Lattice decomposition of modules
aa r X i v : . [ m a t h . R A ] F e b Lattice decomposition of modules
Josefa M. GarcíaPascual JaraLuis M. Merino
Abstract
The first aim of this work is to characterize when the lattice of all submodulesof a module is a direct product of two lattices. In particular, which decompositionsof a module M produce these decompositions: the lattice decompositions . In a first étage this can be done using endomorphisms of M , which produce a decomposi-tion of the ring End R ( M ) as a product of rings, i.e., they are central idempotentendomorphisms. But since not every central idempotent endomorphism produces alattice decomposition, the classical theory is not of application. In a second step wecharacterize when a particular module M has a lattice decomposition; this can bedone, in the commutative case in a simple way using the support, Supp ( M ) , of M ;but, in general, it is not so easy. Once we know when a module decomposes, welook for characterizing its decompositions. We show that a good framework for thisstudy, and its generalizations, could be provided by the category σ [ M ] , the smallestGrothendieck subcategory of Mod − R containing M . Introduction
Let M be a (unitary) right R –module over a (unitary) ring R ; it is well known that de-compositions of M , as a direct sum of two submodules, are parameterized by idempotent February 3, 2021J. M. García ( [email protected] ) Department of Applied Mathematics. University of Granada. E–18071Granada. Spain.P. Jara (Corresponding author) ( [email protected] ) Department of Algebra and IEMath–GR (Instituto deMatemáticas). University of Granada. E–18071 Granada. Spain.L. M. Merino ( [email protected] ) Department of Algebra and IEMath–GR (Instituto de Matemáticas). Uni-versity of Granada. E–18071 Granada. Spain.2020
Mathematics Subject Classification:
Key words: module, ring, lattice, lattice decomposition, Grothendieck category. S = End R ( M ) . Thus, if M = N ⊕ H , there exists e ∈ End R ( M ) suchthat N = e ( M ) , and H = ( − e )( M ) . In general, e is not necessarily central in S , henceit does not produce a decomposition of S in a direct product of two rings. In this paperwe deal with some special decompositions of modules so that the lattice L ( M ) , of allsubmodules of M , will be a direct product of two lattices: the lattice decomposition of M .This kind of decompositions are of interest as if M = N ⊕ H is a lattice decomposition,then every submodule X ⊆ M can be expressed as a direct sum, X = ( X ∩ N ) ⊕ ( X ∩ H ) ,and this property has great importance in order to study the structure of M .From this point of view, we first recall that a lattice decomposition defines an element N ∈ L ( M ) which is distributive; this means that for any X , Y ∈ L ( M ) , the sublattice of L ( M ) generated by N , X and Y is distributive: a well known notion in lattice theory; inaddition, it is complemented, as M = N ⊕ H . Elements of this kind have good propertiesas members of the lattice L ( M ) .We exploit the existence of complemented distributive submodules of a right R –module M . In order to characterize them first we deal with endomorphisms. Thus we show acharacterization of those central idempotent endomorphisms in End R ( M ) that define lat-tice decompositions of M . In the particular case of modules over a commutative ring,these idempotent endomorphisms are those that belong to the closure of R in End R ( M ) ,with respect to the finite topology. In addition, we show that every complemented dis-tributive submodule N ⊆ M is stable under any endomorphism f ∈ End R ( M ) .The behaviour of complemented distributive submodules is also studied, thus it is shownthat for any complemented submodule N ⊆ M , and any index set I we get a comple-mented distributive submodule N ( I ) ⊆ M ( I ) . This property, together with the knowncharacterization of distributive submodules as those submodules N ⊆ M such that forevery submodule H ⊆ M the factor modules N / ( N ∩ H ) and H / ( N ∩ H ) have no non–zero isomorphic subfactors allow us to extend the theory to categories which are defineddirectly from M , as the category σ [ M ] . Indeed, we recover a decomposition theory forthese categories showing that there exists a closed relationship between decompositionof σ [ M ] , as a product of two subcategories, and lattice decompositions of M as a right R –module.The paper is organized in sections. In the first section we recall the notions of product oflattices and the consequences of the existence of a lattice decomposition. In sections twoand three we study distributive submodules of a right R –module, and show that simplesubfactors are decisive to characterize complemented distributive submodules. In partic-ular, if the base ring R is commutative, a direct sum decomposition M = N ⊕ H is a latticedecomposition if, and only if, Supp ( N ) ∩ Supp ( H ) = ∅ . One of the main aims is to relatecomplemented distributive submodules N ⊆ M and central idempotent endomorphisms.Thus we show, in examples, that not every such idempotent endomorphism defines a dis-tributive submodule, and in Theorem (3.11.) we show that it is necessary and sufficient2hat this endomorphism stabilizes every submodule. In consequence, these idempotentendomorphisms are close to multiplication by elements of R , and in Proposition (3.15.)we show that they must belong to the closure of R in End R ( M ) with respect to the finitetopology, whenever R is commutative. In order to extend these results to categories, insection four we establish, Proposition (4.1.), showing that direct powers preserve com-plemented distributive submodules. The strong relationship between decompositions ofthe category σ [ M ] and lattice decompositions of the module M is also studied.References to undefined terms can be find either in the following papers: [ ] , [ ] and [ ] , or in the books: [ ] , [ ] and [ ] . Let L , L be lattices, and define in the cartesian product L × L the operations: ( a , a ) ∧ ( b , b ) = ( a ∧ b , a ∧ b ) and ( a , a ) ∨ ( b , b ) = ( a ∨ b , a ∨ b ) ,then ( L × L , ∧ , ∨ ) is a lattice, and the canonical projections p i : L × L −→ L i , i =
1, 2,are lattice maps. In addition, ( L × L , { p , p } ) is the product of L and L in the categoryof lattices and lattice maps.Examples of lattices appear in many different contexts; we are interested in those latticesthat appear in module theory, i.e., if R is a (unitary) ring and M a right (unitary) R –module, in the lattice L ( M ) of all submodules of M , and in the particular problem ofcharacterizing when L ( M ) is the product of two lattices.For any right R –module M the lattice L ( M ) has extra properties in addition to those thatdefine a lattice, for instance:(1) L ( M ) is bounded , i.e., there exists a bottom element, 0 ⊆ M , and a top one, M .(2) L ( M ) is modular , i.e., for any N , N , N ⊆ M such that N ⊆ N , we have ( N + N ) ∩ N = N + ( N ∩ N ) .If L ( M ) is a product of lattices, using that L ( M ) is bounded, the following easy resultsholds. Lemma 1.1. If L ( M ) is the product of two lattices, say L ( M ) = L × L , with projec-tions { p , p } , M = ( M , M ) ,and 0 = ( , 0 ) ,then L satisfiesthefollowingproperties:(1) L isaboundedlattice withbottom 0 and top M .(2) The map q : L −→ L ( M ) ,defined q ( X ) = ( X , 0 ) ,isaone–to–onelatticemap.(3) Thereisalatticeisomorphismbetween L and { H ⊆ M | H ⊆ M } .(4) The map h : L −→ L ( M ) ,defined h ( X ) = ( X , M ) ,isa one–to–onelattice map.35) Thereisalatticeisomorphismbetween L and { L ⊆ M | M ⊆ L } .The same propertieshold for the lattice L . In particular, M is the direct product of M and M , and thereisalattice isomorphism L ( M ) ∼ = L ( M ) × L ( M ) .P ROOF (1). For any X ∈ L we have ( , 0 ) = ≤ ( X , 0 ) , hence 0 = p ( , 0 ) ≤ p ( X , 0 ) = X , and ( X , M ) ≤ M = ( M , M ) ; therefore, X = p ( X , M ) ≤ p ( M , M ) = M .(2) and (3). It is clear that q is a lattice map, and the announced isomorphism is givenby q .(4) and (5). It is clear that h is a lattice map, and the announced isomorphism is givenby h .Observe that M ∩ M = M + M = M , hence M = M × M . ƒ A right R –module M has a lattice decomposition whenever L ( M ) is a product of twonontrivial lattices.It is clear that not every decomposition of a module M as a direct product gives a latticedecomposition of L ( M ) in a product of lattices. See the following example. Example 1.2.
Consider the abelian group M = Z × Z , the lattice of subgroups of M isnot a product of two nontrivial lattices; in particular, L ( M ) is not the product L ( Z ) ×L ( Z ) . Example 1.3.
A commutative ring A has a lattice decomposition if, and only if, A is theproduct of two nontrivial ideals. Indeed, if L ( A ) = L × L , there exist ideals a , a ⊆ A such that A = a × a . Otherwise, if A = a × a , there are idempotent elements a i ∈ a i , i =
1, 2, such that 1 = a + a . For any ideal a ⊆ A we have a = a a × a a , and anisomorphism L ( A ) ∼ = L ( a ) × L ( a ) . Example 1.4.
This result for non–commutative rings does not hold. Let us consider afield K and the matrix ring M ( K ) of all square matrices of order 2. The ideals a = (cid:129) K K ‹ and a = (cid:129) K K ‹ satisfy M ( K ) = a ⊕ a . Otherwise, each a i is a simple right M ( K ) –module, hence L ( a i ) = { a i } , but L ( M ( K )) is not the product L ( a ) × L ( a ) because, for any 0 = a ∈ K , the right ideal (cid:129) a ‹ M ( K ) is not in this product. SeeCorollary (3.12.) to determine when a ring R have a lattice decomposition as right R –module.Our aim in the next section shall be to show some characterizations of modules having alattice decomposition. 4 Distributive submodules
Let M be a right R –module. If N ⊆ M is a submodule, there exists a short exact sequence0 → N → M → M / N →
0, and maps L ( N ) L ( M ) L ( M / N ) i ∗ / / p ∗ (cid:30) (cid:30) p ∗ o o i ∗ ^ ^ Defined by:(1) i ∗ ( X ) = X , for every X ⊆ N ; it is a lattice homomorphism.(2) i ∗ ( Y ) = Y ∩ N , for every Y ⊆ M ; it satisfies i ∗ ( Y ∧ Y ) = i ∗ ( Y ) ∧ i ∗ ( Y ) , but it is not alattice homomorphism unless N satisfies ( Y + Y ) ∩ N = ( Y ∩ N ) + ( Y ∩ N ) for any Y , Y ⊆ M .(3) p ∗ ( Y ) = ( Y + N ) / N , for every Y ⊆ M ; it satisfies p ∗ ( Y ∨ Y ) = p ∗ ( Y ) ∨ p ∗ ( Y ) , but itis not a lattice homomorphism unless N satisfies ( Y ∩ Y ) + N = ( Y + N ) ∩ ( Y + N ) for any Y , Y ⊆ M .(4) p ∗ ( Y / N ) = Y , for every Y / N ⊆ M / N ; it is a lattice homomorphism.Thus, in the above diagram all maps are lattice maps if, and only if, N satisfies conditionsin (2) and in (3). In [ ] an element in a lattice satisfying property in (3) is called a distributive element , and if it satisfies property in (2), a dual distributive element ,proving in [
5, Theorem III.2.6 ] that an element in a modular lattice is distributive if andonly if it is dual distributive if, and only if, the sublattice generated by N , Y and Y , inthe former notation, is distributive.We call a submodule N ⊆ M distributive whenever ( Y ∩ Y ) + N = ( Y + N ) ∩ ( Y + N ) forany Y , Y ⊆ M , or equivalently if ( Y + Y ) ∩ N = ( Y ∩ N ) + ( Y ∩ N ) for any Y , Y ⊆ M ,and observe that this is also equivalent to the condition that the sublattice of L ( M ) ,generated by N , Y and Y , is distributive.Our aim in this section is to characterize distributive submodules of a module. To dothat we need the following definition. Let M be a right R –module, a subfactor of M isa submodule of a homomorphic image of M . Observe that for any subfactor L of a right5 –module M , and any submodule K ⊆ L we may build a commutative diagram K (cid:0)❅ (cid:31) (cid:31) ❅❅❅❅❅❅❅❅ L ❴✤ (cid:15) (cid:15) ! ! ! ! ❈❈❈❈❈❈❈❈ L / K ❴✤ (cid:15) (cid:15) M / / / / X ! ! ! ! ❈❈❈❈❈❈❈❈ X / K Therefore, if L is a subfactor of M , then K and L / K are also subfactors of M .This situation can be enhanced if we make use of elements of the module M . So, dis-tributive submodules can be also characterized in the following way; where we refer to [
9, Theorem 1.6 ] or [
1, Proposition 1.1 ] for condition (b), and to [ ] for condition (c). Proposition 2.1.
Let M bearight R –module,and N ⊆ M beasubmodule,thefollowingstatements areequivalent:(a) N ⊆ M isadistributivesubmodule.(b) ( N : m ) + ( mR : n ) = R forany m ∈ M and n ∈ N .(c) For everysubmodule H ⊆ M the modules N / ( N ∩ H ) and H / ( N ∩ H ) have no non–zeroisomorphicsubfactors.(d) Foreverysubmodule H ⊆ M ,themodules N / ( N ∩ H ) and H / ( N ∩ H ) havenosimpleisomorphicsubfactors.(e) For any m ∈ M and n ∈ N , the cyclic modules ( n + ( N ∩ mR )) R and mR / ( N ∩ mR ) have no non–zeroisomorphicsubfactors.(f) For any m ∈ M and n ∈ N , the cyclic modules ( n + ( N ∩ mR )) R and mR / ( N ∩ mR ) have no simpleisomorphicsubfactors.P ROOF (a) ⇒ (b). By hypothesis we have N ∩ (( n − m ) R + mR ) = ( N ∩ ( n − m ) R ) + ( N ∩ mR ) ; hence n = x + y , where x ∈ N ∩ ( n − m ) R and y ∈ N ∩ mR .Let a , b ∈ R such that x = ( n − m ) a , hence ma = na − x ∈ N , and y = mb . In addition,we have n ( − a ) = n − na = x + y − na = ( n − m ) a + mb − na = m ( b − a ) ∈ mR . As aconsequence, ( N : m ) + ( mR : n ) = R .(b) ⇒ (a). For any X , Y ⊆ M we always have ( N ∩ X ) + ( N ∩ Y ) ⊆ N ∩ ( X + Y ) . On theother hand, let n = x + y ∈ N ∩ ( X + Y ) , where x ∈ X and y ∈ Y , and consider the pair6 ∈ N and x ∈ M . By hypothesis, we have ( N : x ) + ( x R : n ) = R , there exist a ∈ ( N : x ) , b ∈ ( x R : n ) such that a + b =
1, and we have: n = x a + y a + nb ; since x a , nb ∈ N ∩ X ,hence y a = n − x a − nb ∈ N , whence y a ∈ N ∩ Y . Therefore, n ∈ ( N ∩ X ) + ( N ∩ Y ) .(b) ⇒ (c). For any non–zero subfactor S F of N / N ∩ H , and any subfactor S F of H / ( N ∩ H ) , let us consider the diagram X / / / / ❴✤ (cid:15) (cid:15) X / ( N ∩ H ) ❴✤ (cid:15) (cid:15) f / / / / S F ❴✤ (cid:15) (cid:15) η ∼ = / / S F ❴✤ (cid:15) (cid:15) Y / ( N ∩ H ) g o o o o ❴✤ (cid:15) (cid:15) Y ❴✤ (cid:15) (cid:15) o o o o N / / / / N / ( N ∩ H ) / / / / • • H / ( N ∩ H ) o o o o H o o o o there exists 0 = x ∈ X such that f ( x ) =
0, where x = x + ( N ∩ H ) . Let y ∈ Y suchthat η f ( x ) = g ( y ) , where y = y + ( N ∩ H ) . By the hypothesis ( N : y ) + ( yR : x ) = R ;let 1 = a + b with a ∈ ( N : y ) and b ∈ ( yR : x ) , then x = x ( a + b ) = x a + x b . Onthe other hand, η f ( x a ) = g ( y a ) =
0, whence x a ∈ Ker ( f ) ; since x b ∈ Ker ( f ) , we have x ∈ Ker ( f ) , which is a contradiction.(c) ⇒ (d), (e) ⇒ (f). They are trivial.(f) ⇒ (b). Let x ∈ M and n ∈ N , if ( N : x ) + ( x R : n ) = R , there exists a maximal rightideal m ⊆ R such that ( N : x ) + ( x R : n ) ⊆ m , and for any a ∈ m we have 1 − a / ∈ m , hence1 − a / ∈ ( N : x ) , ( x R , n ) . We proceed as follows:(1) Since 1 − a / ∈ ( N : x ) , then ( − a ) x / ∈ N ∩ x R , and for any a ∈ m we have x = x a in M / ( N ∩ x R ) , i. e., x m $ x R , and x R / ( N ∩ x R ) has a simple subfactor x R / x m ∼ = R / m .(2) Since 1 − a / ∈ ( x R : n ) , then n ( − a ) / ∈ N ∩ x R , and for any a ∈ m we have n = na in M / ( N ∩ x R ) , i. e., n m $ nR , and nR has a simple subfactor nR / n m ∼ = R / m .In any case we have a contradiction. ƒ As a consequence, of the above proposition, if
Mod − R has only, up to isomorphism, onesimple right R –module, for instance if either R has only one maximal right ideal, i.e., R is a local ring, then we have the following proposition; compare with [ ] . Proposition 2.2.
Let R bearingsuchthat Mod − R has,uptoisomorphism,onlyasimplerightmodule,foranypropersubmodule N $ M thefollowingstatementsareequivalent:(a) N ⊆ M isdistributive.(b) N ⊆ mR forany m ∈ M \ N .(c) N iscomparablewitheverynon–zerosubmodule of M .In particular, if 0 = N $ M is a distributive submodule, then Soc ( M ) ⊆ N and it isessential in M . 7 ROOF (a) ⇒ (b). If N ⊆ M is distributive and m ∈ M \ N then NN ∩ mR and mRN ∩ mR have no simple isomorphic subfactors, hence one of them is equal to zero. If mRN ∩ mR = mR ⊆ N , which is a contradiction, hence NN ∩ mR =
0, and N ⊆ mR .(b) ⇒ (c). Let H ⊆ M be a submodule. If H * N , there exists h ∈ H \ N , hence N ⊆ hR ⊆ H .(c) ⇒ (a). Let H ⊆ M be a submodule, then either N ⊆ H , hence NN ∩ H =
0, or H ⊆ N ,hence HN ∩ H = N ⊆ M is essential. If H ⊆ M is simple and H * N , there exists h ∈ H \ N , and N ⊆ hR ⊆ H , so N = H , which is a contradiction. As a consequence, forany simple submodule H ⊆ M we have H ⊆ N , and Soc ( M ) ⊆ N . ƒ A second consequence of the afore–mentioned characterization of distributive submod-ules given in Proposition (2.1.) is the following proposition.
Proposition 2.3.
Let M bea right R –module, N ⊆ M adistributivesubmodule and H ⊆ M asubmodulesuch that N ∩ H = R ( N , H ) = = Hom R ( H , N ) .P ROOF
For any homomorphism f : N −→ H we have Im ( f ) is a common subfactorof N and H , hence Im ( f ) =
0, and f =
0. The same happens for any homomorphism g : H −→ N . ƒ Finally, we observe that distributive submodules are preserved by some module construc-tions.
Proposition 2.4.
Let M bearight R –module,the followingstatements hold:(1) If N ⊆ M is a distributive submodule, for any submodule H ⊆ M the submodule ( N + H ) / H ⊆ M / H isdistributive.(2) Foreveryfamilyofdistributivesubmodules { N i ⊆ M | i ∈ I } thesum P i N i ⊆ M isadistributivesubmodule.(3) If N , N ⊆ M are distributivesubmodules,then N ∩ N ⊆ M isdistributive.P ROOF (1). Since N ⊆ M is distributive, for any m ∈ M and any n ∈ N we have ( N : m ) + ( mR : n ) = R . The result follows from the following inclusions ( N : m ) ⊆ (cid:129) N + HH : m ‹ and ( mR : n ) ⊆ ( mR : n ) ,where x = x + H for any x ∈ M . 82) and (3). They are well known for finite join and meet of distributive elements of alattice. It is not difficult to see that in the case of sum it can be extended to the infinitecase. ƒ If A is a commutative ring and Σ ⊆ A a multiplicatively closed subset, then we have: Proposition 2.5.
Let M bean A –module;if N ⊆ M isdistributive,then Σ − N ⊆ Σ − M isdistributive.P ROOF
We apply Proposition (2.1.(b)). Let m ∈ Σ − M , and n ∈ Σ − N . We have theequalities: ( Σ − N : m ) = { as ∈ Σ − A | exists t ∈ Σ , such that mat ∈ N } = Σ − ( N : m ) , and ( m Σ − A : n ) = Σ − ( mA : n ) .Since ( N : m ) + ( mA : n ) = A , then ( Σ − N : m ) + ( m Σ − A : n ) = Σ − A , and Σ − N ⊆ Σ − M is distributive. ƒ In particular, if p ⊆ A is a prime ideal, and consider Σ = A \ p , then we have: Corollary 2.6.
Let M be an A –module and p ⊆ A be a primeideal. If N ⊆ M is distribu-tive,then N p ⊆ M p isdistributive.Let us consider the following example. Example 2.7.
Let M = Z the cyclic abelian group of eight elements. Since L ( Z ) isa distributive lattice, then every submodule is distributive, hence Soc ( Z ) = Z $ Z are proper distributive submodules. Otherwise, Z has no nontrivial direct summands,hence L ( Z ) has no a lattice decomposition, see next section.This means that the existence of distributive submodules does not imply a lattice decom-position. On the other hand, for every lattice decomposition M = M ⊕ M we shall provethat M and M are distributive submodules. Let M = N ⊕ H be a decomposition, and let us denote i : N −→ M and i : H −→ M the inclusions and q : M −→ N , q : M −→ H the projections. If p : M −→ M / N is the9rojection, there is an isomorphism f : M / N ∼ = H such that f p = q ; thus, we have adiagram involving the lattices: L ( N ) L ( M ) L ( H ) i ∗ / / i ∗ (cid:30) (cid:30) i ∗ o o i ∗ ^ ^ Being i ∗ and i ∗ lattice homomorphisms. On the other hand, i ∗ and i ∗ are ∧ –homomor-phisms , i.e., i ∗ j ( Y ∧ Y ) = i ∗ j ( Y ) ∧ i ∗ j ( Y ) , for every Y , Y ⊆ M , j =
1, 2, and they arelattice homomorphisms whenever N , or equivalently H , is a distributive submodule. Inthis case, L ( M ) is the direct product of L ( N ) and L ( H ) ; and for every Y ⊆ M we have Y = i ∗ i ∗ ( Y ) ∨ i ∗ i ∗ ( Y ) .The first result is a direct consequence of the characterizations of distributive submod-ules, in Proposition (2.1.), which are direct summands, i.e., complemented distributivesubmodules . Lemma 3.1.
Let M = N ⊕ H be adirectsum, the followingstatementsare equivalent:(a) N ⊆ M isdistributive.(b) N and H have noisomorphicsimplesubfactors.(c) Ann ( n ) + Ann ( h ) = R forany n ∈ N and h ∈ H .(d) Forany submodule X ⊆ M wehave X = ( X ∩ N ) + ( X ∩ H ) = ( X + N ) ∩ ( X + H ) .(e) H ⊆ M isdistributive.If M is a right R –module satisfying the equivalent statements in the above lemma we saythat M = N ⊕ H is a lattice decomposition of M .In this context, if A is a commutative ring, we have the following result that characterizescomplemented distributive submodules. Corollary 3.2.
Let A be a commutative ring, M an A –module, and M = N ⊕ H be adecompositionina directsum, thefollowingstatementsare equivalent:(a) N ⊆ M isadistributivesubmodule.(b) Supp ( N ) ∩ Supp ( H ) = ∅ .(c) H ⊆ M isadistributivesubmodule.P ROOF (a) ⇒ (b). By Lemma (3.1.), for any n ∈ N and any h ∈ H we have Ann ( n ) + Ann ( h ) = A . If p ∈ Supp ( N ) ∩ Supp ( H ) , there exist n ∈ N and h ∈ H such that ( Rn ) p = ( Rh ) p =
0, hence Ann ( n ) ⊆ p and Ann ( h ) ⊆ p , which is a contradiction.(b) ⇒ (a). If N is not distributive, by Lemma (3.1.), N and H have isomorphic simplesubfactors, hence Supp ( N ) ∩ Supp ( H ) = ∅ , which is a contradiction. ƒ As a consequence, if in addition A is a noetherian ring, then lattice decomposition isinherited by injective hulls. 10 orollary 3.3. Let A beacommutativenoetherianring, M an A –module,and M = N ⊕ H bea latticedecomposition,then E ( M ) = E ( N ) ⊕ E ( H ) isalatticedecomposition.P ROOF
It is a direct consequence of the well known fact that Supp ( N ) = Supp ( E ( N )) for any A –module N . ƒ Also we have the following straightforward result.
Lemma 3.4.
Let M be a right R –module such that L ( M ) is a direct product of two lat-tices,say L ( M ) = L × L ,thereexist M , M ⊆ M such that(1) M = M ⊕ M .(2) M and M are distributivesubmodules.(3) L i ∼ = L ( M i ) ,forevery i =
1, 2.(4) L ( M ) = [ M ] × [ M ] .(5) There exists an idempotent endomorphism e ∈ End R ( M ) such that e ( M ) = M , and ( − e )( M ) = M . Inaddition, e | M = id M , and ( − e ) | M = id M .The existence of a non trivial idempotent endomorphism in End R ( M ) is necessary, butit is not sufficient to get a lattice decomposition, i. e., not every idempotent endomor-phism e ∈ End R ( M ) defines a lattice decomposition of L ( M ) . Let us illustrate it by someexamples. Example 3.5.
Let us consider the abelian group M = Z × Z . It is clear that M has notnon trivial distributive submodules, but End ( M ) has non trivial idempotents. Indeed,the ring End ( Z × Z ) = M ( Z ) has six non trivial idempotent endomorphisms (cid:129) ‹ , (cid:129) ‹ , (cid:129) ‹ , (cid:129) ‹ , (cid:129) ‹ and (cid:129) ‹ , but no one of them defines a lattice de-composition. Example 3.6.
In the positive we have: If we consider the abelian group M = Z × Z ,then End ( Z × Z ) ∼ = End ( Z ) × End ( Z ) ; this ring decomposes, and there is a non trivialidempotent that produces a lattice decomposition of M .The next example shows that in a lattice decomposable module not every idempotentendomorphism provides a lattice decomposition.11 xample 3.7. Let us consider the abelian group M = Z × Z × Z . The lattice of allsubgroups is: 〈 e , e , f 〉 rrrrrrrrrr ◆◆◆◆◆◆◆◆◆◆◆ 〈 e , f 〉 〈 e , f 〉 〈 e + e , f 〉 〈 e , e 〉 ✈✈✈✈✈✈✈✈✈ ▲▲▲▲▲▲▲▲▲▲ ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ 〈 f 〉 ▲▲▲▲▲▲▲▲▲▲ ♣♣♣♣♣♣♣♣♣♣♣♣ 〈 e 〉 ❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦ 〈 e 〉 ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ 〈 e + e 〉 ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ 〈 〉 ❍❍❍❍❍❍❍❍❍ rrrrrrrrrr ❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥ The decomposition given by N = 〈 e , e 〉 , N = 〈 f 〉 corresponds to the idempotent centralendomorphism e ∈ End ( M ) defined by e ( e ) = e ( e ) = e ( f ) = − e ( e ) = ( e ) = ( f ) = f and it defines the lattice decomposition of L ( M ) represented in the above diagram. Inaddition, e defines a lattice decomposition of the ring S = End R ( M ) . Question 3.8.
Does every central idempotent endomorphism in End R ( M ) induce a lat-ticedecompositionof L ( M ) ?The answer is no, as the following example shows. Example 3.9.
Let us consider M = Z ( ) × Z ( ) , where Z ( ) and Z ( ) are the localizationof Z at 2 Z and 3 Z , respectively. We claim Hom ( Z ( ) , Z ( ) ) =
0. Indeed, for any f ∈ Hom ( Z ( ) , Z ( ) ) , let f ( ) = ad , then f ( ) = bc and it satisfies: 3 bc = ad , and 3 bd = ac . Byhypothesis 3 ∤ c , hence 3 | a . Similarly, if f ( t ) = bc , then 3 t d b = ac , and 3 t | a for every t ∈ N , which implies a = ( M ) = End ( Z ( ) ) × End ( Z ( ) ) ; hence, in End ( M ) there exist centralidempotent elements. Since Z ( ) and Z ( ) have a non zero isomorphic submodule incommon, then Z ( ) ⊆ M is not distributive. Remark 3.10.
Observe that any central idempotent element e ∈ End R ( M ) = S inducesa complemented distributive two–sided ideal eS ⊆ S , with complement ( − e ) S , see12orollary (3.12.) below. The above example shows that if e ∈ End R ( M ) is a centralidempotent element, then the submodule e ( M ) ⊆ M is not necessarily distributive, evenif S has a lattice decomposition.In conclusion, the question is: how we may describe the idempotent elements in End R ( M ) that produce lattice decomposition? The next theorem provides the answer. Theorem 3.11.
Let M be a right R –module, N ⊆ M be a direct summand, and let e ∈ End R ( M ) beanidempotentsuchthat e ( M ) = N ,thefollowingstatementsareequivalent:(a) N iscomplementeddistributive.(b) e isacentraland e ( X ) ⊆ X forany submodule X ⊆ M .P ROOF If N ⊆ M is a complemented distributive submodule with complement H , thenfor every submodule X ⊆ M we have X = ( N ∩ X )+( H ∩ X ) = ( e ( M ) ∩ X )+(( − e )( M ) ∩ X ) ,that expressed in terms of the endomorphism e , implies: e ( X ) = e ( e ( M ) ∩ X ) + e (( − e )( M ) ∩ X ) ⊆ e ( M ) ∩ X ⊆ X ,Hence a necessary condition on the endomorphism e to get a complemented distributivesubmodule is e ( X ) ⊆ X , for any submodule X ⊆ M . This is also a sufficient condition;indeed, if e ( X ) ⊆ X (or equivalently ( − e )( X ) ⊆ X ), then, for any element x ∈ X wehave x = e ( x ) + ( − e )( x ) , where e ( x ) ∈ e ( M ) ∩ X and ( − e )( x ) ∈ ( − e )( M ) ∩ X .Let M be a right R –module M , for any complemented distributive submodule N ⊆ M with complement H , following Cohn’s theory in [ ] , for any homomorphism f : N −→ H we define a submodule Γ ( f ) = { ( x , f ( x )) ∈ M | x ∈ N } . Since Γ ( f ) is a complementof N , and it is unique, it follows that Γ ( f ) = Γ ( ) = N . In particular, End R ( N , H ) = R ( M ) ∼ = End R ( N ) × End R ( H ) . As a consequence End R ( N ) is a directsummand ideal of End R ( M ) , and there exists a central idempotent e ∈ End R ( M ) such thatEnd R ( N ) = e End R ( M ) . Therefore, the idempotent endomorphism that defines the latticedecomposition is central. ƒ An idempotent endomorphism e ∈ End R ( M ) is named fully invariant if e ( X ) ⊆ X for anysubmodule X ⊆ M . Corollary 3.12.
Let R be a ring, and e ∈ R idempotent, the following statements areequivalent:(a) eR ⊆ R isacomplementeddistributiverightideal.(b) e iscentral.(c) R = Re × R ( − e ) isadirectproductofrings.In this case R has a lattice decomposition in a direct product of two (twosided) ideals: Re and R ( − e ) , and conversely; any direct sum decomposition R = A ⊕ B , with two(two–sided)ideals, isalattice decomposition.13 emark 3.13. Observe that if N ⊆ M is a complemented distributive submodule, de-fined by the idempotent endomorphism e ∈ End R ( M ) , then End R ( M ) e ⊆ End R ( M ) is acomplemented distributive two–sided ideal; therefore a ring with unity e .In particular, we have the next result that enhances [
9, Proposition 4.3 ] and [
1, Proposi-tion 1.3 ] . Lemma 3.14.
Every complemented distributive submodule N ⊆ M is stable under any f ∈ End R ( M ) .P ROOF
By hypothesis there exists e ∈ End R ( M ) , central idempotent, such that N = e ( M ) .For every endomorphism f ∈ End R ( M ) we have f ( N ) = f e ( M ) = e f ( M ) ⊆ e ( M ) = N . ƒ If we analysed the question if every central idempotent endomorphism e ∈ End R ( M ) de-fines a complemented distributive submodule e ( M ) ⊆ M . We found that Example (3.9.)gave a negative answer. A necessary and sufficient condition in order to get e ( M ) ⊆ M a complemented distributive submodule, as we have seen before, is that for every sub-module X ⊆ M we have e ( X ) ⊆ X . In consequence, for every element m ∈ M we have e 〈 m 〉 ⊆ 〈 m 〉 , and there exists a m ∈ R such that em = ma m .Let us explore this condition in order to get more examples of complemented distributivesubmodules.Of particular interest is the situation in which A is a commutative ring. In this case wehave: Proposition 3.15.
Let A be a commutative ring and N ⊆ M be a submodule of the A –module M , thefollowingstatements are equivalent:(a) N ⊆ M isacomplementeddistributivesubmodule.(b) There exists a central idempotent endomorphism e ∈ End A ( M ) such that e ( M ) = N and e belongs to the closure of A / Ann ( M ) ⊆ End A ( M ) in the finite topology,i.e., the topology with subbase of neighbourhoods of any f ∈ End A ( M ) given by B ( { m , . . . , m t } , f ) = { g ∈ End A ( M ) | g ( m i ) = f ( m i ) , i =
1, . . . , t } .P ROOF (a) ⇒ (b). Let N ⊆ M be a complemented distributive submodule and e ∈ End A ( M ) be a central idempotent element such that e ( M ) = N , and e ( X ) ⊆ X for everysubmodule X ⊆ M . If H = ( − e )( M ) for any m ∈ M there are n ∈ N and h ∈ H suchthat m = n + h , and there exists a ∈ A such that n = e ( m ) = ma = ( n + h ) a , hence ha = n ( − a ) = m , . . . , m t ∈ M with m i = n i + h i and e ( m i ) = m i a i . Let us consider new elements: m i j = n i + h j ; observe that there are elements a i j ∈ A such that e ( m i j ) = m i j a i j and a ii = a i ; therefore n i j ( − a i j ) = h i j a i j =
0, for any i , j ∈ {
1, . . . , t } .14f we fix i , we have the elements m i = n i + h , . . . , m it = n i + h t , that satisfy: n i ( − a i · · · a it ) = n i − n i ( a i · · · a it ) = h j ( a i · · · a it ) = e ( m i j ) = m i j ( a i · · · a it ) Thus we may assume a i = · · · = a it , and all of them are equal to the product a i · · · a it j for the former a i j . Since this can be done for every index i , then we may assume a i = · · · = a it for every index i .In this new context, if we fix j , we have elements m j = n + h j , . . . , m t j = n t + h j , thatsatisfy: n i ( − a j ) · · · ( − a t j ) =
0, for every index i , ( − a j ) · · · ( − a t j ) = − P i a i j + P i < i a i j a i j + · · · + ( − ) t a j · · · a t j ,let us denote x = P i a i j − P i < i a i j a i j + · · · + ( − ) t + a j · · · a t j , h j x = e ( m i j ) = m i j x .Thus we may assume a j = · · · = a t j , and all of them are equal to the element x definedjust before using the former a i j . Since this can be done for every index j , then we mayassume a j = · · · = a t j = x for every index j . In consequence, we have found that e ( m i j ) = m i j x , and in particular x ∈ B ( { m , . . . , m t } , e ) ∩ A . Therefore, e belongs to theclosure of A / Ann ( M ) in the finite topology of End A ( M ) .(b) ⇒ (a). It is consequence of Theorem (3.11.). ƒ Corollary 3.16.
For any commutativering A and any finitelygenerated A –module M ,if N ⊆ M is a complemented distributive submodule with idempotent endomorphism e ,thereexists a ∈ A such that e ( m ) = ma forany m ∈ M .Even in this particular case we may characterize complemented distributive submodules. Corollary 3.17.
Let A beacommutativeringandlet M beafinitelygenerated A –module;every complementeddistributive submodule N ⊆ M determinesan idempotentelementin A / Ann ( M ) . Andconversely,everynonzeroidempotentin A / Ann ( M ) definesanonzerocomplementedand distributivesubmoduleof M . Example 3.18.
Let us consider the abelian group M = Z × Z × Z in the Example (3.7.);we have Ann ( M ) = Z , and the non trivial idempotents of Z / Z are 4 and 3, which definethe subgroups 2 M = 〈 f 〉 and 3 M = 〈 e , e 〉 .15 Decomposition of categories. Examples
In this section we will apply the above result to study the decomposition of the categoryof right R –modules and the category σ [ M ] defined by a right R –module M . The category Mod − R Each lattice decomposition of R , as right R –module, is defined by a central idempotentelement e ∈ R , hence R = Re × R ( − e ) , being Re and R ( − e ) rings (and (twosided)ideals). Therefore, we have a decomposition of the module category as Mod − R ∼ = Mod − eR × Mod − ( − e ) R . The category σ [ M ] The lattice decomposition theory of a right R –module M is closely linked to the structureof the M module and also to the structure of submodules of the modules it generates;there is a category that studies precisely these modules: the category σ [ M ] . The category σ [ M ] , as defined in [ ] , is the full subcategory of Mod − R whose objects are all theright R –modules isomorphic to modules subgenerated by M , i.e., submodules of factorsof direct sums of copies of M .Our aim is to study under which circumstances the category σ [ M ] is a direct product oftwo categories N and H .Let us assume F : σ [ M ] ∼ = N × H , without losing of generality we may assume N and H are subcategories of σ [ M ] , the objects of N × H are pairs ( X , Y ) , where X is andobject of N and Y an object of H , and F ( M ) = ( N , H ) satisfying M = N ⊕ H .First we need a technical result, which will be useful in later developments. Proposition 4.1.
Let N ⊆ M beacomplementeddistributivesubmodule,foreveryindexset I wehave N ( I ) ⊆ M ( I ) isacomplementeddistributivesubmodule. Thereciprocalalsoholds.P ROOF
It is obvious that if M = N ⊕ H , then M ( I ) = N ( I ) ⊕ H ( I ) . Otherwise, if S is asimple subfactor of N ( I ) , there exists a finite subset F ⊆ I such that S is a subfactor of N ( F ) , hence S is a subfactor of N . Since N and H have no isomorphic simple subfactors, N ( I ) y H ( I ) have no isomorphic simple subfactors, hence N ( I ) ⊆ M ( I ) is a complementeddistributive submodule. ƒ Every submodule Z ⊆ M corresponds to a pair ( X , Y ) , object of N × H satisfying X ⊆ N , Y ⊆ H and Z = X ⊕ Y , hence N ⊆ M is a distributive submodule of M with complement H . 16onversely, for any complemented and distributive submodule N ⊆ M , with complement H ⊆ M , and any index set I , we have N ( I ) ⊆ M ( I ) is a complemented and distributivesubmodule with complement H ( I ) , see Proposition (4.1.), and any submodule X ⊆ M ( I ) can be written as X = ( X ∩ N ( I ) ) ⊕ ( X ∩ H ( I ) ) , hence M ( I ) X ∼ = N ( I ) X ∩ N ( I ) ⊕ H ( I ) X ∩ H ( I ) ∼ = N ( I ) + XX ⊕ H ( I ) + XX , and N ( I ) X ∩ N ( I ) ∼ = N ( I ) + XX ⊆ MX is a complemented and distributive submodule.As a consequence, every submodule Y ⊆ M ( I ) X can be written as Y = (cid:18) Y ∩ N ( I ) + XX (cid:19) ⊕ (cid:18) Y ∩ N ( I ) + XX (cid:19) , where Y ∩ N ( I ) + XX is an object of σ [ N ] , and Y ∩ H ( I ) + XX is an objectof σ [ H ] . Therefore there is a category isomorphism σ [ M ] ∼ = σ [ N ] × σ [ H ] . Comparewith [
11, Proposition 2.2 ] and [
13, 2.4 ] . Theorem 4.2.
With the above assumptions. Let M be a right R –module, the followingstatements areequivalent:(a) σ [ M ] isadirectproductoftwocategories, σ [ M ] ∼ = σ [ N ] × σ [ H ] .(b) M has acomplementedand distributivesubmodule N ⊆ M ,withcomplement H .As a consequence of Corollary (3.3.), if A is a noetherian commutative ring, M an A –module, and E ( M ) its injective hull, for any complemented distributive submodule N ⊆ M we have that E ( N ) ⊆ E ( M ) is distributive. This result does not necessarily hold in anon–commutative framework as the following example shows, see [ ] . Example 4.3.
Let K be a field and R = (cid:129) K K K ‹ be a ring. The maximal right ideals of R are P = (cid:129) K K ‹ and Q = (cid:129) K K ‹ ; hence there are, up to isomorphism, two differentsimple right R –modules. Let us consider the right R –module N = (cid:129) K
00 0 ‹ , which isisomorphic to R / Q , and the cyclic right R –module E = (cid:129) K K ‹ , generated by (cid:129) ‹ .Since E is injective and the inclusion N ⊆ E is essential, then E is the injective hull of N .In addition we have an isomorphism E / N ∼ = R / P .Consider now the right R –module M = N ⊕ ( E / N ) . Since both factors are simple right R –module, M has a lattice decomposition; N ⊆ M is a complemented distributive submod-ule. Otherwise, E ( M ) = E ( N ) ⊕ E ( E / N ) = E ⊕ ( E / N ) , and E ⊆ E ( M ) is not a distributivesubmodule. Indeed, E ( M ) has no nontrivial complemented distributive submodules.17 eferences [ ] A. Barnard,
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