample, the Z-module Zp∞. However, such modules do not show up as underlying modules of k-algebras.(We shall see in §3 that this can be deduced, though not trivially, from the Hopkins-Levitzki Theorem, whichsays that left Artinian rings are also left Noetherian.) The result of the next section can be thought of as ageneralization of this fact.I am grateful to J. Krempa for pointing out two misstatements in the first version of this note, toK. Goodearl, D. Herbera, T. Y. Lam and L. W. Small for references to related literature, and to the referee ofa previous version of this note for an alternative proof of the main theorem, noted at the end of §1.1. Our main resultIn the proof of the following theorem, it is interesting that everything before the next-to-last sentenceworks, mutatis mutandis, if A and B are both assumed Noetherian rather than Artinian, though the finalconclusion is clearly false in that case. (On the other hand, the argument does not work at all if one of Aand B is assumed Artinian, and the other Noetherian.)Theorem 1. Suppose k is a commutative ring, and f : A × B → C a bilinear map of k -modules which isnondegenerate, in the sense that for every nonzero a ∈ A, the induced map f(a, −) : B → C is nonzero,and for every nonzero b ∈ B, the induced map f (−, b) : A → C is nonzero.Then if A and B are Artinian, they both have finite length.Proof. If elements a ∈ A, b ∈ B satisfy f(a, b) = 0, we shall say they annihilate one another. (The conceptof an element c of the base ring k annihilating an element x of A, B, or C will retain its usual meaning,2010 Mathematics Subject Classification. Primary: 13E10, 16P20; Secondary: 13C05, 13E05, 15A63, 16D20, 16P40.Key words and phrases. Nondegenerate bilinear map, Artinian module, Noetherian module, bimodule.http://arXiv.org/abs/1108.2520 . After publication of this note, updates, errata, related references etc., if found, will berecorded at http://math.berkeley.edu/~gbergman/papers/.12 GEORGE M. BERGMANc x = 0.) For subsets X ⊆ A, respectively Y ⊆ B, we define the annihilator sets(1)X⊥= {b ∈ B | (∀ x ∈ X) f(x, b) = 0} ⊆ B,Y⊥= {a ∈ A | (∀ y ∈ Y ) f(a, y) = 0} ⊆ A.We see that these are submodules of B and A respectively, that the set of annihilator submodules in B(respectively, in A) forms a lattice (in the order-theoretic sense) under inclusion, and that these two latticesof annihilator submodules are antiisomorphic to one another, via the maps U 7→ U⊥. (This situation is anexample of a “Galois connection” [2, §5.5], but I will not assume familiarity with that formalism.)When A and B are Artinian, these lattices of submodules of A and B both have descending chaincondition; so since they are antiisomorphic, they also have ascending chain condition. Hence all their chainshave finite length. Let us choose a maximal (i.e., unrefinable) chain of annihilator submodules of A,(2) {0} = A0⊆ A1⊆ . . . ⊆ An= A.This yields a maximal chain of annihilator submodules of B,(3) B = B0⊇ B1⊇ . . . ⊇ Bn= {0},where(4)Bi= A⊥i, Ai= B⊥i.It is easy to see that for each i < n, f induces a k -bilinear map(5) fi: (Ai+1/Ai) × (Bi/Bi+1) → C,via(6) fi(a + Ai, b + Bi+1) = f(a, b) (a ∈ Ai+1, b ∈ Bi).I claim that under fi, every nonzero element of Ai+1/Aihas zero annihilator in Bi/Bi+1, and viceversa. For if we had a counterexample, say a ∈ Ai+1/Ai, then its annihilator would be a proper nonzeroannihilator submodule of Bi/Bi+1, and by (6) this would lead to an annihilator submodule of B strictlybetween Biand Bi+1, contradicting the maximality of the chain (3).From this we can deduce that every nonzero element of Ai+1/Aiand every nonzero element of Bi/Bi+1have the same annihilator in k. Indeed, if c ∈ k annihilates the nonzero element x ∈ Ai+1/Ai, then forevery y ∈ Bi/Bi+1, c y will annihilate x, hence must be zero; so every c ∈ k annihilating one nonzeromember of A annihilates all of B, and dually.It is now easy to see that the common annihilator of all nonzero elements of these two modules is aprime ideal Pi⊆ k, making k/Pian integral domain, such that Ai+1/Aiand Bi/Bi+1are k/Pi-modules.Moreover, taking any nonzero element of either of our modules, say x ∈ Ai+1/Ai, we have kx∼=k/Piasmodules, so since Ai+1/Aiis Artinian, so is k/Pi.But an Artinian integral domain is a field; so Ai+1/Aiand Bi/Bi+1are vector spaces over the fieldk/Pi, so the fact that they are Artinian means that they have finite length. Thus, A and B admit finitechains (2), (3) of submodules with factor-modules of finite length; so they are each of finite length. Here is an illuminating alternative proof of Theorem 1, sketched by the referee of an earlier version ofthis note, which uses the theory of “secondary representations” of modules over a commutative ring [11,Appendix to §6]. Let k, A, B, C, f be as in the hypothesis of our theorem.Because A is Artinian, it has, by [11, Theorem 6.11], a representation A = A1+ · · · + An, where eachAmis a Pm-secondary module for some prime ideal Pm⊆ k. This means that(7)for every c ∈ Pm, some power ciannihilates Am,and(8) for every c ∈ k − Pm, we have c Am= Am.Now for each m, the fact that Amis Artinian clearly implies that there exists a finitely generatedsubideal P0⊆ Pmhaving the same annihilator in Amthat Pmhas. From this one can show by inductionthat for every i ≥ 0, the ideal (P0)ilikewise has the same annihilator in Amas (Pm)idoes [6, Lemma3, p.54]. (Key steps: if {0} = (P0)ix, write this {0} =