aa r X i v : . [ m a t h . A C ] N ov A NOTE ON PROJECTIVE MODULES
HOSSEIN FARIDIAN
Abstract.
This expository note delves into the theory of projective modules parallelto the one developed for injective modules by Matlis. Given a perfect ring R , wepresent a characterization of indecomposable projective R -modules and describe aone-to-one correspondence between the projective indecomposable R -modules andthe simple R -modules.
1. Introduction
Throughout this note, R denotes an associative ring with identity. All modules areassumed to be left and unitary.There is a well-known theorem in the representation theory of finite-dimensional alge-bras over a field which provides a 1-1 correspondence between the isomorphism classesof indecomposable projective modules over the algebra on the one hand, and the isomor-phism classes of simple modules over the algebra on the other hand. More specifically,let k be a field and A a finite-dimensional k -algebra. Then every simple A -module is aquotient of some indecomposable projective A -module which is unique up to isomorphism.Conversely, for every indecomposable projective A -module P , there is a simple A -module,unique up to isomorphism, that is a quotient of P by some maximal submodule; see[Le]. For a generalization of this result to Artin algebras, see [ARS, Corollary 4.5], andto perfect rings, see [Sh, Proposition 5]. The purpose of this note is to present a noveldescriptive proof of this theorem in the following more comprehensive form; see Theorems3.5 and 3.8. Theorem 1.1.
Let R be a perfect ring, and P a nonzero R -module. Then the followingassertions are equivalent: (i) P is an indecomposable projective R -module. (ii) P is a sum-irreducible projective R -module. (iii) P is the projective cover of its every nonzero quotient module. (iv) P is the projective cover of R/ m for some maximal left ideal m of R .Further if R is commutative, then the above assertions are equivalent to the following one: (v) P ∼ = R m for some maximal ideal m of R . Theorem 1.2.
Let R be a perfect ring. There is a one-to-one correspondence betweenthe indecomposable projective R -modules and the simple R -modules. Mathematics Subject Classification.
Key words and phrases.
Indecomposable module; perfect ring; projective module; sum-irreduciblemodule.
2. Preliminaries
We begin with reminding the classical notion of a projective cover.
Definition 2.1.
Let R be a ring, and M an R -module. Then:(i) A submodule N of M is said to be superfluous , written as N ⊆ sup M , if whenever N + L = M for some submodule L of M , then we have L = M .(ii) By a projective cover of M , we mean a projective R -module P together with anepimorphism π : P → M such that ker π ⊆ sup P .(ii) R is said to be perfect if every R -module has a projective cover.Enochs and Jenda have defined the general notion of a cover as follows; see [EJ, Defi-nition 5.1.1]. Definition 2.2.
Let R be a ring, M an R -module, and A a class of R -modules. By an A -cover of M , we mean an R -module A ∈ A together with an R -homomorphism ϕ : A → M that satisfy the following conditions:(i) Any R -homomorphism ψ : B → M with B ∈ A factors through ϕ , i.e. there isan R -homomorphism f : B → A that makes the following diagram commutative: BA Mψϕf (ii) Any R -homomorphism f : A → A that makes the diagram AA Mϕϕf commutative is an automorphism.The classical definition of a projective cover amounts to the modern definition asrecorded in the following result for the sake of bookkeeping.
Lemma 2.3.
Let R be a ring, M an R -module, P a projective R -module, and π : P → M an R -homomorphism. Let P ( R ) denote the class of projective R -modules. Then thefollowing assertions are equivalent: (i) π : P → M is a projective cover of M . (ii) π : P → M is a P ( R ) -cover of M . Proof.
See [Xu, Theorem 1.2.12]. (cid:3)
In light of Lemma 2.3, the projective cover of a given R -module M is unique up toisomorphism if it exists, in the sense that if π : P → M and π ′ : P ′ → M are two NOTE ON PROJECTIVE MODULES 3 projective covers of M , then there is an isomorphism P ′ ∼ = −→ P that makes the followingdiagram commutative: P ′ P Mπ ′ π ∼ =Accordingly, we denote a choice of a projective cover of M by π M : P R ( M ) → M wheneverone exists. Definition 2.4.
Let R be a ring. Then:(i) A nonzero R -module M is said to be indecomposable if whenever M = N ⊕ N for some submodules N and N of M , then we have either N = 0 or N = 0.(ii) An R -module M is said to be sum-irreducible if whenever M = N + N for somesubmodules N and N of M , then we have either N = M or N = M .By [AF, Theorem 28.4], one has the following lemma: Lemma 2.5.
Let R be a ring, and J ( R ) denote its Jacobson radical. Then the followingassertions are equivalent: (i) R is perfect. (ii) Every descending chain of principal right ideals of R stabilizes. (iii) Every flat R -module is projective. (iv) R/ J ( R ) is semisimple and J ( R ) is left T-nilpotent. Recall that a subset A of a ring R is said to be left T-nilpotent if for every sequence a , a , ... ∈ A , there is a natural number n such that a a · · · a n = 0.
3. Main Results
In this section, we delve into the structure of indecomposable projective R -modules. Lemma 3.1.
Let R be a perfect ring, and P a nonzero projective R -module. Then thefollowing assertions are equivalent: (i) P is indecomposable. (ii) Given any proper submodule N of P , there is an isomorphism P ∼ = −→ P R ( P/N ) that makes the following diagram commutative: P R ( P/N ) P P/Nπ
P/N π ∼ = where π is the natural epimorphism. (iii) P is sum-irreducible. H. FARIDIAN
Proof. (i) ⇒ (ii): Let N be a proper submodule of P . By Lemma 2.3, the R -homomorphism π P/N : P R ( P/N ) → P/N is a P ( R )-cover, so we can find an R -homomorphism f : P → P R ( P/N ), and also using the projectivity of P R ( P/N ), wecan find an R -homomorphism g : P R ( P/N ) → P that make the following diagram com-mutative: P R ( P/N ) P P/N π P/N πg f
We thus have π P/N f g = πg = π P/N . Another use of Lemma 2.3 implies that f g is anautomorphism, so f is surjective. The short exact sequence0 → ker f → P f −→ P R ( P/N ) → P ∼ = ker f ⊕ P R ( P/N ). But P R ( P/N ) = 0, so the hypothesis implies thatker f = 0, whence f : P → P R ( P/N ) is an isomorphism.(ii) ⇒ (iii): Suppose that N and N are two submodules of P such that P = N + N . If N is proper, then the hypothesis implies that there is an isomorphism f : P → P R ( P/N )that makes the following diagram commutative: P R ( P/N ) P P/N π P/N πf We thus obtain the following commutative diagram with exact rows:0 N P P/N
00 ker π P/N P R ( P/N ) P/N ππ P/N ∼ = f Since ker π P/N ⊆ sup P R ( P/N ), we deduces that N ⊆ sup P . It follows that N = P , so P is sum-irreducible.(iii) ⇒ (i): Suppose that P = N ⊕ N for some submodules N and N of P . Thehypothesis implies that either N = P or N = P . It follows that either N = 0 or N = 0,so P is indecomposable. (cid:3) Corollary 3.2.
Let R be a perfect ring, and M an R -module. Then M is sum-irreducibleif and only if P R ( M ) is indecomposable. Proof.
Let π M : P R ( M ) → M be a projective cover of M . Suppose that M is sum-irreducible, and let P R ( M ) = P + P for some submodules P and P of P R ( M ). Then M = π M (cid:0) P R ( M ) (cid:1) = π M ( P ) + π M ( P ) . NOTE ON PROJECTIVE MODULES 5
Without loss of generality, we can conclude that M = π M ( P ). It follows that P R ( M ) = P + ker π M . But ker π M ⊆ sup P R ( M ), so we infer that P R ( M ) = P . It follows that P R ( M ) is sum-irreducible. Therefore, by Lemma 3.1, P R ( M ) is indecomposable.Conversely, suppose that P R ( M ) is indecomposable. Hence by Lemma 3.1, P R ( M ) issum-irreducible. Assume that M = N + N . It is easy to see that P R ( M ) = π − M ( N ) + π − M ( N ) + ker π M . But ker π M ⊆ sup P R ( M ), so we infer that P R ( M ) = π − M ( N ) + π − M ( N ). Without lossof generality, we can conclude that P R ( M ) = π − M ( N ). This shows that M = N , so M is sum-irreducible. (cid:3) Corollary 3.3.
Let R be a perfect ring, and P a nonzero projective R -module. Then P is indecomposable if and only if P ∼ = P R ( S ) for some simple R -module S . Proof.
Suppose that P is indecomposable. By [AF, Proposition 17.14], P has a maximalsubmodule N . Then by Lemma 3.1, P ∼ = P R ( P/N ). It is clear that S := P/N is a simple R -module.Conversely, suppose that P ∼ = P R ( S ) for some simple R -module S . Obviously, S issum-irreducible. Now, Corollary 3.2 implies that P R ( S ) is indecomposable. (cid:3) Lemma 3.4.
Let R be a commutative perfect ring. Then the following assertions hold: (i) Spec( R ) = Max( R ) . (ii) Supp R ( R m ) = { m } for every m ∈ Max( R ) . (iii) R m is an indecomposable R -module for every m ∈ Max( R ) . Proof. (i): Let p ∈ Spec( R ), and 0 = a + p ∈ R/ p . By Lemma 2.5, the descending chain( a ) ⊇ ( a ) ⊇ · · · of principal ideals of R stabilizes, i.e. there is an integer n ≥ a n ) = ( a n +1 ) = · · · . In particular, a n = ra n +1 for some r ∈ R . It follows that(1 − ra ) a n = 0. As a / ∈ p , we get 1 − ra ∈ p , which yields that ( r + p )( a + p ) = 1 + p , i.e. a + p ∈ ( R/ p ) × . Hence R/ p is a field, so p ∈ Max( R ).(ii): Let m , n ∈ Max( R ) be such that m = n . Set J u ( R ) := \ v ∈ Max( R ) \{ u } v , for any u ∈ Max( R ). By Lemma 2.5, the ring R/ J ( R ) is semisimple. This implies that R is semilocal, and so for any i ≥
1, there are elements a i ∈ J m ( R ) i \ m and b i ∈ J n ( R ) i \ n .Set c i := a i b i for every i ≥
1. It is clear that c i ∈ J ( R ) for every i ≥
1. By Lemma 2.5, J ( R ) is T-nilpotent. It follows that there is an integer n ≥ c c · · · c n = 0.Set a := a a · · · a n and b := b b · · · b n . Then it is obvious that a / ∈ m , b / ∈ n , and ab = 0.This shows that ( R m ) n = 0, so Supp R ( R m ) = { m } .(iii): Suppose that R m = N ⊕ N ′ for some R -submodules N and N ′ of R m . Localizingat m , we get R m = N m ⊕ N ′ m . But since R m is a local ring, it follows that R m is anindecomposable R m -module. Hence, N m = 0 or N ′ m = 0. Say N m = 0. But, (ii) impliesthat Supp R ( N ) ⊆ { m } . Therefore, N = 0. (cid:3) H. FARIDIAN
Theorem 3.5.
Let R be a perfect ring, and P a nonzero R -module. Then the followingassertions are equivalent: (i) P is an indecomposable projective R -module. (ii) P is a sum-irreducible projective R -module. (iii) P is the projective cover of its every nonzero quotient module. (iv) P ∼ = P R ( R/ m ) for some maximal left ideal m of R .Further if R is commutative, then the above assertions are equivalent to the following one: (v) P ∼ = R m for some maximal ideal m of R . Proof.
Lemma 3.1 yields the equivalence of (i), (ii) and (iii).(i) ⇒ (iv): Corollary 3.3 implies that P ∼ = P R ( S ) for some simple R -module S . Butthen S ∼ = R/ m for some maximal left ideal m of R .(iv) ⇒ (v): Suppose that P ∼ = P R ( R/ m ) for some maximal ideal m of R . By Lemma 3.4, R m is an indecomposable R -module. Moreover, Lemma 2.5 yields that R m is a projective R -module. By Corollary 3.3, we have R m ∼ = P R ( R/ n ) for some maximal ideal n of R . If n = m , then take any a ∈ n \ m . Therefore, the R -homomorphism a R m : R m → R m is anisomorphism. The commutative diagram R m R m P R ( R/ n ) P R ( R/ n ) R/ n R/ n a ∼ = ∼ = aπ R/ n π R/ n a shows that the R -homomorphism a R/ n : R/ n → R/ n is surjective, which is a contradic-tion. Hence, n = m .(v) ⇒ (i): Follows from Lemmas 3.4 and 2.5. (cid:3) The following result may be considered as dual to Matlis Theorem which asserts thatover a commutative noetherian ring R , every injective R -module decomposes uniquelyinto a direct sum of indecomposable injective R -modules; see [Ma]. Corollary 3.6.
Let R be a commutative artinian ring, and P a projective R -module.Then P ∼ = Y p ∈ Spec( R ) P R ( R/ p ) π ( p ,P ) , where π ( p , P ) is the zeroth dual Bass number of P with respect to p , i.e. π ( p , P ) = rank R p / p R p (cid:16) R p / p R p ⊗ R p Hom R ( R p , P ) (cid:17) . NOTE ON PROJECTIVE MODULES 7
Proof.
By [Xu, Theorem 1.2.13], R is perfect, and thus [Xu, Proposition 3.3.1] impliesthat P is a flat cotorsion R -module. It follows from [Xu, Theorem 4.1.15] that P ∼ = Y p ∈ Spec( R ) T p , where T p is the p R p -adic completion of a free R p -module of rank π ( p , P ). But J ( R p ) = p R p is nilpotent, so every R p -module is p R p -adically complete. It follows that T p is a free R p -module of rank π ( p , P ). Now, the result follows from Theorem 3.5. (cid:3) Lemma 3.7.
Let R be a perfect ring, and f : M → N an R -homomorphism. Then thefollowing assertions hold: (i) There is an R -homomorphism ˜ f : P R ( M ) → P R ( N ) that makes the followingdiagram commutative: P R ( M ) P R ( N ) M N ˜ fπ M π N f (ii) If f is an epimorphism, then any R -homomorphism g : P R ( M ) → P R ( N ) thatmakes the diagram in (i) commutative is an epimorphism. (iii) If f is an isomorphism, then any R -homomorphism g : P R ( M ) → P R ( N ) thatmakes the diagram in (i) commutative is an isomorphism. Proof. (i): The existence of ˜ f follows readily from the projectivity of P R ( M ) in light ofthe following diagram: P R ( M ) MP R ( N ) N π M fπ N ˜ f (ii): Suppose that f is an epimorphism and an R -homomorphism g : P R ( M ) → P R ( N )makes the following diagram commutative: P R ( M ) P R ( N ) M Ngπ M π N f The diagram shows that π N g is surjective. Therefore, it can be seen by inspection thatim g + ker π N = P R ( N ) . H. FARIDIAN
But ker π N ⊆ sup P R ( N ), so im g = P R ( N ), i.e. g is surjective.(iii): Suppose that f is an isomorphism and an R -homomorphism g : P R ( M ) → P R ( N )makes the following diagram commutative: P R ( M ) P R ( N ) M Ngπ M π N f By (ii), g is surjective. The short exact sequence0 → ker g → P R ( M ) g −→ P R ( N ) → P R ( M ) = ker g + im g ′ , where g ′ : P R ( N ) → P R ( M ) is an R -homomorphismsuch that gg ′ = 1 P R ( N ) . It is clear that ker g ⊆ ker π M , so P R ( M ) = ker π M + im g ′ . Butker π M ⊆ sup P R ( M ), so P R ( M ) = im g ′ . It follows that g ′ is an isomorphism, so g is anisomorphism. (cid:3) The next result extends the main result of [Le] from finite-dimensional algebras over afield to perfect rings.
Theorem 3.8.
Let R be a perfect ring, and m , n two maximal left ideals of R . Then P R ( R/ m ) ∼ = P R ( R/ n ) if and only if m = n . Hence, there is a one-to-one correspondencebetween the indecomposable projective R -modules and the simple R -modules. Proof.
Let ϕ : P R ( R/ m ) → P R ( R/ n ) be any isomorphism. Suppose to the contrarythat m = n , and take any element a ∈ m \ n . It follows that the R -homomorphism a R/ n : R/ n → R/ n is an isomorphism. In view of Lemma 3.7, the commutative diagram P R ( R/ n ) P R ( R/ n ) R/ n R/ n aπ R/ n π R/ n a shows that the R -homomorphism a P R ( R/ n ) : P R ( R/ n ) → P R ( R/ n ) is an isomorphism.Therefore, the commutative diagram P R ( R/ m ) P R ( R/ m ) P R ( R/ n ) P R ( R/ n ) aϕ ϕa NOTE ON PROJECTIVE MODULES 9 yields that the R -homomorphism a P R ( R/ m ) : P R ( R/ m ) → P R ( R/ m ) is an isomorphism.Now, the commutative diagram P R ( R/ m ) P R ( R/ m ) R/ m R/ m aπ R/ m π R/ m a implies that the R -homomorphism a R/ m : R/ m → R/ m is surjective. But this map iszero, so we arrive at a contradiction. Hence, m = n .The converse is immediate. (cid:3) References [AF] F.W. Anderson and K.R. Fuller,
Rings and categories of modules , Second edition, Graduate Textsin Mathematics , Springer-Verlag, New York, 1992.[ARS] M. Auslander, I. Reiten and S.O. Smalø, Representation theory of Artin algebras , CambridgeStudies in Advanced Mathematics, , Cambridge University Press, Cambridge, 1995.[EJ] E.E. Enochs and O.M.G. Jenda, Relative homological algebra, Volume 1 , Second edition, De GruyterExposition in Mathematics, , Walter de Gruyter GmbH & Co. KG, Berlin, 2011.[Le] T. Leinster, The bijection between projective indecomposable and simple modules , Bull. Belg. Math.Soc. Simon Stevin, (5), (2015), 725-735.[Ma] E. Matlis, Injective modules over Noetherian rings , Pacific J. Math., , (1958), 511-528.[Sh] U. Shukla, On the projective cover of a module and related results , Pacific J. Math. 12 (1962),709-717.[Xu] J. Xu,
Flat covers of modules , Lecture Notes in Mathematics , Springer-Verlag, Berlin, 1996.
H. Faridian, School of Mathematical and Statistical Sciences, Clemson University, SC29634, USA.
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