EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 3, 2018, 751-761 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Generalization of Schur’s Lemma in Ring Representations on Modules over a Commutative Ring Na’imah Hijriati1,2,∗, Sri Wahyuni1, Indah Emilia Wijayanti1 1 Department of Mathematics, Faculty of Mathematics and Natural Sciences, Universitas Gadjah Mada, Yogyakarta, Indonesia 2 Department of Mathematics, Faculty of Mathematics and Natural Sciences, Universitas Lambung Mangkurat, Banjarmasin, Indonesia Abstract. Let R,S be rings with unity, M a module over S, where S a commutative ring, and f : R→ S a ring homomorphism. A ring representation of R on M via f is a ring homomorphism µ : R → EndS(M), where EndS(M) is a ring of all S-module homomorphisms on M . One of the important properties in representation of ring is the Schur’s Lemma. The main result of this paper is partly the generalization of Schur’s Lemma in representations of ring on modules over a commutative ring 2010 Mathematics Subject Classifications: 16G10, 16W20, 12E15 Key Words and Phrases: Representation of ring on module, Generalized Schur’s Lemma, Ring homomorphism 1. Introduction In [2], an abelian group M is called an R-module if there is a ring homomorphism ϕ : R → EndZ(M) where EndZ(M) is a ring of all group endomorphism of M . We know that if there is a ring homomorphism f : R → S, then every S-module is also R-module, where the scalar multiplication on R defined by rm = f(r)m [2]. We generalize the codomain of the ring homomorphism ϕ to the ring of endomorphisms of a module over any ring S. Due to this aim, we need a connection between the ring R and the S-module M . Moreover, We need the commutativity of S to guarantee µ(r) ∈ EndS(M), where µ(r) := µr : M →M,m 7→ f(r)m. Recall the definition of a representation of a ring R on a vector space V over a field F i.e a ring homomorphism ρ : r → EndF (V ), where EndF (V ) is a ring of all linear transformations of V [7]. Analog to the definition of representations of rings on vector spaces, we define representations of rings on modules over a commutative ring as follow : ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3285 Email addresses: naimah.hijriati@mail.ugm.ac.id (N. Hijriati), swahyuni@ugm.ac.id (S. Wahyuni), ind wijayanti@ugm.ac.id (I.E Wijayanti) http://www.ejpam.com 751 c© 2018 EJPAM All rights reserved. N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 752 Definition 1. Let R be a ring with unity, S a commutative ring with unity and f : R → S a ring homomorphism. A representation of a ring R on an S-module M is a ring homomorphism µ : R→ EndS(M), r 7→ µr, (1) where µr ∈ EndS(M) is defined by µr(m) = f(r)m for every r ∈ R and m ∈M . Further- more, this representation µ of ring R on module M is called an f -representation of R and M is called an f -representation module of R. Note that prefix f in ”f -representation” depends on a ring homomorphism f : R→ S we choose. These are some examples of representation of ring. Example 1. Every ring commutative R has a 1d-representation on R-module M , since there exists the identity ring homomorphism 1d : R→ R. Example 2. Let R[G] be a group ring, where R is a commutative ring and G is a finite group. Since there is a ring homomorphism h : R[G]→ R, h( ∑ g∈G agg) = ∑ g∈G ag, (2) a ring homomorphism µ : R[G]→ EndR(M) defined by µ( ∑ g∈G agg) = µ∑ g∈G agg ∈ EndR(M) (3) where µ∑ g∈G agg(m) = h( ∑ g∈G agg)m = ∑ g∈G agm (4) is an h-representation of R[G]. Example 3. Let M ′2(Z) be a ring of all 2 × 2 lower triangle matrices and let Z2 be a module over itself, where the scalar multiplication defined by (a, b)(x, y) = (ax, by) for every (a, b) ∈ Z2 and (x, y) ∈ Z2. The function θ : M ′2(Z)→ EndZ2(Z2) defined by θr(m) = f(r)m is an f -representation of ring M ′2(Z) by a ring homomorphism f : M ′2(Z) → Z2 defined by f( [ a 0 c b ] ) = (a, b). Example 4. Let M2(Z) be an additive group of all 2 × 2 matrices, where entries in Z. We define the scalar multiplication in M2(Z) over Z2 as (a, b) [ u r s t ] := [ au br bs at ] . Hence M2(Z) is a Z2-module. Since g : M ′2(Z) → Z2 defined by g( [ a 0 c b ] ) = (a, 0) is a ring homomorphism, ϕ : M ′2(Z) → EndZ2(M2(Z)) defined by ϕ(r)(m) = g(r)m is a g-representation of ring M ′2(Z). N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 753 Example 5. Let M2(Z) be a Z2-module defined as Example 4. If f, g : Z3 → Z2 are ring homomorphisms defined by f(a, b, c) = (a, b) and g(a, b, c) = (0, c) respectively, then we have an f -representation µ : Z3 → EndZ2(M2(Z)) of Z3 and a g-representation ϕ : Z3 → EndZ2(M2(Z)) of Z3 where µr(m) = f(r)m and ϕr(m) = g(r)m for all r ∈ R, and m ∈M . Example 3, Example 4, and Example 5 can be generalized to any commutative ring R. Furthermore, since µ is a ring homomorphism, so we can find the kernel of µ, i.e. ker(µ) = {r ∈ R | f(r) ∈ Ann(M)}. (5) We know that the element of ker(µ) depend on the properties of f and Ann(M). So we have the following properties: Proposition 1. Let µ : R→ EndS(M) is an f -representation of ring R. If f is injective and M is faithful (Ann(M) = {0S}), then µ is injective. Proof. Let r be any element of ker(µ). Then µr = O ∈ EndS(M). For any m ∈ M , we have µr(m) = O(m) = 0 if and only if f(r)m = 0. So we have f(r) ∈ Ann(M). Since Ann(M) = {0}, f(r) = 0 and r ∈ ker(f). Furthermore since f is injective, r = 0. Thus µ is injective. The converse of Proposition 1 is not always true. This is the counterexample. Example 6. Let f : Z→ Z2, a 7→ (a, 0) (6) be a ring homomorphism and M2(Z) a module over Z2. The scalar multiplication is defined by (a, b) [ u v w x ] = [ au av aw ax ] (7) The f -representation µ : Z→ EndZ2(M2(Z)) is injective, since ker(µ) = {0}. But M2(Z) is not faithful, since Ann(M) = {(0, b) | b ∈ Z}. Remark 1. If S is an integral domain and M is a free torsian S-module, then an f - representation µ of a ring R on a module M is injective if and only if f is injective. There is an important result in representation of rings on vector spaces i.e Schur’s Lemma. Schur’s Lemma showed that the set all morphism of an irreducible representation ρ of a ring R (HomR(ρ, ρ)) on a vector space V over a field F is a skew field, and the necessary condition when ρ is irreducible([7],[8]). In linear algebra, the ring End(V ), where V is a finite dimension vector space over F (dim(V ) = n), is isomorphic to the ring of all n × n matrices Mn(F )[1]. So the definition of the representation of a ring R on a vector spaces V over F can be denoted as a ring homomorphism from ρ : R → Mn(F ). If we generalize the ring of all n × n matrices to the ring of all m × n matrices, Schur’s Lemma still considered [12]. N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 754 Let S be an R-algebra. Then there is a ring homomorphism g : R → S such that sg(r) = g(r)s for all r ∈ R, and s ∈ S (g(r) ∈ Z(S), for all r ∈ R). If M is an S-module, then M is an R-module with scalar multiplication defined by r.m = g(r)m for every r ∈ R and m ∈M , such that we can defined a ring homomorphism ϕ : R→ EndS(M), r 7→ ϕr (8) where ϕr : M → M,m 7→ r.m [8]. Since ring S in Definition 1 is a commutative ring then Z(S) = S and S is an R-algebra and a representation module M is a module over R-algebra S. Auslander introduces the concept of modules over an R-algebra, where the ring R is Artinian by using categorical approach [3],[4], and used the result to study the repre- sentation of finite dimension of algebra [5]. This concept is used in the representation theory of finite F -algebras on vector spaces over a field F , such as the representation theory of quiver [15], and the representation theory of a group ring F [G] where G is a finite group([6]. There are many mathematicians is developed the concept that Auslander given, such as Iyama [10],[11], and Oppermann [14]. From the previous paragraph, we have been knowing that the f -representation module of ring R is a module over R-algebra S. However, base on Example 5 and Proposition 1, the properties of the f -representation do not only depend on f -representation module of R but also on the properties of a ring homomorphism f . In the case of the representation of ring on the module over a commutative ring, the generalization of Schur’s Lemma is considered. In this paper, we investigate how to generalize Schur’s Lemma in representations of rings on modules over a commutative ring. The proof of generalization of Schur’s Lemma in this paper is analog with the proof of Schur’s Lemma in module theory. However, the properties of the representation module M (M is an S-module) of a ring R depend on the properties of a ring S and M as S-module. 2. The Main Result To generalize Schur’s Lemma, we need to see some properties of representation of ring on module over commutative ring such as equivalence of two representations, and morphism between two representations. Let µ be a representation of ring R on an S-module M . A submodule U of a rep- resentation module M is called R-invariant if for any r ∈ R, µr(U) ⊆ U . Since every f -representation module M is an R-module, every submodule of M is an R-invariant. So we have this properties. Remark 2. (i) If an S-module M is an f -representation module of an f -representation µ of ring R, then every submodule of M is R-invariant. (ii) If U is an R-invariant submodule, then we can construct a new f -representa-tion of R, that is a ring homomorphism µ′ : R→ EndS(U) defined by µ′r(a) = µr(a) = f(r)a for any r ∈ R and a ∈ U . N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 755 A non zero R-module M is called irreducible if it has only trivial submodule. The definition of an irreducible representation depend on properties of a representation module that we give as the following : Definition 2. A non-zero f -representation µ : R → EndS(M) of a ring R is said to be irreducible if the only R-invariant submodules of M are zero and M . Let µ be an f -representation of R on an S-module M and ϕ a g-representation of R on S-module N . If there is a module homomorphism T : M → N , then we obtain Tµr : M → N and ϕrT : M → N for every r ∈ R. But it is not necessary Tµr = ϕrT . We give an example to show this fact. Example 7. We consider Example 3 and Example 4. Let T : Z2 → M2(Z) be a Z2- module homomorphism defined by T (a, b) = [ a b b a ] . Since for any A = [ u 0 w v ] ∈ M ′2(Z) and (a, b) ∈ Z2 we have TµA(a, b) = [ ua bv bv ua ] , ϕAT (a, b) = [ ua 0 0 av ] . We conclude TµA 6= ϕAT . Furthermore, if there is a module isomorphism T : M → N such that Tµr = ϕrT for all r ∈ R, then µ and ϕ are called equivalent. The definition of the equivalent of two representations is given in the following definition. Definition 3. Let µ : R → EndS(M) be an f -representation of R and let ϕ : R → EndS(N) be a g-representations of R. Representations µ and ϕ are called equivalent if there is an S-module isomorphism T : M → N , such that Tµr = ϕrT for any r ∈ R. Furthermore µ equivalent to ϕ denoted by µ ∼ ϕ. From Definition 3, we know that two representations µ and ϕ of a ring R are equivalent if there is an S-module isomorphism T : M → N , such that the diagram M M N N T µr ϕr T commutes. The following proposition is the sufficient condition two representations of a ring on modules over a commutative ring are equivalent. N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 756 Proposition 2. Let µ : R → EndS(M) be an f -representation of ring R and ϕ : R → EndS(N) a g-representation of ring R. Representation µ and ϕ are equivalent (µ ∼ ϕ) if and only if there is an S-module isomorphism T : M → N and satisfy f(r)−g(r) ∈ Ann(N) for any r ∈ R. Proof. Suppose that T : M → N is an S-module isomorphism and f(r) − g(r) ∈ Ann(N). Then Im(T ) = N , and for any r ∈ R, n ∈ N (f(r)− g(r))n = 0N ⇔ f(r)n− g(r)n = 0N ⇔ f(r)n = g(r)n (9) So for any m ∈M (Tµr)(m) = Tµr(m) = T (f(r)m) = f(r)T (m) = g(r)T (m) = ϕr(T (m)) = (ϕrT )(m). (10) Thus µ and ϕ are equivalent. Conversely, if µ is equivalent to ϕ, then there is an S-module isomorphism T : M → N , such that Tµr = ϕrT for any r ∈ R. Then we have for any n ∈ N there is m ∈ M such that T (m) = n. So for any r ∈ R, g(r)T (m) = ϕr(T (m)) = (ϕrT )(m) = (Tµr)(m) = T (µr(m)) = T (f(r)m) = f(r)T (m). (11) Hence we have g(r)T (m) = f(r)T (m)⇔ (f(r)−g(r))T (m) = 0. Since T 6= 0, f(r)−g(r) ∈ Ann(T (m)) = Ann(n) for any n ∈ N . Thus f(r)− g(r) ∈ Ann(N). Example 8. Let µ be an f -representation of M ′2(Z) defined in Example 3. Let M∗2 (Z) be an abelian group of all 2 × 2 diagonal matrices and it also a Z2-module, where scalar multiplication defined as Example 4. We define ϕ : M ′2(Z)→ EndZ2(M∗2 (Z)) (12) as a g-representation of M ′2(Z) where g is a ring homomorphism in Example 4. For any A = [ u 0 w v ] ∈ M ′2(Z), we have f(A) − g(A) = (0, v) ∈ Ann(M∗2 (Z)) and there is a Z2-module isomorphism T : Z2 →M∗2 (Z), (a, b) 7→ [ a 0 0 b ] . (13) So we conclude µ ∼ ϕ. N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 757 From Definition 3, if f = g, then for any m ∈ M and r ∈ R we always have Tµr(m) = T (f(r)m) = f(r)T (m) = ϕrT (m). Furthermore, if N is a torsian S-module, then Ann(N) = 0. Based on this facts we have this following : Corollary 1. Let µ : R → EndS(M) be an f -representation of R and ϕ : R → EndS(N) a g-representation of R. (i) If f = g, then µ ∼ ϕ if and only if there is an S-module isomorphism T : M → N . (ii) If N is a free torsian S-module, then µ ∼ ϕ if and only if there is a ring isomorphism from M to N and f = g. Proof. (i) If f = g, then f(r)− g(r) = 0 for any r ∈ R. Hence by Proposition 2 µ ∼ ϕ if and only if there is an S-module isomorphism T : M → N . (ii) If N is a free module, then Ann(N) = {0}. So by Proposition 2, µ ∼ ϕ if and only if there is a ring isomorphism from M to N and f(r) − g(r) ∈ Ann(N) = {0} ⇔ f(r) = g(r) for any r ∈ R i.e f = g. In the paragraph before, we have explained that not every S-module homomorphism T : M → N satisfy Tµr = ϕrT where µ is an f -representation of R on S-module M and ϕ is a g-representation of R on S-module N . If T satisfy Tµr = ϕrT for all r ∈ R, then T is called a morphism from µ to ϕ. Definition 4. Let µ : R→ EndS(M) be an f -representation of R and ϕ : R→ EndS(N) a g-representation of ring R. A morphism from µ to ϕ is an S-module homomorphism T : M → N , such that Tµr = ϕrT for all r ∈ R. Proposition 3. Let µ : R→ EndS(M) be an f -representation of R and ϕ : R→ EndS(N) a g-representation of ring R. An S-module homomorphism T : M → N is a morphism from µ to ϕ if f(r)− g(r) ∈ Ann(Im(T )). Proof. Suppose f(r) − g(r) ∈ Ann(Im(T )). Then for any n ∈ Im(T ) there is m ∈ M such that T (m) = n, and (f(r)− g(r))n = 0⇔ f(r)n = g(r)n. So for any m ∈M (Tµr)(m) = T (µr(m)) = T (f(r)m) = f(r)T (m) = g(r)T (m) = ϕr(T (m)) = (ϕrT )(m). (14) Hence T is a morphism from µ to ϕ. Example 9. Let µ be an f -representation of M ′2(Z) defined in Example 3 and ϕ a g- representation of M ′2(Z) defined in Example 4. For any [ a 0 c b ] ∈M ′2(Z) we have f( [ a 0 c b ] )− g( [ a 0 c b ] ) = (0, b) (15) N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 758 (i) Let T : R2 → M2(R) defined by T (a, b) = [ a 0 0 a ] be an S-module homomorphism. Since an annihilator of S-module homomorphism is a set {(0, x) ∈ Z2 | x ∈ Z}, then T is a morphism from µ to ϕ. (ii) Let T ′ : R2 → M2(R) defined by T ′(a, b) = [ a b b a ] is an S-module homomorphism. Because an annihilator of T ′ is only (0, 0) and there is [ 0 0 0 b ] ∈M ′2(Z), where b 6= 0 such that f( [ 0 0 0 b ] )− g( [ 0 0 0 b, ] ) = (0, b) 6= (0, 0) (16) then T ′ is not a morphism from µ to ϕ. Let µ : R → EndS(M) be an f -representation of R and let ϕ : R → EndS(N) be a g-representation of R. The set of all morphism from µ to ϕ is denoted HomR(µ, ϕ). From Definition 4, we know that every T ∈ HomR(µ, ϕ) is an element in HomS(M,N). Remark 3. Let µ : R→ EndS(M) be an f -representation of R and ϕ : R→ EndS(N) a g-representation of R. (i) if f = g, then every S-module homomorphism T : M → N is morphism from µ to ϕ (ii) if T ∈ HomR(µ, ϕ) is an S-module isomorphism, then µ ∼ ϕ. (iii) The identity map id : M →M is always an element in HomR(µ, µ). Proposition 4. Let µ : R→ EndS(M) be f -representation of R and ϕ : R→ EndS(N) a g-representation of ring R. Then HomR(µ, ϕ) is a module over S, and it is a submodule of HomS(M,N). Proof. To prove HomR(µ, ϕ) is an S-module, we must show HomR(µ, ϕ) is an abelian additive group and it is closed under scalar multiplication over S. Let T, T1 and T2 be any elements of HomR(µ, ϕ), and let s be any element of S, then we have T, T1, T2 ∈ HomS(M,N), and from [1] sT, T1 + T2 ∈ HomS(M,N). So By Proposition 3 to prove HomR(µ, ϕ) is an S-module, it is enough to prove f(r) − g(r) ∈ Ann(Im(T1 + T2)) and f(r)− g(r) ∈ Ann(Im(sT )) for any r ∈ R. (i) Since T1, T2 ∈ HomR(µ, ϕ), then by Proposition 3 we have f(r)−g(r) ∈ Ann(Im(Ti)) i.e (f(r)− g(r))Ti(m) = 0 for any m ∈M , i = 1, 2. Let n be any element in Im(T1+T2), then there is m ∈M such that (T1+T2)(m) = n if and only if T1(m) + T2(m) = n. Hence we have (f(r)− g(r))n = (f(r)− g(r))(T1(m) + T2(m)) = (f(r)− g(r))T1(m) + (f(r)− g(r))T2(m) = 0 (17) N. Hijriati, S. Wahyuni, I.E. Wijayanti / Eur. J. Pure Appl. Math, 11 (3) (2018), 751-761 759 (ii) Analog to (1), if T ∈ HomR(µ, ϕ), then for any m ∈M f(r)− g(r)T (m) = 0. So we have for any n ∈ Im(sT ), there is m ∈ M such that (sT )(m) = sT (m) = n and we have (f(r)− g(r))sT (m) = s(f(r)− g(r))T (m) = 0 (18) From (1) and (2) so we have f(r)−g(r) ∈ Ann(Im(T1+T2)) and f(r)−g(r) ∈ Ann(Im(sT )). In other words HomR(µ, ϕ) is an S-module. Furthermore, since HomR(µ, ϕ) ⊆ HomS(M,N), then HomR(µ, ϕ) is an S-submodule of HomS(M,N). Now we give the generalization of Schur’s theorem which is the main result of this paper as follows Proposition 5. Let µ : R → EndS(M) be an irreducible f -representation of R and ϕ : R→ EndS(N) an irreducible g-representation of ring R. We have the following : (i) If µ � ϕ, then HomR(µ, ϕ) = 0. (ii) If µ = ϕ, then HomR(µ, µ) is a skew field. Furthermore, if S is a principle ideal domain (PID), where K is a fractional field of S and K ′ is an extension field of K, and M is a free S-module with finite dimension, then there is scalar α ∈ K ′, such that µ = αI. Proof. (i) Suppose that HomR(µ, ϕ) 6= 0. Then there is T 6= 0 in HomR(µ, ϕ). Since kerT is a submodule of M , by Remarks 2 ker(T ) is an R-invariant submodule of M and hence either ker(T ) = M or ker(T ) = 0. Because µ is irreducible and T 6= 0, then kerT = 0. So T is injective. Furthermore, since Im(T ) is also submodule of N , by Remarks 2 Im(T ) is an R-invariant submodule of N , so Im(T ) = 0 or Im(T ) = N . If Im(T ) = 0, then T = 0. So It must be Im(T ) = N , i.e T is surjective. Thus T is invertible. Hence by Corollary 1 µ ∼ ϕ. This is the contrapositive of what we want to prove. (ii) By using Proposition 4 and Schur’s Lemma in [13], then HomR(µ, µ) is a skew field. Let T ∈ HomR(µ.µ) and T 6= 0. If M is a free S-module with finite dimension, then there is α ∈ K ′ which is eigenvalue of T . Since S is a principle ideal domain, S is Dedekind domain and by Proposition 16.3.14 in [9], S is integrally closed. Hence, if α ∈ K, then α ∈ R. By definition eigenvalue αI − T is not invertible. 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