aa r X i v : . [ m a t h . A C ] D ec G-GRADED IRREDUCIBILITY AND THE INDEX OFREDUCIBILITY
CHENG MENG
Abstract.
Let R be a commutative Noetherian ring graded by a torsionfreeabelian group G . We introduce the notion of G -graded irreducibility and provethat G -graded irreducibility is equivalent to irreducibility in the usual sense.This is a generalization of Chen and Kim’s result in the Z -graded case. Wealso discuss the concept of the index of reducibility and give an inequality forthe indices of reducibility between any radical non-graded ideal and its largestgraded subideal. introduction Let R be a commutative Noetherian ring, M an R -module, and N an R -submoduleof M . It’s known that N has an irreducible decomposition, that is, N is an inter-section of irreducible submodules in M . When R, M, N are all graded with respectto a torsionfree abelian group G , we can talk about G -graded irreducible submod-ules of M and irreducible decomposition of N in M in the category of G -gradedmodules. It’s natural to ask whether these two irreducibilities are the same. Moreprecisely, we want to know whether graded irreducibility implies irreducibility inthe nongraded sense. It’s well known that irreducibility implies being primary; in[1, IV.3.3.5] we know being graded primary is the same as being primary. Chenand Kim proved in [3] that the two irreducibilities are the same in the Z -gradedcase. In this paper we extend this result to the case of any G -grading where G is atorsionfree abelian group. In particular, as a consequence, a G -graded irreducibledecomposition is an irreducible decomposition in the usual sense, and both indexesof reducibility, defined for G-grading and in the usual sense, will be the same. Fi-nally we estimate the indexes of reducibility of a nongraded ideal and its largestgraded subideal. We prove one inequality in the radical case and show by examplethat it fails in general case.In all the sections below we make the following assumptions unless otherwisestated: Assumption 1.1. R is a commutative Noetherian ring, M is a finitely generated R -module. When we say R and M are graded without mentioning the group usedfor grading, we are assuming that they are G -graded for a torsionfree abelian group G . The identity element of G is denoted by 0. The reason for these assumptions are as follows.
Date : December 18, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Graded module, graded irreducibility, index of reducibility.
When R is Noetherian and N ⊂ M are Noetherian modules, we have a finiteirredundant irreducible decomposition for N. So the index of reducibility definedbelow will make sense.The torsionfree property is essential. In fact, in the G -graded case where G hastorsion, the definition of prime ideals, primary ideals, associated primes will bedifferent. An example is the group algebra k [ Z ] = k [ x ] / ( x − x + 1) , ( x −
1) are all nongraded, so here we need a different definition for gradedassociated primes. Such definitions can be found in [6]. In the torsionfree case,a graded prime ideal is just a prime ideal that is graded; and the same holds forgraded primary submodules and graded associated prime ideals.2. preliminaries
We recall the following standard definitions.
Definition 2.1.
Let ( G, +) be an abelian group. A ring R is said to be G -graded ifthere is a family of additive subgroups R g such that R = ⊕ g ∈ G R g and R g R h ⊂ R g + h for any g, h ∈ G . For a G -graded ring R , an R -module M is G -graded if there is afamily of additive subgroups M g such that M = ⊕ g ∈ G M g and R g M h ⊂ M g + h forany g, h ∈ G . Definition 2.2.
Let R be a Noetherian ring, M an R -module, N an R -submoduleof M . Then(1) The submodule N is called irreducible if whenever N , N are two submoduleof M satisfying N ∩ N = N we have N = N or N = N .(2) Suppose moreover that R, M, N are G -graded. Then N ⊂ M is called G -graded-irreducible, or simply graded-irreducible when G is clear, if whenever N , N are two G -graded submodule of M satisfying N ∩ N = N we have N = N or N = N .(3) The submodule N is called primary if M/N has only one associated prime. Ifthis prime is p we say N is p -primary. The set of associated primes of a module M is denoted by Ass ( M ).The above definitions hold for N ⊂ M if and only if they hold for 0 ⊂ M/N .The following property is well known.
Property 2.3.
An abelian group is torsionfree if and only if it can be totally or-dered.Proof.
The if part is trivial. Now for the converse, if G is torsionfree, we can embed G into some Q -vector space, order the basis element, and give the lexicographicorder on the vector space and restrict this order to G . (cid:3) So now we can equip each torsionfree abelian group with a total order. We havethe following property.
Property 2.4.
Let R be a graded ring satisfying (1.1). Then(1) Let p be a graded proper ideal in R such that if f, g are homogeneous ele-ments in R , f g ∈ p , then f ∈ p or g ∈ p . Then p is a prime ideal. -GRADED IRREDUCIBILITY AND THE INDEX OF REDUCIBILITY 3 (2) Let M be a graded R -module, N be a graded submodule of M . Then everyassociated prime of M/N is graded and is the annihilator of a homogeneouselement. In particular, N ⊂ M is primary if and only if M/N has onlyone associated G -graded prime.(3) If N , M are as in (2), then there is a graded primary decomposition.(4) If N , M are as in (2) and N ⊂ M is graded irreducible, then M/N is gradedprimary, hence primary with a unique graded associated prime.Proof.
For (1), See [7, A.II.1.4]. For (2), See [7, A.II.7.3] or [5, Prop 3.12]. For (3),See [7, A.II.7.11], or [5, Ex 3.5]. (4) is a corollary of (3). (cid:3)
The following definition comes from [7, A.I.4] ,[2, definition 1.5.13], and [3] inthe Z -graded case. In [2], [3] ” G -graded local” is called ”*local”. Definition 2.5. A G -graded maximal ideal of R is a G -graded ideal m which ismaximal with respect to inclusion in all G -graded ideals properly contained in R .A G -graded ring R is called G -graded local if it has a unique G -graded maximalideal. A G -graded field is a G -graded ring k such that all the nonzero homogeneouselements in k are invertible. Remark . A G -graded ideal m is a G -graded maximal ideal if and only if R/ m is a G -graded field. In particular, if k = 0 is a G -graded ring, then it’s a gradedfield if and only if it has only two G -graded ideals, namely 0 and k , if and only if 0is a G -graded maximal ideal of k . Definition 2.7.
Let R be a G -graded ring. For an ideal I ⊂ R which is notnecessarily G -graded, as in [2] and [3], we define I ∗ to be the ideal of R generatedby all the homogeneous elements in I . Remark . Assume (1.1). Since G is torsionfree, Property 2.4(1) yields that p ∗ is a graded prime ideal contained in p when p is a prime ideal of R. In particular,every G -graded maximal ideal m in R is a prime ideal of R , because m is containedin some (not necessarily graded) maximal ideal n , and it follows that m = n ∗ bydefinition. So a graded field k must be a domain. Therefore it makes sense to talkabout the rank of a k -module if k is a graded field. Definition 2.9.
Let R be G -graded, M a G -graded R -module, and p a G -gradedprime of R . The homogeneous localization of M at p , denoted by M ( p ) , is W − M where W is the multiplicative set of all homogeneous elements not in p .If R is graded, p is a graded prime of R , then R ( p ) is graded local. Lemma 2.10.
Let R be a Noetherian ring, p a prime ideal of R , M a finitelygenerated R -module, and N an R -submodule of M which is p -primary in M . Then N is irreducible in M if and only if N p is irreducible in M p . Moreover, if R, M, N, p are all G -graded, then N is graded-irreducible in M if and only if N ( p ) is graded-irreducible in M ( p ) .Proof. See [3, Lemma 2]. The proof is for the Z -graded case but can be applied tothe G -graded case. (cid:3) The structure of modules over graded fields
It’s well known that if k is a field, then every vector space over k is free. Herewe prove a similar result when k is a graded field. CHENG MENG
Definition 3.1.
Let R be a G -graded ring. We define the support of R , denotedby Supp ( R ), to be { g ∈ G : R g = 0 } .If R is a domain, then Supp ( R ) is a subsemigroup of G. If k is a graded field,then Supp ( k ) is a subgroup of G.The following two theorems have more general versions using the notions ofstrongly graded rings and graded division rings, see [7, A.I.3 and A.I.4]. We presentexplicit proofs in our particular case. Theorem 3.2.
Let G be a group, and k a G -graded field. Then k is a field, and k g ∼ = k as a k -vector space for any g ∈ Supp ( k ) .Proof. Every nonzero element in k has an inverse in k , so k is a field. Take anynonzero u ∈ k g . Then the multiplication by u is a k isomorphism of k → k g withan inverse which is the multiplication by u − . (cid:3) Theorem 3.3.
Let G be a torsionfree abelian group, and k a G -graded field. Thenany G -graded k -module M is free over k .Proof. Let G ′ = Supp ( k ). Let S be a set of representatives in G of all the cosetsin G/G ′ . Then M = ⊕ ( M g ) g ∈ G ∼ = ⊕ s ∈ S ⊕ (( M sh ) h ∈ G ′ ) as k -module. So aftershifting it suffices to prove ⊕ ( M h ) h ∈ G ′ is a free k -module for any s and any graded k -module M . Now M is a k -vector space. Let { e i , i ∈ I } be a basis for M andchoose a basis u g in k g for each degree g ∈ G . Then in each degree, u g ∗ e i is a k -basis for M g . This means that M = ⊕ i ∈ I ke i . Hence M is a free k -module. (cid:3) We want to restrict to the case where G is finitely generated using the Noetheriancondition. We can do this in the case of the group algebra.
Theorem 3.4.
Let G be a finite generated torsionfree abelian group, say Z n . Thenevery G -graded field k is isomorphic to k [ G ′ ] as a graded field, where G ′ is thesupport of k.Proof. Since G ′ is a subgroup of G , it is still a finitely generated torsionfree abeliangroup, say ⊕ mi =1 Z e i .For each i take a nonzero element a i in k e i . Then for any h = n e + n e + · · · + n m e m ∈ G ′ , a n a n · · · a n m m is a nonzero element in k h , thus k h = k ∗ a n a n · · · a n m m . This means that k = k [ a , a , ..., a m ] ∼ = k [ G ′ ]. (cid:3) Remark . The conclusion of the theorem above is not true in general for tor-sionfree abelian groups which are not finitely generated. In fact, we have to finda basis for all the nonzero components of the graded field or graded module; andthere is no guarantee that one can find a collection of bases, labeled by the group,that is closed under multiplication if the group is not finitely generated. There aredifferent isomorphic classes of such graded fields corresponding to the cohomologyclasses in H ( G ′ , k ∗ ); see [8, Ex 1.5.10].From the following theorem we see that if k is a Noetherian group algebra, thenits support is finitely generated. So we may assume G is finitely generated in thiscase. Theorem 3.6.
Let k = k [ G ] be a group algebra over a field k , G any abeliangroup, not necessarily torsionfree. Then k is Noetherian if and only if G is finitelygenerated. -GRADED IRREDUCIBILITY AND THE INDEX OF REDUCIBILITY 5 Proof.
The if direction is obvious since in the finitely generated case the groupalgebra is the localization of a quotient of a polynomial ring over a field. Nowsuppose G is not finitely generated. Consider any finitely generated ideal I . Thegenerators live in finitely many degrees. Let H be the subgroup generated by thesedegrees, then I must be G/H -graded for some finitely generated H . Now considerthe map π : k → k , Σ a i g i → Σ a i . The kernel J is an ideal in k [ G ] generated by( e g − g ∈ G . If H is a subgroup such that J is G/H -graded then we must have G = H . Thus J cannot be finitely generated. (cid:3) G-graded irreducibility implies irreducibility
In this section we prove our main result, that is, a graded irreducible submoduleof a graded module is irreducible.
Definition 4.1.
Let ( R, m , k ) be a local ring or a graded-local ring. Let M be an R -module. The socle of M , denoted by soc ( M ), is (0 : M m ). When M is graded, soc ( M ) is also graded. In both cases soc ( M ) is a free k-module. Lemma 4.2.
Let R be a Noetherian ring, M a finitely generated R -module, and N ⊂ M a submodule. Suppose:(1) ( R, m , k ) is local and M/N is Artinian or (2)
R, M and N are all G -graded, ( R, m , k ) is graded-local, M/N is m -primary. Then N ⊂ M is irreducible (resp. graded-irreducible) if and only if soc ( M/N ) has rank 1.Proof. We may assume N = 0 after replacing M with M/N . Now suppose the rankof soc ( M ) is at least two. Then there exist a k-basis of soc ( M ). Take the first twobasis element: e , e ∈ soc ( M ) ⊂ M then Re ∩ Re = 0. So 0 is not irreducible.Now suppose the rank of soc ( M ) is one, say, soc ( M ) is Re ∼ = k as an R-module.Then for any N , N ∈ M , Ass ( N ) = Ass ( N ) = Ass ( M ) = { m } because Ass ( M )consists of one prime and Ass ( N ) , Ass ( N ) are nonempty subsets of Ass ( M ). Now soc ( N ) = 0 , soc ( N ) = 0 so we can take nonzero e ∈ soc ( N ) , e ∈ soc ( N ). Theymust all lie in soc ( M ) = Re . Now k is a domain, hence 0 is an irreducible k -submodule of k , hence 0 is an irreducible R -submodule of the R -module k . So Re ∩ Re = 0 in Re ∼ = k . So 0 is irreducible. In the graded case, just take all themodules to be graded and elements to be homogeneous. (cid:3) Theorem 4.3.
Let R be G -graded, M be a graded R -module, N be a graded primarysubmodule of M , Then N is graded irreducible if and only if N is irreducible.Proof. We know that
M/N has a unique associated prime, denoted by p , and it’sgraded under assumption (2). Also, we may assume N = 0 after replacing M with M/N . The ”if” direction is trivial. Now let 0 be graded irreducible in M . Then0 is an R ( p ) -submodule in M ( p ) which is graded irreducible by Lemma 2.10. Then soc ( M ( p ) ) ∼ = R ( p ) / p R ( p ) . Now (0 : M ( p ) m ( p ) ) p = (0 : M p m p ) because m is finitelygenerated. So soc ( M p ) ∼ = R p / p R p . This means 0 is irreducible in M p . So 0 isirreducible in M by Lemma 2.10. (cid:3) We have proved that graded-irreducibility is the same as being graded and irre-ducible. Now we give the following definitions from [3] and [4], generalized to the G -graded case. In [3] they are called the index of irreducibility and denoted by r M ( N ) (resp. r gM ( N )). Definition 4.4.
Let N ⊂ M be R -modules. Then CHENG MENG (1) The index of reducibility of N in M is ir M ( N ) = min { r : N = ∩ ri =1 N i , N i irreducible R -submodules of M } .(2) When M and N are graded, the graded index of reducibility of N in M is ir gM ( N ) = min { r : N = ∩ ri =1 N i , N i graded-irreducible R -submodules of M } .When M is clearly understood we simply denote them by ir ( N ) (resp. ir g ( N )).Here is the G -graded version of [3, Theorem 7]. The proof is identical. Theorem 4.5.
Let R be a G -graded ring, M a G -graded module where G is abelianbut not necessarily torsionfree. Then the following are equivalent.(1) Every graded-irreducible submodule of M is irreducible.(2) For every graded submodule N of M , ir M ( N ) = ir gM ( N ) .(3) Every graded submodule N of M is a finite intersection of irreducible gradedsubmodules of M .In particular these equivalent conditions all hold if G is torsionfree abelian. Theorem 4.6.
Let ( R, m , k ) be graded local, N ⊂ M are graded R -modules suchthat M/N is m -primary. Then ir M ( N ) = rank k soc ( M/N ) .Proof. Localizing at m . We have ir M ( N ) = rank k m soc ( M/N ) m . Notice that therank of soc ( M/N ) will not change after localizing. (cid:3) The relation between the index of reducibility of I and I ∗ Let R be a graded ring and I be an ideal of R which is not necessarily graded.We want to compare ir R ( I ) and ir R ( I ∗ ). Let’s consider a special case: G = Z and R be the coordinate ring of a cone C in an affine variety A n , then R is G -graded.In this case the operation I → I ∗ has a geometric interpretation. Suppose I isa radical ideal corresponding to a closed subset X in C , and X is not supportedat the origin. That is, X = V ( I ) is the vanishing set of I . There is a naturalprojection π : A n − { (0) } → P n − . π restricts to two maps: C − { } → P n − and X − { } → P n − . Then I ∗ is a radical ideal, and its vanishing set is π ( X − { } )in P n − , because the maximal homogeneous ideal in I corresponds to the minimalclosed subset containing π ( X − { } ).For a morphism f : Y → Y ′ between varieties, if Z ⊂ Y is an irreducibleclosed subset, then f ( Z ) is also irreducible. So if all the irreducible components arereflected in the variety as a set, ir R ( I ) should be greater or equal to ir R ( I ∗ ), becauseevery irreducible component of I or V ( I ) map to an irreducible subset containedin V ( I ∗ ). The equality holds if and only if different irreducible components do notcollapse to contain each other. Let M in ( J ) denote the minimal prime over an ideal J , and | S | denote the cardinality of a set S . We have the following theorem: Theorem 5.1.
Let R be a G -graded ring where G is torsionfree abelian. Let I bean R -ideal which is radical but not G -graded. Then ir ( I ) ≥ ir ( I ∗ ) . The equalityholds if and only if the *-map induces a bijection between M in ( I ) and M in ( I ∗ ) .Proof. The irreducible decomposition of I is I = ∩ ( p i ) i ∈ Min ( I ) and this decompo-sition is irredundant. So ir ( I ) = | M in ( I ) | . Now taking star commutes with inter-section, so I ∗ = ∩ ( p ∗ i ) i ∈ Min ( I ) . Note that I ∗ is also radical so ir ( I ∗ ) = | M in ( I ∗ ) | .After deleting some p ∗ i it becomes an irredundant irreducible decomposition. So ir ( I ) ≥ ir ( I ∗ ). The equality holds if and only if no prime ideal is deleted, so those -GRADED IRREDUCIBILITY AND THE INDEX OF REDUCIBILITY 7 p ∗ i are just all the minimal primes of I ∗ , so the *-map induces a bijection between M in ( I ) and M in ( I ∗ ). (cid:3) In general there is no control on the difference between ir ( I ) and ir ( I ∗ ). Example 5.2.
Let R = k [ x, y ] which is Z -graded. Let I = ( x, ( y − a )( y − a ) · · · ( y − a r )) where a , a , ..., a r are pairwise distinct and all nonzero elements in k . Then I ∗ = ( x ). In this case we see that I has r components which collapse to becomeone component of I ∗ .In general, there is no fixed inequality between ir ( I ) and ir ( I ∗ ). Here are twoexamples where I and I ∗ are both m -primary for a graded prime ideal m . Example 5.3.
Let R = k [ x, y ] be G = Z graded where k is a field. Let I =( x , xy, y , x − y ), then I ∗ = ( x , xy, y ). They are both m = ( x, y )-primary. R/m = k . soc ( I ) = kx, soc ( I ∗ ) = kx + ky . So ir ( I ) = 1 < ir ( I ∗ ). Example 5.4.
Let
R, m, G be as above, I = ( x , x y , y , x y − y x ). I ∗ =( x , x y , y ). They are still m -primary. Now soc ( I ∗ ) = kx y + kxy , soc ( I ) = kx y + k ( x − xy ) + k ( x y − y ). So ir ( I ) = 3 > ir ( I ∗ ). Acknowledgements
The author would like to thank Professor Giulio Caviglia for introducing thisproblem and showing a proof in the Z -graded case. The author would like to thankProfessor William Heinzer for reading an early draft. References [1] Bourbaki, N. Alg`ebre Commutative, Chapitres 1 `a 4. Springer-Verlag Berlin Heidelberg,2006.[2] Winfried Bruns and Jurgen Herzog, Cohen-Macaulay rings, volume 39 of CambridgeStudies in Advanced Mathematics. Cambridge University Press, Cambridge, 1993.[3] Justin Chen and Youngsu Kim, Graded-irreducible modules are irreducible.Communications in Algebra, 04 May 2017, Vol.45(5), p.1907-1913. DOI:10.1080/00927872.2016.1226864.[4] Nguyen Tu Cuong, Pham Hung Quy and Hoang Le Truong, On the index of reducibilityin Noetherian modules. Journal of Pure and Applied Algebra, October 2015, Vol.219(10),pp.4510-4520.[5] David Eisenbud, Commutative algebra with a view towards algebraic geometry. Volume150 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1995.[6] Shiv Datt Kumar and Srinivas Behara,Uniqueness of graded primary decomposition ofmodules graded over finitely generated abelian groups.Communications in Algebra, 01July 2011, Vol.39(7), p.2607-2614.[7] C. Nastasescu and F. Van Oystaeyen. Graded ring theory. North-Holland Publishing,1982.[8] C. Nastasescu and F. Van Oystaeyen, Methods of graded rings. Springer Berlin Heidel-berg : Imprint: Springer — 2004.
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