# On the Extensions of Zassenhaus Lemma and Goursat’s Lemma to Algebraic Structures

Authors: Fanning Meng; Junhui Guo
Journal: Journal of Mathematics
 (2022)

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## Abstract

The Jordan–Hölder theorem is proved by using Zassenhaus lemma which is a generalization of the Second Isomorphism Theorem for groups. Goursat’s lemma is a generalization of Zassenhaus lemma, it is an algebraic theorem for characterizing subgroups of the direct product of two groupsG1×G2, and it involves isomorphisms between quotient groups of subgroups of G1 and G2. In this paper, we first extend Goursat’s lemma to R-algebras, i.e., give the version of Goursat’s lemma for algebras, and then generalize Zassenhaus lemma to rings, R-modules, and R-algebras by using the corresponding Goursat’s lemma, i.e., give the versions of Zassenhaus lemma for rings, R-modules, and R-algebras, respectively.


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## Body

## 1. Introduction

The Fundamental Homomorphism Theorem (or the First Isomorphism Theorem) provided by Noether [1] in 1927 shows that every homomorphism gives rise to an isomorphism and that quotient groups are merely constructions of homomorphic images. Even it has simple form, it expresses the important properties of quotient group. Noether emphasized the fundamental importance of this fact, and it has been widely used in the field of universal algebra and to prove the existence of some natural isomorphisms. The Diamond Isomorphism Theorem (or the Second Isomorphism Theorem) which is the consequence of the Fundamental Homomorphism Theorem is formulated in terms of subgroups of the normalizer and relates two quotient groups involving products and intersections of subgroups. After that, many researchers generalized the Second Isomorphism Theorem to other structures, such as Tamaschke [2, 3] generalized it to Schur semigroups [4] and Endam and Vilela [5] extended it to B-algebras introduced by Neggers and Kim [6, 7]. Sequentially, the First Isomorphism Theorem and Second Isomorphism Theorem are generalized to rings, vector spaces, R-modules, and R-algebras, respectively. In this paper, we suppose that the readers are familiar with the structures of groups, rings, R-modules, and R-algebras.In 1934, Zassenhaus [8] found a new and beautiful proof of the Jordan–Hölder theorem via Zassenhaus lemma. Zassenhaus lemma (also called Butterfly lemma) is well known in group theory as a generalization of the Second Isomorphism Theorem for groups. Subsequently, many researchers extended Zassenhaus lemma to other structures, such as Teh [9, 10] extended it to universal algebras via employment of the graph theory, Wyler [11] extended it to categories under conditions, Ngaha Ngaha [12] considered the Second Isomorphism Theorem and Zassenhaus lemma in star-regular categories and described Zassenhaus lemma in the category of commutative Hopf algebras, and Prathomjit et al. [13] extended Zassenhaus lemma and the Schreier refinement theorem to the case of gyrogroups and then used these results to prove the Jordan–Hölder theorem for gyrogroups.On the other hand, as we know, there is a natural question in group theory, that is, how many subgroups do exist for the general groupG? If we want to count the number of subgroups of G, we should first express the form of subgroups of G. In 1889, Goursat [14] gave Goursat’s lemma which is an algebraic theorem for characterizing subgroups of the direct product of two groups G1×G2, and involves isomorphisms between quotient groups of subgroups of G1 and G2. After that, many researchers expressed the subgroups in certain classes of groups; for instance, Usenko [15] introduced a reduced crossed homomorphism and used it to describe the subgroups of a semidirect product of groups and characterized the subsemidirect products and semidirect products with a given structure of normal subgroups. Bauer et al. [16] obtained a description of the subgroups of a direct product G1×G2×⋯×Gn of a finite number of groups by proving a generalization of Goursat’s lemma. In 2009, Anderson and Camillo [17] proved Zassenhaus lemma by using Goursat’s lemma, which means that Zassenhaus lemma is a corollary of Goursat’s lemma. From Goursat’s lemma, one can recover a more general version of Zassenhaus lemma. According to the above, in this paper, our main aim is to extend Goursat’s lemma to R-algebras and give the expression of the form of subalgebras of general R-algebras. Further, we generalize Zassenhaus lemma to rings, R-modules, and R-algebras by using the corresponding Goursat’s lemma, respectively, and show that the corresponding results are the generalization of the Second Isomorphism Theorem for rings, R-modules, and R-algebras, respectively.In fact, Goursat’s lemma (see Theorem1) gave a way to describe the subgroups of a direct product G1×G2 which involves isomorphisms between quotient groups of subgroups of G1 and G2, i.e.,(1)H=h1,h2∈H12×H22|fHh1H11=h2H21is a subgroup of G1×G2, where Hi1andHi2 are subgroups of Gi such that Hi1⊲Hi2 for i=1,2, and fH:H12/H11⟶H22/H21 is a group isomorphism. Subsequently, Anderson and Camillo (see Theorem 2) described the subrings of a direct product R1×R2 which involves isomorphisms between quotient rings of subrings of R1 and R2, i.e.,(2)T=r1,r2∈T12×T22|fTr1+T11=r2+T21is a subring of R1×R2, where Ti1andTi2 are subrings of Ri such that Ti1 is an ideal of Ti2 for i=1,2, and fT:T12/T11⟶T22/T21 is a ring isomorphism. For R-modules, Dickson (see Theorem 3) described the submodules of a direct product M1×M2 which involves isomorphisms between quotient R-modules of submodules of M1 and M2, i.e.,(3)M=m1,m2∈M12×M22|fMm1+M11=m2+M21is a submodule of M1×M2, where Mi1andMi2 are R-submodules of Mi such that Mi1⊆Mi2 for i=1,2, and fM:M12/M11⟶M22/M21 is a R-module isomorphism.Since anR-algebra has the ring structure and R-module structure, as a generalization, we consider Goursat’s lemma for R-algebras (see Corollary 1) and describe the subalgebras of a direct product A1×A2 which involves isomorphisms between quotient R-algebras of subalgebras of A1 and A2, that is,(4)A=a1,a2∈A12×A22|fAa1+A11=a2+A21is a subalgebra of A1×A2, where Ai1andAi2 are subalgebras of Ai such that Ai1 is algebraic ideal of Ai2 for i=1,2, and fA:A12/A11⟶A22/A21 is an R-algebra isomorphism. Indeed, Lambek [18] gave Goursat’s characterization of the subgroups of the direct product of two groups (also for a general class of algebras) under conditions by using graph theory. In this paper, we give a different form for the subalgebras of R-algebras. However, for other structures, such as Lie algebras, quantum cluster algebras [19], B-algebras, and free differential algebras [20], we still do not know their Goursat’s characterization.Furthermore, Anderson and Camillo [17] used Goursat’s lemma to prove Zassenhaus lemma for groups (see Theorem 4) which is stated as follows:(5)N1H1∩H2N1H1∩N2≅N2H1∩H2N2N1∩H2,where H1,N1,H2,andN2 are subgroups of G such that N1⊲H1 and N2⊲H2. As a generalization, we consider Zassenhaus lemma for rings (see Theorem 5), R-modules (see Theorem 6), and R-algebras (see Theorem 7), respectively, and obtain the following results:(1)
IfR1,I1,R2,andI2 are subrings of a ring R such that Ii is an ideal of Ri for i=1,2, then(6)I1+R1∩R2I1+R1∩I2≅I2+R1∩R2I2+I1∩R2.(2)
IfM1,N1,M2,andN2 are submodules of R-module M satisfying N1⊆M1,N2⊆M2 for commutative ring R with identity, then(7)N1+M1∩M2N1+M1∩N2≅N2+M1∩M2N2+N1∩M2.(3)
IfA is an R-algebra and B1,C1,B2,andC2 are subalgebras of A such that Ci is an algebraic ideal of Bi for i=1,2, then(8)C1+B1∩B2C1+B1∩C2≅C2+B1∩B2C2+C1∩B2.This paper is organized as follows. In Section2, we first complete the proof of Goursat’s lemma for R-modules and then extend Goursat’s lemma to R-algebras, i.e., give the version of Goursat’s lemma for algebras and give the form of subalgebras of R-algebras, which is different from Lambek’s form [18]. Further, since Zassenhaus lemma is used to prove the Jordan–Hölder theorem and it is also the generalization the Second Isomorphism Theorem for groups, as a generalization, we extend Zassenhaus lemma to rings, R-modules, vector spaces, and R-algebras in terms of algebraic ideal in Section 3, i.e., give the versions of Zassenhaus lemma for rings, R-modules, and R-algebras, respectively.


## 2. Goursat’s Lemma for Groups, Rings,R-Modules, and R-Algebras

Anderson and Camillo [17] gave an exposition of Goursat’s lemma for groups and rings, and Dickson [21] gave Goursat’s lemma for R-modules without proof, respectively. In this section, we complete the proof of Goursat’s lemma for R-modules and give an exposition of Goursat’s lemma for R-algebras as a corollary.Theorem 1.
(Goursat’s lemma for groups, Theorem 4 in [17]). Let G1andG2 be groups.(1)
LetH be a subgroup of G1×G2, and(9)H11=a∈G1|a,1∈H,H12=a∈G1|a,b∈Hforsomeb∈G2,H21=b∈G2|1,b∈H,H22=b∈G2|a,b∈Hforsomea∈G1.Then,Hi1andHi2 are subgroups of Gi with Hi1⊲Hi2 for i=1,2, and the map(10)fH:H12H11⟶H22H21,aH11↦bH21is an isomorphism, wherea,b∈H. Moreover, if H⊲G1×G2, then Hi1,Hi2⊲Gi and Hi2/Hi1⊆CGi/Hi1, the center of Gi/Hi1.(2)
LetHi1andHi2 be subgroups of Gi with Hi1⊲Hi2 for i=1,2, and let f:H12/H11⟶H22/H21 be an isomorphism. Then,(11)H=a,b∈H12×H22|faH11=bH21is a subgroup ofG1×G2. Furthermore, suppose Hi1,Hi2⊲Gi and Hi2/Hi1⊆CGi/Hi1 for i=1,2, then H⊲G1×G2.(3)
The constructions given in (1) and (2) are inverses to each other.Theorem 2.
(Goursat’s lemma for rings, Theorem 11 in [17]). Let R1andR2 be rings.(1)
LetT be a subring of R1×R2, and(12)T11=r∈R1|r,0∈T,T12=r∈R1|r,s∈Tforsomes∈R2,T21=s∈R2|0,s∈T,T22=s∈R2|r,s∈Tforsomer∈R1.Then,Ti2 is a subring of Ri and Ti1 is an ideal of Ti2 for i=1,2. Moreover, the map fT:T12/T11⟶T22/T21 defined by fTr+T11=s+T21 for r,s∈T is a ring isomorphism.(2)
Suppose thatTi2 is a subring of Ri and Ti1 is an ideal of Ti2 for i=1,2, and fT:T12/T11⟶T22/T21 is a ring isomorphism. Then,(13)T=r,s∈T12×T22|fTr+T11=s+T21is a subring ofR1×R2.(3)
The construction given in (1) and (2) is inverse to each other.
In Theorem 4 in [17], the authors stated Goursat’s lemma for R-modules without proof (also see [21, 22]). In the following, we provide a proof of Goursat’s lemma for R-modules by using the submodule criterion [23].Theorem 3.
(Goursat’s lemma forR-modules). Let R be a commutative ring with identity and M1andM2 are R-modules.(1)
LetM be a submodule of M1×M2, and(14)M11=m1∈M1|m1,0∈M,M21=m2∈M2|0,m2∈M,M12=m1∈M1|m1,m2∈Mforsomem2∈M2,M22=m2∈M2|m1,m2∈Mforsomem1∈M1,thenMi1andMi2 are submodules of Mi with Mi1⊆Mi2 for i=1,2, and the map fM:M12/M11⟶M22/M21 given by fMm1+M11=m2+M21 is a R-module isomorphism, where m1,m2∈M.(2)
Suppose thatMi1andMi2 are submodules of Mi with Mi1⊆Mi2 for i=1,2 and the map fM:M12/M11⟶M22/M21 is a R-module isomorphism, then(15)M=m1,m2∈M12×M22|fMm1+M11=m2+M21,is a submodule ofM1×M2.(3)
The construction given in (1) and (2) is inverse to each other.Proof.
(1) From Theorem1, it is obvious that Mi1⊆Mi2⊆Mi for i=1,2. Since M is a submodule of M1×M2, we have(16)m11,0+rm11′,0=m11+rm11′,0∈M,for any m11,m11′∈M11 and any r∈R. This means that m11+rm11′∈M11. Thus, M11 is a submodule of M1 following the submodule criterion [23]. On the other hand, for any m12,m12′∈M12, there exist m2,m2′∈M2 such that m12,m2,m12′,m2′∈M. Hence,(17)m12,m2+rm12′,m2′=m12+rm12′,m2+rm2′∈M,for any r∈R. It follows that m12+rm12′∈M12, and then M12 is a submodule of M1. Similarly, using the submodule criterion, we can obtain that M21andM22 are submodules of M2 easily.
For the mapfM:M12/M11⟶M22/M21 given by fMm1+M11=m2+M21, where m1,m2∈M, it is clear that M12/M11andM22/M21 are R-modules and fM is an additive group isomorphism by Theorem 1. So, it is enough to prove that fMrm1+M11=rfMm1+M11 for any r∈R and m1∈M12. Since rm1,m2=rm1,rm2∈M for any m1∈M12,m2∈M22, and r∈R, we have fMrm1+M11=fMrm1+M11=rm2+M21=rfMm1+M11. Hence, fM is a R-module isomorphism.
(2) Suppose thatMi1andMi2 are submodules of Mi with Mi1⊆Mi2 for i=1,2, and the map fM:M12/M11⟶M22/M21 is a R-module isomorphism. Since M=m1,m2∈M12×M22|fMm1+M11=m2+M21, we have fMm1+M11=m2+M21,fMm1′+M11=m2′+M21 for any m1,m2,m1′,m2′∈M, and(18)fMm1+M11+rm1′+M11=fMm1+rm1′+M11=fMm1+M11+rfMm1′+M11=m2+M21+rm2′+M21=m2+rm2′+M21.
This means thatm1+rm1′,m2+rm2′=m1,m2+rm1′,m2′∈M. Therefore, M is a submodule of M1×M2 by the submodule criterion.
(3) From the proofs of (1) and (2), we can easily obtain (3).
It is well known that anR-algebra A has the ring structure and R-module structure concurrently, and the operations of these two structures are compatible, i.e., rxy=rxy=xry for any x,y∈A and r∈R. We firstly introduce some definitions about R-algebras for the commutative ring R with identity.Definition 1.
(see [23–25]). Let R be a commutative ring with identity. An R-algebra is a ring A with identity together with a ring homomorphism f:R⟶A mapping 1R to 1A such that the subring fR of A is contained in the center of A, i.e., fr commutes with every element of A for each r∈R. A subalgebra of an R-algebra A is a subring of A and a submodule of A. A left (respectively, right, two sided) algebraic ideal of an R-algebra A is a left (respectively, right, two sided) ideal of the ring A and a submodule of A.
IfAandB are two R-algebras, an R-algebra homomorphism (respectively, isomorphism) is a ring homomorphism (respectively, isomorphism) φ:A⟶B mapping 1A to 1B such that φr·a=r·φa for all r∈R and a∈A.
It is easy to check that ifA is an R-algebra, then A is a ring with identity and has a natural left and right R-module structure defined by r·a=a·r=fra, where fra is just the multiplication in the ring A. Every ring with identity is actually an Z-algebra.
SinceR-algebra has the ring structure and module structure by Definition 1, according to Theorem 2 and Theorem 3, we can obtain Goursat’s lemma for R-algebras easily.Corollary 1.
(Goursat’s lemma forR-algebras). Let R be a commutative ring with identity and A1andA2 be R-algebras.(1)
LetA be a subalgebra of A1×A2, and(19)A11=a∈A1|a,0∈A,A12=a∈A1|a,b∈Aforsomeb∈A2,A21=b∈A2|0,b∈A,A22=b∈A2|a,b∈Aforsomea∈A1,thenAi1andAi2 are subalgebras of Ai such that Ai1 is an algebraic ideal of Ai2 for i=1,2, and the map(20)fA:A12A11→A22A21,a+A11↦b+A21,is anR-algebra isomorphism, where a,b∈A.(2)
Suppose thatAi1andAi2 are subalgebras of Ai such that Ai1 is an algebraic ideal of Ai2 for i=1,2, and fA:A12/A11⟶A22/A21 is an R-algebra isomorphism, then(21)A=a,b∈A12×A22|fAa+A11=b+A21is a subalgebra ofA1×A2.(3)
The construction given in (1) and (2) is inverse to each other.Example 1.
LetA=Z2x/x3−1, which is an Z2-algebra. We want to find out all the subalgebras of A×A.Step 1.
Find out all the subalgebrasAi’s (1≤i≤6) and their corresponding algebraic ideals in A. Here, we omit the case when the corresponding algebraic ideal is Ai itself.Case 1.
The subalgebraA1=0,1,x,x2,1+x,1+x2,x+x2,1+x+x2 with 8 elements of A, and the algebraic ideals of A1 are as follows:(22)J11=0,1+x,1+x2,x+x2,J12=0,1+x+x2,J13=0.
Thus,(23)A1J11=0=1+x=1+x2=x+x2,1=x=x2=1+x+x2,A1J12=0=1+x+x2,1=x+x2,x=1+x2,x2=1+x,A1J13=R10=A1,and A1/J11 has 2 elements, A1/J12 has 4 elements, and A1 has 8 elements.Case 2.
The subalgebraA2=0,1,x+x2,1+x+x2 with 4 elements of A, and the algebraic ideals of A2 are(24)J21=0,x+x2,J22=0,1+x+x2,J23=0,A2J21=0=x+x2,1=1+x+x2,A2J22=0=1+x+x2,1=x+x2,A2J23=A20=A2,and A2/J2i has 2 elements for i=1,2, and A2 has 4 elements.Case 3.
For the subalgebrasA3=0,1+x,1+x2,x+x2,A4=0,x+x2,A5=0,1+x+x2,andA6=0,1, their algebraic ideals are only 0 and Ai for i=3,4,5,6.Step 2.
According to Corollary1, we first construct the isomorphism and then give the subalgebras of A×A.Case 4.
(with 2 elements). Here, we only discuss one case. We have an isomorphismA2/J21≅A4/0, and the subalgebra of A×A is(25)0,0,x+x2,0,1,x+x2,1+x+x2,x+x2.Case 5.
(with 4 elements).(a)
We have an isomorphismf:A1/J12⟶A3/0 defined by f0=0,f1=x+x2,fx=1+x,fx2=1+x2, and the subalgebra of A×A is(26)0,0,1+x+x2,0,1,x+x2,x+x2,x+x2,x,1+x,1+x2,1+x,x2,1+x2,1+x,1+x2.(b)
We have an isomorphismf:A1/J12⟶A1/J12 which is an identity or defined by f0=0,f1=1,fx=x2,andfx2=x, and the subalgebra of A×A for the identity case is(27)0,0,0,1+x+x2,1+x+x2,0,1+x+x2,1+x+x2,1,1,1,x+x2,x+x2,1,x+x2,x+x2,x,x,x,1+x2,1+x2,x,1+x2,1+x2,x2,x2,x2,1+x,1+x,x2,1+x,1+x,and the subalgebra ofA×A for the other case is(28)0,0,0,1+x+x2,1+x+x2,0,1+x+x2,1+x+x2,1,1,1,x+x2,x+x2,1,x+x2,x+x2,x,x2,x,1+x,1+x2,x2,1+x2,1+x,x2,x,x2,1+x2,1+x,x,1+x,1+x2.(c)
For the isomorphismf:A3/0⟶A3/0 which is an identity or defined by f0=0,fx+x2=x+x2,f1+x=1+x2,andf1+x2=1+x, the subalgebra of A×A for the identity case is(29)0,0,x+x2,x+x2,1+x,1+x,1+x2,1+x2,and the subalgebra ofA×A for the other case is(30)0,0,x+x2,x+x2,1+x,1+x2,1+x2,1+x.(d)
For the isomorphismf:A2/0⟶A2/0 which is an identity or defined by f0=0,f1=1,fx+x2=1+x+x2,andf1+x+x2=x+x2, the subalgebra of A×A for the identity case is(31)0,0,1,1,x+x2,x+x2,1+x+x2,1+x+x2,and the subalgebra ofA×A for the other case is(32)0,0,1,1,x+x2,1+x+x2,1+x+x2,x+x2.Case 6.
(with 8 elements). The isomorphismf:A1/0⟶A1/0 is an identity, and the subalgebra of A×A is(33)0,0,1,1,x,x,x2,x2,x+x2,x+x2,1+x,1+x,1+x2,1+x2,1+x+x2,1+x+x2.


## 3. The Zassenhaus Lemma for Groups, Rings,R-Modules, and R-Algebras

In group theory, it is well known that the four basic isomorphism theorems (Fundamental Homomorphism Theorem, Diamond Isomorphism Theorem, Freshman Theorem, and Correspondence Theorem) about homomorphism and their structure are very useful in the study of groups. All of these theorems have analogues in other algebraic structures, such as rings, vector spaces,R-modules, and R-algebras. Furthermore, for the Second Isomorphism Theorem, the set-theoretic version of the product formula (i.e., if G1andG2 are subgroups of G, then there is a natural bijection between G1/G1∩G2 and G1G2/G2), establishes a bijection which is the same as the bijection of the Second Isomorphism Theorem, but without the conditions of normality, and the bijection is purely at the set-theoretic level. In 1934, Zassenhaus gave Zassenhaus lemma which is a generalization of the Second Isomorphism Theorem for groups (see Lemma 5.10 in [26] and Lemma 4.52 in [27]). In this section, we give Zassenhaus lemma for rings, R-modules, and R-algebras, which is also a generalization of the Second Isomorphism Theorem for rings, R-modules, and R-algebras, respectively.Theorem 4.
(Zassenhaus lemma for groups, Lemma 4.52 in [27]). Suppose that G is a group with subgroups H1,N1,H2,andN2 such that N1⊲H1 and N2⊲H2, then(34)N1H1∩N2⊲N1H1∩H2,N2N1∩H2⊲N2H1∩H2,and there is an isomorphism(35)N1H1∩H2N1H1∩N2≅H1∩H2H1∩N2N1∩H2≅N2H1∩H2N2N1∩H2.Remark 1.
Zassenhaus lemma for groups implies the Second Isomorphism Theorem for groups, i.e., ifH and N are subgroups of a group G with N⊲G, then(36)NHN≅HH∩N.
In fact, we can letH=H1∩H2 and N=N1H1∩N2. Thus, according to Theorem 4, we have(37)N1H1∩H2N≅HH1∩N2N1∩H2.
Indeed, for anyxy∈H1∩N2N1∩H2 where x∈H1∩N2⊆H1∩H2 and y∈N1∩H2⊆H1∩H2, since xy=yy−1xy∈N1H1∩N2, we have xy∈H1∩H2∩N1H1∩N2=H∩N. Conversely, for any n1x∈H1∩H2∩N1H1∩N2=H∩N where n1∈N1 and x∈H1∩N2⊆H1∩H2, there exists h2∈H2 such that n1x=h2∈H1∩H2, i.e., n1=h2x−1∈H2. It follows that n1x=xx−1n1x∈H1∩N2N1∩H2. Thus,(38)H1∩N2N1∩H2=H1∩H2∩N1H1∩N2=H∩N.
Further, it is easy to obtain that(39)N1H1∩H2=N1H1∩N2H1∩H2=NH.
Therefore, (37) can be written as NH/N≅H/H∩N.
Following Lemma 4.52 in [27], we can also set H1=G,N1=N,H2=H, and N2=N∩H and then obtain that NH/N≅H/H∩N.Example 2.
Consider the symmetric groupS4 with 24 elements which are as follows:(40)1,12,13,14,23,24,34,1234,1324,1423,123,124,132,134,142,143,234,243,1234,1243,1324,1342,1423,1432.
Let(41)N1=1,1324,1234,1423,N2=1,123,132,134,143,124,142,234,243,1234,1324,1423,H1=1,12,34,1324,1423,1234,1324,1423,then N2H1=N2N1 by applying for Theorem 4. In fact, according to the definitions of N1,N2,andH1, we have(42)H1∩N2=1,1234,1324,1423,and H1,N1,andN2 are subgroups of S4 with N1⊲H1. Let H2=S4, then N2⊲H2. Thus, (43)N1H1∩H2=N1H1=H1,N1H1∩N2=1,12,34,1324,1423,1234,1324,1423=H1,N2H1∩H2=N2H1,N2N1∩H2=N2N1.
By Theorem4, we have N2H1/N2N1=H1/H1, which means that N2H1=N2N1.
As a generalization of the Second Isomorphism Theorem for rings, by using Theorem2, we obtained Zassenhaus lemma for rings stated as follows.Theorem 5.
(Zassenhaus lemma for rings). LetR1,I1,R2,andI2 be subrings of a ring R such that Ii is an ideal of Ri for i=1,2, then I1+R1∩I2 (respectively, I2+I1∩R2) is an ideal of I1+R1∩R2 (respectively, I2+R1∩R2), and(44)I1+R1∩R2I1+R1∩I2≅I2+R1∩R2I2+I1∩R2.Proof.
LetT=a+c,b+c∈R1×R2|a∈I1,b∈I2,c∈R1∩R2, then T is a subring of R1×R2. Indeed, for any a1+c1,b1+c1,a2+c2,b2+c2∈T, where ai∈I1,bi∈I2,ci∈R1∩R2 for i=1,2, since R1∩R2 is also a subring of R and Ii is an ideal of Ri for i=1,2, we have(45)a1+c1,b1+c1−a2+c2,b2+c2∈T,a1+c1,b1+c1a2+c2,b2+c2∈T,which implies that T is a subring of R1×R2. According to Theorem 2, let(46)T12=r∈R1|r,s∈Tforsomes∈R2,T22=s∈R2|r,s∈Tforsomer∈R1,then T12=I1+R1∩R2,T22=I2+R1∩R2, and Ti2=Ii+R1∩R2 is a subring of Ri for i=1,2 clearly. Further, let(47)T11=r∈R1|r,0∈T,T21=s∈R2|0,s∈T,then T11=I1+R1∩I2,T21=I2+I1∩R2, and Ti1 is an ideal of Ti2 for i=1,2. Following Theorem 2, we have T12/T11≅T22/T21, which shows that(48)I1+R1∩R2I1+R1∩I2≅I2+R1∩R2I2+I1∩R2.Remark 2.
The Zassenhaus lemma for rings implies the second isomorphism theorem for rings, i.e., ifS is a subring of a ring R and I an ideal of R, then(49)S+I/I≅S/S∩I.
In fact, letI1=I, R1=R, I2=I∩S, and R2=S, since I+I∩S=I, I∩S+S=S, and I∩S+I∩S=I∩S, we have(50)I1+R1∩R2=I1+S=I+S,I1+R1∩I2=I1+I2=I+I∩S=I,I2+R1∩R2=I2+S=I∩S+S=S,I2+I1∩R2=I2+I∩S=I∩S+I∩S=I∩S.
According to Theorem5, we have S+I/I≅S/S∩I.ForR-modules, we know that a module is a mathematical object in which things can be added together commutatively by multiplying coefficients and in which most of the rules of manipulating vectors hold. A module is abstractly very similar to a vector space although in modules coefficients are taken in rings that are much more general algebraic objects than the fields used in vector spaces. R-modules can be thought of as generalizations of vector spaces and abelian groups. We will also see that they can be regarded as “representations” of a ring.For the Second Isomorphism Theorem forR-modules, by using Goursat’s lemma for R-modules, we have the following generalized Zassenhaus lemma.Theorem 6.
(Zassenhaus lemma forR-modules). Let R be a commutative ring with identity and M an R-module. Suppose that M1,N1,M2,andN2 are submodules of M satisfying N1⊆M1,N2⊆M2, then N1+M1∩N2 (respectively, N2+N1∩M2) is a submodule of N1+M1∩M2 (respectively, N2+M1∩M2), and(51)N1+M1∩M2N1+M1∩N2≅N2+M1∩M2N2+N1∩M2.Proof.
LetL=a+c,b+c∈M1×M2|a∈N1,b∈N2,c∈M1∩M2; since M1∩M2 is also a submodule of M and Ni is a submodule of Mi for i=1,2, we have(52)a1+c1,b1+c1+ra2+c2,b2+c2=a1+ra2+c1+rc2,b1+rb2+c1+rc2∈L,for any r∈R,ai∈N1,bi∈N2,ci∈M1∩M2,i=1,2, which implies that L is a submodule of M1×M2 by the submodule criterion [23]. Following Theorem 3, let(53)M11=m1∈M1|m1,0∈L,M21=m2∈M2|0,m2∈L,M12=m1∈M1|m1,m2∈Lforsomem2∈M2,M22=m2∈M2|m1,m2∈Lforsomem1∈M1,as in the proof of Theorem 5, we have(54)M11=N1+M1∩N2,M12=N1+M1∩M2,M21=N2+N1∩M2,M22=N2+M1∩M2,and Mi1andMi2 are submodules of Mi with Mi1⊆Mi2 for i=1,2. Thus, by Theorem 3, we have(55)N1+M1∩M2N1+M1∩N2≅N2+M1∩M2N2+N1∩M2.Remark 3.
Zassenhaus lemma for modules implies the Second Isomorphism Theorem for modules, i.e., ifK1andK2 are submodules of a R-module M, then K1+K2/K1≅K2/K1∩K2. In fact, let N1=K1, M1=M, N2=K1∩K2, and M2=K2; since K1+K1∩K2=K1,K2+K1∩K2=K2,K1∩K2+K1∩K2=K1∩K2, we have(56)N1+M1∩M2=K1+K2,N1+M1∩N2=K1+K1∩K2=K1,N2+M1∩M2=K1∩K2+K2=K2,N2+N1∩M2=K1∩K2+K1∩K2=K1∩K2.
Following Theorem6, we have K1+K2/K1≅K2/K1∩K2.
Similarly, we can easily obtain Zassenhaus lemma for vector spaces, which is also a generalization of the Second Isomorphism Theorem for vector spaces; that is, ifV is a vector space and V1andV2 are linear subspaces of V, then V1+V2/V2≅V1/V1∩V2. By Theorem 6, we have the following corollary.Corollary 2.
(Zassenhaus lemma for vector spaces). LetV be a vector space and U1,V1,U2,andV2 are linear subspaces of V satisfying V1⊆U1,V2⊆U2, then V1+U1∩V2 (respectively, V2+V1∩U2) is a subspace of V1+U1∩U2 (respectively, V2+U1∩U2), and(57)V1+U1∩U2V1+U1∩V2≅V2+U1∩U2V2+V1∩U2.
For the Second Isomorphism Theorem forR-algebras, by using Corollary 1, we have the following generalized Zassenhaus lemma.Theorem 7.
(Zassenhaus lemma forR-algebras). Suppose that A is an R-algebra and B1,C1,B2,andC2 are subalgebras of A such that Ci is an algebraic ideal of Bi for i=1,2, then C1+B1∩C2 (respectively, C2+C1∩B2) is an algebraic ideal of C1+B1∩B2 (respectively, C2+B1∩B2), and(58)C1+B1∩B2C1+B1∩C2≅C2+B1∩B2C2+C1∩B2.Proof.
SinceB1∩B2 is a subring and a submodule of A, it tells that B1∩B2 is a subalgebra of A. Let Γ=a+c,b+c∈B1×B2|a∈C1,b∈C2,c∈B1∩B2; since Ci is an algebraic ideal of Bi for i=1,2, then Γ is a subring and a submodule of B1×B2 from Theorem 5 and Theorem 6, which implies that Γ is a subalgebra of B1×B2. Following Theorem 3, let(59)Γ11=γ1∈B1|γ1,0∈Γ,Γ21=γ2∈B2|0,γ2∈Γ,Γ12=γ1∈B1|γ1,γ2∈Γforsomeγ2∈B2,Γ22=γ2∈B2|γ1,γ2∈Γforsomeγ1∈B1,then we have(60)Γ11=C1+B1∩C2,Γ12=C1+B1∩B2,Γ21=C2+C1∩B2,Γ22=C2+B1∩B2,and Γi1andΓi2 are subalgebras of Bi such that Γi1 is an algebraic ideal of Γi2 for i=1,2 according to Theorem 5 and Theorem 6. Thus, by Theorem 3, we have(61)C1+B1∩B2C1+B1∩C2≅C2+B1∩B2C2+C1∩B2.Remark 4.
Zassenhaus lemma forR-algebra implies the Second Isomorphism Theorem for R-algebra, i.e., if A is an R-algebra and BandJ are subalgebras of A such that J is an algebraic ideal of A, then(62)J+BJ≅BB∩J.
In fact, letC1=J, B1=A, C2=J∩B, and B2=B; since J+J∩B=J,B+J∩B=B,J∩B+J∩B=J∩B, we have(63)C1+B1∩B2=J+B,C1+B1∩C2=J+J∩B=J,C2+B1∩B2=J∩B+B=B,C2+C1∩B2=J∩B+J∩B=J∩B.
Following Theorem7, we have J+B/J≅B/B∩J.Example 3.
LetC1=1+x=1+xfx|fx∈Zx, B1=Zx which is an Z-algebra, C2=x2, and B2=x2,5=x2fx+5gx|fx,gx∈Zx, then Ci is an algebraic ideal of Bi for i=1,2, and(64)C1+B1∩B2=1+x+x2,5=1+x,x2,5,C1+B1∩C2=1+x+x2=1+x,x2,C2+B1∩B2=x2+x2,5=x2,5,C2+C1∩B2=x2+1+x∩x2,5.
According to Theorem7, we have(65)1+x,x2,51+x,x2≅5Z≅x2,5x2+x2,5∩1+x.



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*Source: 7705500-2022-02-03.xml*