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TEXed April 15, 2019
BILINEAR MAPS ON ARTINIAN MODULES
GEORGE M. BERGMAN
Abstract.
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 abovestatement to balanced bilinear maps of bimodules over noncommutative rings, while the question is raisedwhether other such generalizations are true. Rings and algebras will be understood to be 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 Z p ∞ . However, such modules do not show up as underlying modules of k -algebras.(We shall see in § Our main result
In the proof of the following theorem, it is striking 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 A and 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 ∈ k annihilating an element x of A, B, or C will retain its usual meaning, c x = 0 . ) Forsubsets Y ⊆ B, respectively X ⊆ A, we define the annihilator sets(1) Y ⊥ = { a ∈ A | ( ∀ y ∈ Y ) f ( a, y ) = 0 } ⊆ A,X ⊥ = { b ∈ B | ( ∀ x ∈ X ) f ( x, b ) = 0 } ⊆ B. We see that these are submodules of A and B respectively, that the set of annihilator submodules in A (respectively, in B ) 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 U ⊥ . (This situation is anexample of a “Galois connection” [1, § 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) { } = A ⊆ A ⊆ . . . ⊆ A n = A. This yields a maximal chain of annihilator submodules of B, (3) B = B ⊇ B ⊇ . . . ⊇ B n = { } , Mathematics Subject Classification.
Primary: 13E10, 16P20; Secondary: 13E05, 15A63, 16D20, 16P40.
Key words and phrases.
Nondegenerate bilinear map, Artinian module, Noetherian module, bimodule.This preprint is readable online at http://math.berkeley.edu/~gbergman/papers/ . where(4) B i = A ⊥ i , A i = B ⊥ i . It is easy to see that for each i, f induces a k -bilinear map(5) f i : ( A i +1 /A i ) × ( B i /B i +1 ) → C, via(6) f i ( a + A i , b + B i +1 ) = f ( a, b ) ( a ∈ A i +1 , b ∈ B i ) . I claim that under f i , every nonzero element of A i +1 /A i has zero annihilator in B i /B i +1 , and viceversa. For if we had any counterexample, say a ∈ A i +1 /A i , then its annihilator would be a proper nonzeroannihilator submodule of B i /B i +1 , and this would lead to an annihilator submodule of B strictly between B i and B i +1 , contradicting the maximality of the chain (3).From this we can deduce that every nonzero element of A i +1 /A i and every nonzero element of B i /B i +1 have the same annihilator in k. Indeed, if c ∈ k annihilates the nonzero element x ∈ A i +1 /A i , then forevery y ∈ B i /B i +1 , cy will annihilate x, hence must be zero; so every c ∈ k annihilating one nonzeromember of A annihilates all of B, and dually.It follows that the common annihilator of all nonzero elements of these two modules is a prime ideal P i ⊆ k, making k/P i an integral domain, such that A i +1 /A i and B i /B i +1 are k/P i -modules. Moreover,taking any nonzero element of either of our modules, say x ∈ A i +1 /A i , we have kx ∼ = k/P i as modules, sosince A i +1 /A i is Artinian, so is k/P i . But an Artinian integral domain is a field; so A i +1 /A i and B i /B i +1 are vector spaces over the field k/P i , 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. (cid:3) Some immediate consequences
We start with a trivial consequence of our theorem.
Corollary 2.
If in the statement of Theorem 1, we assume only one of the nondegeneracy conditions, namelythat for all nonzero a ∈ A, the induced map f ( a, − ) : B → C is nonzero ( respectively, that for all nonzero b ∈ B, the induced map f ( − , b ) : A → C is nonzero ) , we can still conclude that A ( respectively, B ) hasfinite length.Assuming neither nondegeneracy condition, we can still conclude that A/B ⊥ and B/A ⊥ have finitelength.Proof. Without any nondegeneracy assumption, note that f induces a nondegenerate bilinear map(7) A/B ⊥ × B/A ⊥ → C. Since
A/B ⊥ and B/A ⊥ are again Artinian, we can apply Theorem 1 to (7) and conclude that both thesefactor-modules have finite length.On the other hand, the two nondegeneracy conditions in the first assertion of the corollary are, respectively,equivalent to A = A/B ⊥ and to B = B/A ⊥ ; so combining one or the other of these with the above result,we get the first asserted conclusion. (cid:3) We can now recover a result of A. Facchini, C. Faith and D. Herbera.
Corollary 3 ([3, Proposition 6.1]) . The tensor product A ⊗ k B of two Artinian modules over a commutativering k has finite length.Proof. Letting A ⊥ and B ⊥ be annihilators with respect to the tensor multiplication ⊗ : A × B → A ⊗ k B, the preceding corollary tells us that A/B ⊥ and B/A ⊥ have finite length. But A ⊗ k B can also be regardedas the tensor product of these factor modules; and a tensor product of modules of finite length over acommutative ring has finite length. (cid:3) Let us next apply Theorem 1 to the multiplication of a k -algebra R. This does not require R to beassociative, so we shall make no such assumption. In the study of nonassociative algebras, it is often notnatural to require a unit; but without one, nondegeneracy of the multiplication is not automatic; so in thestatement below we get this nondegeneracy by dividing out by an appropriate annihilator ideal. ILINEAR MAPS ON ARTINIAN MODULES 3 (Caveat: below, we name that ideal Z ( R ) , though it is not the center of R. This notation, from [2], isbased on the phrase “zero multiplication”, and also on the use of that symbol in the theory of Lie algebras,where it does coincide with the center.)
Corollary 4.
Let R be a not-necessarily-associative algebra over a commutative ring k, and let (8) Z ( R ) = { x ∈ R | xR = Rx = { } } . Then if
R/Z ( R ) is Artinian as a k -module, it is also Noetherian as a k -module.Hence, if Z ( R ) = { } ( in particular, if R has a unit element ) , then if R is Artinian as a k -module, itis also Noetherian as a k -module.Proof. Let(9) Z l ( R ) = { x ∈ R | xR = { } } , Z r ( R ) = { x ∈ R | Rx = { } } . Thus, Z ( R ) = Z l ( R ) ∩ Z r ( R ) . The multiplication of R induces a nondegenerate bilinear map of k -modules R/Z l ( R ) × R/Z r ( R ) → R. Since
R/Z l ( R ) and R/Z r ( R ) are homomorphic images of R/Z ( R ) , they areArtinian over k, hence by Theorem 1 they are Noetherian over k. Hence
R/Z ( R ) , which embeds in R/Z l ( R ) × R/Z r ( R ) , is also Noetherian.The assertion of the final sentence clearly follows. (cid:3) This shows, as mentioned at the start of this note, that groups such as Z p ∞ cannot be the additive groupsof rings. There are other groups, such as Q / Z , which one feels should be excluded for similar reasons, thoughthey are not themselves Artinian. This leads us to formulate the following consequence of Theorem 1. Corollary 5.
Suppose k is a commutative ring, and f : A × B → C a nondegenerate bilinear map of k -modules.Then if B ( respectively A ) is locally Artinian ( i.e., if every finitely generated submodule thereof is Ar-tinian ) , then every Artinian submodule of A ( respectively B ) has finite length.Proof. Let A be an Artinian submodule of A, and consider the annihilators in A of all finitely generatedsubmodules of B. This family of annihilators is clearly closed under finite intersections; so by descendingchain condition on submodules of A , the intersection of this family is itself such an annihilator, say ofthe finitely generated submodule B ⊆ B. But by nondegeneracy of f, that intersection is { } ; so B has trivial annihilator in A . Now assuming B locally Artinian, B will be Artinian, so we can applyCorollary 2 to the restricted map f : A × B → C, and conclude that A has finite length. By symmetry,we also have the corresponding implication with the roles of A and B interchanged. (cid:3) One has obvious analogs of Corollaries 2, 3 and 4 for this result. From the last of these, we see thatthe Z -module Q / Z , which is locally Artinian and has Artinian submodules of infinite length, cannot be theadditive group of a unital (associative or nonassociative) ring.3. Relation to the Hopkins-Levitzki Theorem
Going back to Corollary 4, can the case of that result where R is an associative unital k -algebra bededuced from the well-known fact that a left Artinian ring is also left Noetherian – the Hopkins-LevitzkiTheorem [6, Theorem 4.15(A)] ?Well, if, as assumed in that corollary, R is Artinian as a k -module, then it is certainly Artinian as a left R -module, hence if it is associative and unital, the Hopkins-Levitzki Theorem says it is Noetherian as a left R -module. Can we somehow get from this that it is Noetherian as a k -module?We can, using the following striking result.(10) (Theorem of Lenagan and Herbera [9, Theorem on p.2044].) If R and S are rings, and R M S a bimodule which is left Noetherian and right Artinian, then M is also left Artinian and rightNoetherian.Indeed, from (10) we deduce Corollary 6 (to (10)) . If R is a k -algebra, and R M an R -module of finite length ( as an R -module ) , andif, moreover, M is Artinian or Noetherian as a k -module, then it has finite length as a k -module.Proof. Regard M as a bimodule R M k . Then (10) or its left-right dual, applied to this bimodule, gives thedesired conclusion. (cid:3)
GEORGE M. BERGMAN
So in the associative unital case of Corollary 4, once we know by the Hopkins-Levitzki Theorem that R R has finite length as an R -module, the above result gives an alternative proof of the conclusion of thatcorollary.(Notes on the background of (10): In [9], (10) is described as a result of T. Lenagan, with a new proofcommunicated to the author by D. Herbera. Lenagan [8] had indeed proved the hard part of (10), that if M is both Artinian and Noetherian on one side, and Noetherian on the other, then it is also Artinian onthe latter side; and this is what is called Lenagan’s Theorem in most sources, e.g, [7]. However, Herbera’sversion in [9] tacitly supplies the additional argument showing that if M is merely assumed Artinian onone side and Noetherian on the other, then it is also Noetherian on the former side. It is that part, andnot the part proved by Lenagan, that we needed for our alternative proof of the associative unital case ofCorollary 4. Incidentally, Lenagan formulated his result for 2-sided ideals, but as noted in [7, p. 332, sentenceafter Theorem 11.30], his proof carries over verbatim to bimodules.)For some results on when not-necessarily-unital associative right Artinian rings must be right Noetherian,and related questions, see [4] and papers cited there.Returning to the relation between Corollary 4 and the Hopkins-Levitzki Theorem, we cannot hope to gothe other way, and obtain the latter from the former: We can’t get started, since the Artinian assumptionon R as a left R -module does not, in general, by itself give such a condition on R as a k -module. Thissuggests that we look for some result that can be applied directly to the right and left R -module structuresof R ; say a generalization of Theorem 1 to a result on balanced bilinear maps(11) f : S A R × R B T → S C T of bimodules over associative rings.I have not been able to find such a generalization. Let us take a brief look at the situation.4. Some counterexamples, and a question
Given a map (11), the annihilator of every element of A is a T -submodule of B, and the annihilatorof every element of B is an S -submodule of A ; so one might hope for a result assuming the Artinianproperty for S A and B T . But this does not work: if we take a field k, two k -algebras S and T, anynonzero Artinian left S -module A, and an Artinian but non-Noetherian right T -module B (for instance, S = T = k [ t ] , A = B = k (( t )) /k [[ t ]] , equivalently, k ( t ) /k [ t ] ( t ) ) , then we see that the canonical map(12) ⊗ : S A k × k B T → S ( A ⊗ k B ) T is a counterexample: it is nondegenerate, and S A and B T are Artinian, but they are not both Noetherian.If, instead, we assume the Artinian condition on A R and R B, counterexamples are harder to find; butwe can get them using the following construction. Lemma 7.
Let k be a field, R a k -algebra, and R M any nonzero left R -module. Then there exists a k -algebra R having (i) a left module R B whose lattice of submodules is obtained, up to isomorphism, from the lattice of submod-ules of R M by adjoining one new element at the bottom, (ii) a simple right module A R , and (iii) a nondegenerate R -balanced k -bilinear map f : A R × R B → k. Proof.
From R M, we shall construct B as a k -vector-space. We shall then define R as a certain k -algebraof linear endomorphisms of B, and A as a certain k -vector space of linear functionals B → k, closed underright composition with the actions of members of R. The map f : A × B → C will be the function thatevaluates members of A at members of B. Here are the details.Writing ω for the set of natural numbers, let B ⊆ M ω be the vector space of those sequences x = ( x i ) i ∈ ω which have the same value at all but finitely many i. Let R be the k -algebra of k -linear maps B → B spanned by(13) those maps which act by multiplication by an element r ∈ R simultaneously on all coordinates;i.e., by rx = ( rx i ) i ∈ ω , and(14) those maps which act by projecting to the sum of finitely many of our copies of M, then mappingthis sum into itself by an arbitrary finite-rank k -vector-space endomorphism. ILINEAR MAPS ON ARTINIAN MODULES 5
It is easy to see that every nonzero R -submodule of B contains the submodule B fin of all elements havingfinite support in ω, that an R -submodule B ′ containing B fin is determined by the values that elements of B ′ assume “almost everywhere”, and that the set of these values can be, precisely, any submodule of M. Thus, the lattice of submodules of B containing B fin is isomorphic to the lattice of submodules of M ; sothe full lattice of submodules of B consists of this and a new bottom element, the zero submodule.Let A be the set of all k -linear functionals a on B that depend on only finitely many coordinates (i.e.,for which there exists a finite subset I ⊆ ω such that a factors through the projection of B ⊆ M ω to M I ) . This set is easily seen to be closed under right composition with elements of R ; hence we may regard A asa right R -module, and evaluation of elements of A on elements of B gives an R -balanced k -bilinear map f : A × B → k. Further, for any a ∈ A − { } and b ∈ B − { } , we can clearly find a u ∈ R of the sortdescribed in (14) which carries b to an element not in ker( a ) , so that 0 = f ( a, ub ) = f ( au, b ) . In particular, f has the two properties defining nondegeneracy (statement of Theorem 1).It remains to show that A R is simple. Given a ∈ A − { } and a ′ ∈ A, we see that there will exist y ∈ B with finite support such that a ( y ) = 1 . Choosing such a y, define u : B → B by u ( x ) = a ′ ( x ) y. It is easyto check that u has the form (14), hence lies in R, and that it satisfies au = a ′ , proving simplicity. (cid:3) Taking for R M in the above lemma any Artinian non-Noetherian module over a k -algebra R , we get,as desired, a nondegenerate balanced bilinear map (11) with A R and R B Artinian (and A R Noetherian),but with R B non-Noetherian.However, neither the examples obtained using (12) nor those gotten as above satisfy all four possibleArtinian conditions on A and B. So we ask
Question 8. If (11) is a nondegenerate balanced bilinear map, and if all of S A, A R , R B and B T areArtinian, must these modules also be Noetherian? If the answer is positive, one could look at intermediate cases, e.g., where three of the above Artinianconditions are assumed. (The case of (12) where S is our given field k, S A = k k, and B T is an Artiniannon-Noetherian right module over a k -algebra T, shows that the assumption that all the above modules except R B are Artinian is not sufficient to prove B T Noetherian; though it does not say whether thoseconditions are sufficient to make S A and/or A R Noetherian.)Some results with the desired sort of conclusion, but with hypotheses of a stronger sort than thosesuggested above, are proved in [9].Alongside Question 8 and its close relatives, one might look for results with the weaker conclusion that A or B have ascending chain condition on sub bimodules . References [1] George M. Bergman,
An Invitation to General Algebra and Universal Constructions,
422 pp. Readable at http://math.berkeley.edu/~gbergman/245 . MR :18001. (Updated every few years. MR review is of 1998 version.)[2] George M. Bergman and Nazih Nahlus,
Homomorphisms on infinite direct product algebras, especially Lie algebras , J. Alg. (2011) 67–104.[3] Carl Faith and Dolors Herbera,
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Math. Z. (1986) 43–52. MR :16037.[5] Jan Krempa,
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Colloq. Math. (2005) 123–135. http://journals.impan.gov.pl/cm/Inf/102-1-11.html MR :16034.[6] T. Y. Lam, A first course in noncommutative rings , Springer GTM, v.131, 1991. MR :16001.[7] T. Y. Lam,
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Springer GTM, v.189, 1999. MR :16001.[8] T. H. Lenagan,
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Comm. Algebra (2002) 1039–1047. Corrigendum at ibid. (2003) 4647–4649. MR :16003, 1995558. University of California, Berkeley, CA 94720-3840, USA
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