We show that if K[G] is a semiprime group algebra of polycyclic- by-finite group G and if K[G] has no direct summands that are domains, then K[G] is a CS-ring if and only if K[G] is here
Definitions, Notations, and Basic Results On Modules and Rings
Throughout this dissertation, unless otherwise specified, R will denote a ring with unity, and all modules are right unital When the ring is clear from the context,
14 anR-homomorphism will be expressed as homomorphism For two rightR-modules
M and N,Hom R (M, N) denotes the additive group of all homomorphisms fromM intoN End R (M) denotes the ring of all endomorphisms of M.
A submodule N of M is said to be an essential submodule of M, denoted by
N ⊆ e M, if for every nonzero submoduleL of M,L∩N 6= 0 We say that M is an essential extension of N, ifN is an essential submodule of M For two submodules
A and B of a module M, B is said to be a complement ofA in M if B is maximal with respect to the property thatA∩B = 0 By Zorn’s Lemma, such a complement
B always exists A submodule C of M is called a complement in M if C is a complement to a submodule of M A submodule C of M is calledclosed in M if it is the maximal essential extension of itself inM, that is, wheneverLis a submodule of M with C ⊆ e L then C =L Closed submodules are precisely complements in
M For modules A ⊆ C ⊆ M, C is said to be an essential closure of A in M if
C is a maximal essential extension of A in M, equivalently, C is minimal among closed submodules ofM containingA An essential closureC ofA inM is a closed submodule ofM.
A module is said to be decomposable if it is a direct sum of two nonzero sub- modules A nonzero module is said to be indecomposable if it is not decomposable. Clearly, a ring R is indecomposable (as a right module over itself) if and only if
R has no nontrivial idempotents A nonzero module M is said to be uniform if every nonzero submodule ofM is essential in M, equivalently, if every two nonzero submodules of M intersect nontrivially M is said to be a uniserial module if all submodules of M are linearly ordered by inclusion A ring R is said to be a right
(left) valuation ring if R R ( R R) is a uniserial module A nonzero module M is called simple if (0) and M are the only submodules of M A module M is called semisimple if every submodule ofM is a direct summand of M.
Let A be a subset of a ring R Then right (left) annihilator of A in R will be denoted by r.ann R (A) (l.ann R (A)), that is r.ann R (A) = {r ∈ R|Ar = 0} (l.ann R (A) = {r ∈R|rA = 0}) It is clear that r.ann R (A) (r.ann R (A)) is a right (left) ideal ofR We will denote the right (left) annihilator of an elementa in R by r.ann R (a) (l.ann R (a)) The right singular ideal of a ring R, denoted by Z(R R ), is defined as: Z(R R ) ={r ∈R|rE = 0 f or some essential right ideal E of R} The (Jacobson) radical of a ring R, denoted by J(R), is defined to be the intersection of all maximal right ideals of R, equivalently, the intersection of all maximal left ideals of R U(R) stands for the multiplicative group of units of R R is called directly finite if for a, b∈ R, ab= 1 implies ba= 1 Clearly, if R has no nontrivial idempotents thenR is directly finite R is called an Utumi ring if its right maximal quotient ring coincides with its left maximal quotient ring For any a ∈ R, l a will denote the left multiplication bya.
An idealP in a ring R is called a prime ideal if P 6=R and, for any two ideals
A and B inR, ifAB ⊆P, then A ⊆P or B ⊆P An ideal I in a ring R is called asemiprime ideal if for any ideal A of R, if A 2 ⊆I, then A⊆I A ringR is called a prime (resp., semiprime) ring, if (0) is a prime (resp., semiprime) ideal.
A module M is called N-projective if for each submodule X of N and each homomorphism f :M →N/X there exists a homomorphism g :M →N such that the following diagram commutes
A moduleM is said to beprojective ifM isN-projective for everyR-moduleN, equivalently, if for any epimorphism of modulesf :A→B, and any homomorphism g : M → B, there exists a homomorphism h : M → A such that g = f ◦h It is known that, a module M is projective if and only if any epimorphismf : N →M splits A moduleF is calledfree if it admits afree basis, that is, a set{f i :i∈I} ⊆F such that every element ofF can be written uniquely as a finite linear combinations of thef i ’s It is known that M is projective if and only if it is a direct summand of a free module.
−→M n−1 −→ ã ã ã ã is said to be exact if image of f n+1 is equal to the kernal of f n for all n An exact sequence of the form
0−→M 3 −→ M 2 −→M 1 −→0 is called a short exact sequence An exact sequence of the form
0−→P n −→P n−1 −→ ã ã ã −→ P 0 −→M −→0 where each P n is a projective module, is called a finite projective resolution of M. The projective dimension of M, denoted pd R (M), is defined to be the minimal integern with
If no such integer exists, equivalentlyM has no finite projective resolution, then we define pd R (M) =∞ The right global dimension of a ring R, denoted r.gl.dim(R), is defined to be the least upper bound of the projective dimensions of all right
R-modules The left global dimension of a ring R is defined similarly.
A module M is called N-injective if for each submodule X of N and each ho- momorphism f : X → M, f can be lifted from X to N That is, there exists a homomorphism g :N →M such that the following diagram commutes
M is called quasi-injective if it is M-injective and injective if it is N-injective for every moduleN Baer’s Criterion states that anR-moduleM is an injective module if and only if M is R R -injective A ring R is called right self-injective if the right module R R is injective Left self-injective rings are defined, similarly.
A moduleM is called a CS (or extending, see [11]) module if every submodule of
M is essential in a direct summand of M M is called finitelyP
-CS if every direct sum of finite copies of M is CS.
We consider the following conditions for a module M :
(C2) Every submodule isomorphic to a direct summand ofM is itself a direct sum- mand ofM.
(C3) Whenever X and Y are direct summands of M with X ∩Y = 0, X ⊕Y is a direct summands ofM.
A CS module M is called continuous (resp quasi-continuous) if M satisfies (C2) (resp (C3)) In the literature, quasi-continuous modules are also known as π-injective [14] It is known that a module M is quasi-continuous (= π-injective) if and only if for any two submodules M 1 and M 2 of M with M 1 ∩M 2 = 0, each projection π i : M 1 ⊕M 2 −→ M i , i = 1, 2, can be extended to an endomorphism of M A ring R is said to be a right CS (resp continuous, quasi-continuous) if the right module R R is CS (resp continuous, quasi-continuous) Left CS (con- tinuous, quasi-continuous) rings are defined, similarly If R is both right and left
CS (resp continuous, quasi-continuous), then R is called a CS (resp continuous, quasi-continuous) ring.
In general we have the following implications.
Injective =⇒Quasi-injective =⇒Continuous =⇒ Quasi-continuous =⇒ CS
In [19], Harada and Tozaki introduced the concept of almost N-projectives M is called almostN-projective if for each submoduleX ofN and each homomorphism f :M →N/X, either there existsg such that diagram (a) commutes or there exists h such that diagram (b) commutes, where
N −→ π N/X −→ 0 N −→ π N/X −→ 0 π : N → N/X is the canonical projection onto N/X, N 1 is a direct summand of N, and i : N 1 → N is the inclusion map Later in [4], Baba defined a dual concept known as almost N-injectives M is called almost N-injective if for each submodule X ofN and each homomorphism f :X →M, either there existsg such that diagram (1) commutes or there existshsuch that diagram (2) commutes, where
N 1 is a nonzero direct summand of N, and π : N → N 1 is a projection onto N 1. Henceforth, these diagrams will be referred to as diagram (1) and diagram (2), respectively M is called almost self-injective if M is almost M-injective A ring
R is called right almost self-injective if it is almost self-injective as a right module over itself Left almost self-injective rings are defined, similarly A ring R is called principally injective if for any a ∈ R, any homomorphism f : aR −→ R can be extended to an endomorphism ofR R
Introduction to Group Rings
In this section we will introduce the notion of Group Rings and state some known results Throughout this dissertation, unless otherwise stated,K will denote a field and G a multiplicative group The group algebra of G over K, denoted by K[G], is defined to be the associative K-algebra ⊕ g∈G Kg, with G as a basis, and multiplication defined by using the multiplication in G Typical elements of K[G] are finite formal sums of the form
Thus, for two elements P g∈G a g g and P g∈G b g g of K[G], we define addition and multi- plication as following
Moreover, we can define scalar multiplication of an element fromK[G] by elements of K as following k X g∈G a g g
By identifying each g ∈G with 1g and each k ∈K with k1, we can verify that with the above operations K[G] is an associative K-algebra, with G as a basis In general, for a ring R, R[G] with the above operations is called the group ring of G over R If R is a commutative ring, then R[G] is also called the group algebra of
G over R In particular, Z[G] (resp Q[G]) is called the integral (resp rational) group algebra The center of K[G] will be denoted by Z(K[G]) For α = P g∈G a g g, the support of α, denoted Supp α, is defined to be
P n i=1 α i g i ) ∗ P n i=1 α i g −1 i defines an anti- automorphism of K[G] of order 2 From this we deduce that K[G] enjoys similar right and left properties In particular, we observe that K[G] is right self-injective (right artinian, right noetherian, right hereditary) if and only if K[G] is left self- injective (resp left artinian, left noetherian, left hereditary) In view of this we will omit right or left prefix while working with self-injective (artinian, noetherian, hereditary) group algebras.
A groupGis calledlocally finite if every finitely generated subgroup ofGis finite. For a group G, o(G) will stand for the order of G and H C G will denote that H is a normal subgroup of G The FC subgroup of G, denoted ∆(G), is defined to be the subset of Gin which the conjugacy class of each element is finite, equivalently,
∆(G) = {g ∈ G | [G : C G (g)] < ∞, where C G (g) is the centralizer of g in G}. The subset consisting of all elements of finite order of ∆(G) is a subgroup and it is denoted by ∆ + (G), that is, ∆ + (G) = {g ∈ G | [G : C G (g)] < ∞ and o(g) < ∞}.
We note ∆ + (G) =∪N, where N C G and o(N) < ∞ It is known that ∆(G) and
∆ + (G) are characteristic subgroups ofG In particular, ∆(G)CGand ∆ + (G)CG.
D ∞ as usual stands for the infinite dihedral group generated by two elementsa and bwith a of infinite order, bof order 2 and ba=a −1 b.
A group G is said to be polycyclic-by-finite if G has a finite subnormal series
=G o CG 1 C ã ã ã CG n =G such that each quotient G i /G i−1 is either finite or cyclic The number of infinite cyclic quotients which appear in the above series is called the Hirsch number of G, denoted by h(G) This number is invariant for the group (see [36]) For example,
D ∞ = {< a, b >| O(a) = ∞, O(b) = 2, ba = a −1 b} is a polycyclic-by-finite group since it has a finite subnormal series
C< a >CD ∞ and we may note that, h(D ∞) = 1.
For a subgroup H of a group G, the right (or left) ideal of K[G] generated by the set{1−h|h ∈H} will be denoted by ω(H) That is,ω(H) = P h∈H
K[G](1−h)) Clearly, if H C G, then ω(H) is a two sided ideal of K[G] ω(G) is called the augmentation ideal of K[G] and will be denoted by ω(K[G]) Thus, for a group algebra K[G], ω(K[G]) = P g∈G
(1−g)K[G] Obviously, ω(H) = ω(K[H])K[G] If H C G, then a K-algebra epimorphism f : K[G] →
K[G/H] can be defined as an extension of the canonical epimorphism of G onto
It is known that ker(f) = ω(H) Thus, if H C G, then K[G]/ω(H) ∼= K[G/H].
In particular, K[G]/ω(K[G]) ∼= K, which means that ω(K[G]) is a maximal ideal of K[G] It is known that a subgroup H of a group G is infinite if and only if r.ann K[G](ω(H)) = 0 (equivalently,l.ann K[G](ω(H)) = 0).
A twisted group algebra K t [G] is an associative K-algebra which has a basis {g, g ∈G} and in which the multiplication is defined distributively: g 1 g 2 =γ(g 1,g 2)g 1 g 2 , g 1, g 2 ∈G whereK o is the set of all nonzero elements of K and γ :G×G→K 0 is a function.
A twisted group algebraK t [G] with a twistingγ, a functionγ :G×G→K 0 , is said to be diagonally equivalent to the group algebra K[G] if we can make a diagonal change of basis by replacing eachg by ∼ g =δ(g)g for some δ(g)∈K 0 with γ(g 1 , g 2) =δ(g 1 g 2)δ(g 1) −1 δ(g 2) −1 for allg 1 , g 2 ∈G In this case K t [G] and K[G] are identical.
For a twisted group algebra K t [G] with a twisting γ the following hold for allx, y, z ∈G γ(x, y)γ(xy, z) =γ(y, z)γ(x, yz).
More basaic information on twisted group algebra can be found in ([36], 1.2).
(1) By choosing γ(g, g 0 ) = 1 for all g, g 0 ∈ G, we get the ordinary group algebraK[G].
(2) LetG={1, x, y, z}be the Klein four group andR the field of real numbers Let γ :G×G→R o be defined as below: γ 1 x y z
ThenR t [G] =R1⊕Rx⊕Ry⊕Rz is a twisted group algebra with multiplication defined as above In fact R t [G] is the quaternion algebra which is a division ring.
Preliminary Results
In this section some preliminary results are presented which will be used and cited throughout this dissertation.
We begin with a well-known result on group algebra.
Theorem 1.3.1 ([36], Theorem 3.2.8) The group algebra K[G] is self-injective if and only if G is finite.
More generally,R[G] is a right self-injective group ring if and only ifRis a right self-injective ring andG is a finite group [38].
The following theorem due to Farkas characterizes principally self-injective group algebras.
Theorem 1.3.2 (c.f.Farkas, [13]) The following properties are equivalent:
The following three theorems characterize prime and semiprime group algebras.
Theorem 1.3.3 ([36], Theorem 4.2.10) Let K[G] be a group algebra Then the following are equivalent:
(iii) G has no nonidentity finite normal subgroup.
Theorem 1.3.4 ([36], Theorem 4.2.12) Let K[G]be a group algebra over a field K of characteristic 0 Then
Theorem 1.3.5 ([36], Theorem 4.2.13) Let K[G]be a group algebra over a field K of characteristic p > 0 Then the following are equivalent:
(iv) G has no finite normal subgroup H with p|o(H).
(v) ∆(G) has no element of orderp.
For a polycyclic-by-finite group Gwe have the following lemmas.
Lemma 1.3.1 ([36], Corollary 10.2.8) Let G be a polycyclic-by-finite group Then K[G] is noetherian.
It is an open question whetherGis polycyclic-by-finite whenK[G] is noetherian.
Lemma 1.3.2 ([36], Lemma 10.2.10) Let G be a polycyclic-by-finite group, and let
H and N be subgroups with N CG Then H and G/N are both polycyclic-by-finite groups with Hirsch numbers satisfying h(H) ≤ h(G) and h(N) +h(G/N) = h(G).
Furthermore,h(H) =h(G) if and only if [G:H]0, thenp-o(H) sinceK[G] is semiprime (Theorem 1.3.5) Hence in either case we have o(H) is invertible in K.
Lete=o(H) −1 P h∈H h Then e is a central idempotent inK[G] Now
Assume K[G] is a CS group algebra Since e is a centeral idempotent in K[G] and K[G] is a CS ring, eK[G] is a CS ring Hence K[G/H] is a prime CS-group algebra which is not a domain with G/H polycyclic-by-finite So by Lemma 2.2.1 and Theorem 2.2.1, G/H ∼=D ∞ and char(K)6= 2.
(c) =⇒ (d) Let H be as above Then K[H] is semisimple artinian since K[G] is semiprime and H is a finite normal subgroup of G (Theorem 1.3.8 and Theo- rem 1.3.5) Thus, gl.dimK[H] = 0 (Theorem 1.3.4 (1)) Also G/H ∼= D ∞ and gl.dimK[D ∞] < ∞ since char(K) 6= 2 (Theorem 1.3.7) So by (Theorem 1.3.6) gl.dimK[G] < ∞ Hence gl.dimK[G] = h(G) = h(D ∞) +h(H) = 1 + 0 = 1 (Remark 1.3.1 and Lemma 1.3.2) Thus K[G] is hereditary.
In Theorem 2.2.2 we have that semiprime CS group algebra of polycyclic-by- finite group without domains as summands is hereditary The following example shows that this result is not true for arbitrary rings.
Example 2.2.1 LetV be an infinite dimensional right vector space over a division ring D Let R = End D (V) It is known that R is a regular primitive right self- injective ring, hence right CS, which is not semisimple artinian ring If R is right hereditary, then for every right ideal I of R, R/I is injective Then by Osofsky’s theorem in [33] R is semisimple artinian ring, a contradiction.
The following lemma will be needed in the next theorem.
Lemma 2.2.2 ([39], Corollary 3.4.10) Let G be a finite group and let K be an algebraically closed field such that char(K)-o(G) Then
The following lemma is a key lemma to prove the next theorem.
Lemma 2.2.3 ([36], Theorem 6.1.9) Let G be a group, and let H CG Suppose
{e 1, e 2 , , e n } is a finite G-orbit of centrally primitive idempotents of K[H] with e 1 K[G]∼=M m (K) Then e=e 1+e 2+ +e n is a central idempotent of K[G] and eK[G]∼=M mn (K t [G 1 /H]) where G 1 ⊇ H is the centralizer of e 1 in G and K t [G 1 /H] is some twisted group ring of G 1 /H.
Now we give the precise structure of the semiprime CS group algebra K[G] of a polycyclic-by-finite group G, when K is an algebraically closed and K[G] has no ring direct summands that are domains.
Theorem 2.2.3 Let K[G]be a semiprime CS group algebra of polycyclic-by-finite group G Suppose K[G] has no ring direct summand which is domain If K is algebraically closed field, then
Proof Let H = ∆ + (G) and e = o(H) −1 P h∈H h Then eK[G] ∼= K[G/H] ∼ K[D ∞] as shown in the proof of Theorem 2.2.2 Since K[G] is semiprime and
H is a finite normal subgroup of G, K[H] is semisimple artinian (Theorem 1.3.8 and Theorem 1.3.5) Also, by Lemma 2.2.2, we have K[H] ∼= ⊕ r i=1 M n i (K) So (1−e)K[H] ∼= ⊕ l i=1 M n i(K), where l ≤ r, after reordering if necessary So there exists a set X = {f 1, f 2, , f l } of centrally primitive orthogonal idempotents in
K[H] such that 1−e =f 1+f 2+ +f l and f i K[H]∼=M n i (K), for every 1≤i≤ l.
Next, we will show thatGpermutes elements ofX by conjugation Letg ∈Gbe an arbitrary element Since H CG and eachf i is a centrally primitive idempotent inK[H],gf i g −1 is a centrally primitive idempotent inK[H], for each 1≤i≤ l Also
{gf 1 g −1 , gf 2 g −1 , , gf l g −1 } is a set of orthogonal idempotents Since H CG and e is a central idempotent in K[G], 1−e=gf 1 g −1 +gf 2 g −1 + +gf l g −1 Put R (1−e)K[G], then⊕ l i=1 f i R =R =⊕ l i=1 gf i g −1 R as ring decompositions Therefore, for every 1 ≤ j ≤ l there exists a unique 1 ≤ i ≤ l such that gf j g −1 R = f i R ([?], Lemma 3.8), which implies gf j g −1 = f i Thus {gf 1 g −1 , gf 2 g −1 , , gf l g −1 } = X. Since g ∈G is arbitrary, it follows that G permutes the set X by conjugation. Let s be the number of all G-orbits in X and {f i 1 , f i 2 , , f i s } a subset of X containing exactly one element from each orbit and let e j = P x∈Gf ij x (the sum of all idempotents in the orbit Gf i j ) Then by Lemma 2.2.3 each e j is a centeral idempotent ofK[G] Since 1−e =e 1 + e 2 + + e s , we have
(1−e)K[G] =e 1 K[G]⊕e 2 K[G]ã ã ã ⊕e s K[G] as a ring direct sum For each j, e j K[G]∼=M n j (K t [G j /H]), whereG j ⊇H is the centralizer ofe j inGand K t [G j /H]is some twisted group ring of G j /H (Lemma 2.2.3) Because G j /H < G/H ∼= D ∞, K t [G j /H] ∼= K[G j /H] (Proposition 2.1.1) Hence
For each j, the index [G : G j ] =| Gf i j |< ∞ and also o(H) < ∞ So G j /H is infinite But infinite subgroups ofD ∞are eihter infinite cyclic or isomorphic toD ∞, and so we obtain
Since K[Z] is a domain, we have the following corollary:
Corollary 2.2.1 LetK[G]be a semiprime CS group algebra of polycyclic-by-finite group G Suppose K[G] has no ring direct summand which is a matrix ring over domain If K is algebraically closed field, then
It is well-known that the group algebra K[G] of a group G over a field K is self-injective if and only if G is a finite group (Theorem 1.3.1), and is principally self-injective if and only if G is a locally finite group (Theorem 1.3.2) Recall a ring R is said to be right continuous if every complement right ideal is a summand and every right ideal isomorphic to a summand is itself a summand Rings in which complement right ideals are summands are called right CS rings A ring R is called a right almost self-injective ring if for each right ideal I of R and for each homomorphism f :I →R either f can be extended to R or there exists a nonzero idempotenet e and a homomorphism g : R →eR such that g ◦f = l e [1] Clearly,
45 every right self-injective ring is right almost self-injective A ring R is called right π-injective (also quasi-continuous) if for every two right idealsA 1 andA 2 of R with
A 1 ∩A 2 = 0, each projection π i : A 1⊕A 2 → A i , i = 1, 2, can be extended to an endomorphism of R.
Definition 3.0.1 We say a ring R satisfies the property (P) if for each right ideal
I of R and for each homomorphism f : I → R there exists u ∈R such that either f =l u or l u ◦f = i I , the identity on I (the later implies that f is one-to-one).
In [2] rings satisfying the property (P) have been called almost self-injective rings as opposed to the general definition given in Chapter 1 due to Baba [4] For rings with no nontrivial idempotent the two notions coincide Obviously, rings satisfy the property (P) are right almost self-injective rings Also, rings satisfy the property (P) are right π-injective rings (Lemma 3.1.1) The purpose of this chapter is to study:
(i) when is a group algebra almost self-injective?
(ii) when is a group algebra continuous?
We show in the first section that if a group algebra satisfies the property (P), then it is self-injective (Theorem 3.1.3); as a consequence of this result, an almost self- injective group algebra K[G] with no nontrivial idempotent is self-injective We further show in the second section that if the group algebraK[G] is continuous then
G is a locally finite group (Theorem 3.2.2) We conclude the second section with a number of examples.
Almost Self-Injective Group Algebra
First we prove some results which will be used in the proof of our main result in this section.
Lemma 3.1.1 Let R be a ring satisfying the property (P) Then R is right quasi- continuous (π-injective).
Proof Let A 1 and A 2 be two nonzero right ideals of R with A 1 ∩A 2 = 0, and π 1 : A 1 ⊕A 2 → A 1 the canonical projection Since ker(π 1) = A 2 6= 0, π 1 is not one-to-one and hence can be extended to an endomorphism of R Therefore, R is right quasi-continuous.
Lemma 3.1.2 Let R be a right almost self-injective ring with no nontrivial idem- potent Then for every a, b ∈ R with r.ann R (a) = 0 and r.ann R (b) 6= 0, Rb ⊂Ra and if r.ann R (a) = 0 and r.ann R (b) = 0, then either Ra⊂Rb or Rb⊂Ra.
Proof Define f : aR → bR by f(ar) = br Clearly, f is a well defined R- homomorphism First assume thatr.ann R (a) = 0 and r.ann R (b)6= 0 Thenf is not one-one SinceR is right almost self-injective,f can be extended toR Hence there existss ∈R such that f =l s on aR Thus b=f(a) =sa Consequently Rb⊂Ra.
Now let r.ann R (a) = 0 and r.ann R (b) = 0 In this case f is one-to-one So either f =l s onaRfor somes ∈Ror there existss∈Rsuch thatl s ◦f =I aR Iff =l s on aR then as aboveRb⊂ Ra If l s ◦f =I aR then a= (l s ◦f)(a) =sf(a) =sb∈Rb.
Lemma 3.1.3 Let R be a ring satisfying the property (P) If R has a nontrivial idempotent element, then R is right self-injective.
Proof Letebe a nontrivial idempotent element of R By Lemma 1.3.6, it is enough to prove R R is both eR-injective and (1−e)R-injective Let A be a nonzero sub- module ofeRand f :A→R be an R-homomorphism Defineg :A⊕(1−e)R →R byg(a+ (1−e)r) =f(a) Then g is an R-homomorphism which is not one-to-one.
Since R satisfies the property (P), there exists a homomorphism h : R → R such that h | A⊕(1−e)R = g But then h | A = f Thus R is eR-injective Similarly, R is (1−e)R-injective.
Lemma 3.1.4 Let R be a right almost self-injective ring with no nontrivial idem- potent element, and let T = P
Ra, where r.ann R (a) = 0 and a is not invertible.Then T is a two-sided ideal of R.
Remark 3.1.1 If there is no element a such that r.ann R (a) = 0 and a is not invertible, then by convention T = 0.
Proof of Lemma 3.1.4 LetT 6= 0 It is enough to show thatar ∈T for each r∈R and for each noninvertible elementa ∈R with r.ann R (a) = 0 So let r ∈R and a be a noninvertible element of R such that r.ann R (a) = 0 If r.ann R (ar) 6= 0, then by Lemma 3.1.2 ar∈T We show that if r.ann R (ar) = 0, then ar is not invertible. Let, if possible, ar is invertible Then there exists x∈R such that xar =arx= 1. Since R has no nontrivial idempotents, R is directly finite Thus rxa=arx= 1, a contradiction because a is not invertible Hencear ∈T.
The following theorem is a key result in the proof of our main result.
Theorem 3.1.1 Let R be a ring satisfying the property (P) Then eitherR is right self-injective or local.
Proof By Lemma 3.1.3 ifRhas a nontrivial idempotent thenRis right self-injective.
So assume R has no nontrivial idempotents Since R satisfies the property (P), by Lemma 3.1.1, R is quasi-continuous and hence R R is uniform Let F = {a ∈ R | r.ann R (a) = 0 and a is not invertible} If F is empty, then a ∈ R is invertible if and only if r.ann R (a) = 0 SinceR R is uniform, Z(R R ) =R\U(R) It follows that
R\U(R) is a two sided ideal Hence R is local If F is not empty, letT = P a∈F
By Lemma 3.1.2,R\U(R)⊂T Now lett∈T We show thattis not invertible By Lemma 3.1.2,t =xc for some c∈F Now if t is invertible, then c is left invertible. SinceRhas no nontrivial idempotents,cis invertible, a contradiction becausec∈F. ThusT =R\U(R) SinceT is a two-sided ideal ofR, it follows thatR is local.
Remark 3.1.2 Since a ring with no nontrivial idempotent satisfies the property (P) if and only if it is right almost self-injective ring, it follows that a right almost self-injective ring with no nontrivial idempotent is local (by Theorem 3.1.1) and right quasi-continuous (by Lemma 3.1.1) So, it is right uniform.
Lemma 3.1.5 ([36], Lemma 10.1.13) Let H be a nonidentity subgroup of a group
G If ω(K[H]) ⊂ J(K[G]), then ω(K[H]) = J(K[H]), where K is a field of char- acteristic p for some prime p, and H is a p-group.
Theorem 3.1.2 Let K[G] be a group algebra satisfying the property (P) Then J(K[G]) =Z(K[G]).
Proof By Theorem 3.1.1, the group algebraK[G] is either self-injective or local IfK[G] is self-injective then J(K[G]) =Z(K[G]).
If K[G] is local, then J(K[G]) = ω(K[G]) By Lemma 3.1.5, G is a p-group.
Thus for everyg ∈G, r.ann K[G](1−g)6= 0, because g is of finite order SinceK[G] satisfyies the property (P),K[G] is uniform (Remark 3.1.2) Consequently, 1−g ∈
Z(K[G]) for every g ∈ G Thus ω(K[G]) ⊂ Z(K[G]) which implies J(K[G]) ω(K[G]) =Z(K[G]).
Now we can prove our main result of this section.
Theorem 3.1.3 Every group algebra K[G] satisfying the property (P) is self- injective and hence G is finite.
Proof Let K[G] be group algebra satisfying the property (P) Then by Theorem
3.1.1, K[G] is either self-injective or local AssumeK[G] is local By Remark 3.1.2
K[G] is almost self-injective and quasi-continuous Hence, K[G] is CS Thus, by
([6], Theorem 4.1), charK =p and G is a locally finite p-group HenceJ(K[G]) is nil and consequently, by ([6], Theorem 3.5 and Corollary 3.7),K[G] is self-injective. Alternative proof: Let L be a right ideal of K[G] and ϕ:L→K[G] be aK[G]- homomorphism Assume that there existsψ:K[G] →K[G] withψ◦ϕ=I L If kerψ
6= 0 then, because K[G] is uniform, kerψ ∩ϕ(L) 6= 0 Let 0 6=ϕ(x) ∈kerψ ∩ϕ(L).
Thus kerψ = 0, that is, r.ann K[G](ψ(1)) = 0 Consequently ψ(1) ∈/ Z(K[G]).Since J(K[G]) = Z(K[G]) by Theorem 3.1.2 and K[G] is local, ψ(1) is invertible.
Define η : K[G] → K[G] by η(α) = ψ(1) −1 α for every α ∈ K[G] Then η is a
K[G]-homomorphism and for every x ∈ L, η(x) = ψ(1) −1 x = ψ(1) −1 (ψ◦ϕ(x)) ψ(1) −1 (ψ(1)ϕ(x)) = ϕ(x) Thus ϕ can be extended to an endomorphism of K[G].
Corollary 3.1.1 Every almost self-injective group algebra K[G] with no nontrivial idempotent is self-injective and hence G is finite.
Continuous Group Algebra
In this section we study continuous group algebras We begin with the following lemma.
Lemma 3.2.1 If a group algebra K[G] satisfies C2, then G is a torsion group.
Proof Letg ∈G and let, if possible, o(g) be infinite Then r.ann K[G](1−g) = 0 l.ann K[G](1−g) SinceK[G] satisfies C2,K[G](1−g) is a direct summand ofK[G]. Hence, 1−g is invertible inK[G] which is a contradiction since 1−g ∈ω(K[G]).
Since a right continuous ring satisfies C2, we have the following corollary.
Corollary 3.2.1 If a group algebra K[G] is continuous, then Gis a torsion group.
Theorem 3.2.1 If G is a torsion group and K[G] is quasi-continuous, then G is a locally finite group.
Proof Let R = K[G] To prove G is locally finite, let H be a finitely generated subgroup ofG We apply induction on the number of generators ofH LetH =hh 1i where h 1 ∈G Then H is finite becauseG is torsion.
Assume that H 0 = hh 1 , h 2 , , h n i is finite and let H = hh 1 , h 2 , , h n , h n+1i. Then ω(H) = K[G](H −1)
Note thatω(H 0) =K[G](H 0−1) =K[G]ω(K[H 0]) SinceH 0is finite,r.ann R (ω(H 0))60 We show that r.ann R (ω(H)) 6= 0 Let, if possible, r.ann R (ω(H)) = 0 Then r.ann R (ω(H 0) +K[G](1−h n+1)) = 0, that is, Hb0 K[G]∩r.ann R (1 −h n+1) = 0.
Since K[G] is quasi-continuous, there exist idempotents e 1 and e 2 in R such that
Hb0 K[G] ⊆ e e 1 R and r.ann R (1 − h n+1) ⊆ e e 2 R Then e 1 R ∩ e 2 R = 0 But then there exists an idempotent e ∈ R such that e 1 R = eR, e 2 R ⊆ (1 − e)R.
Thus l.ann R (e 1 R) = l.ann R (eR) and l.ann R ((1 − e)R) ⊆ l.ann R (e 2 R), that is, R(1−e 1) =R(1−e) andRe⊆R(1−e 2) Now
⊆ ω(H 0) +R(1−h n+1) =ω(H), a contradiction, because 1 ∈/ ω(H) Hence r.ann R (ω(H)) 6= 0 and consequently H is finite.
Remark 3.2.1 Note that for a quasi-continuous group algebra K[G], the groupG need not be torsion For example, the group algebra K[Z] is quasi-continuous.
As a consequence of Corollary 3.2.1 and Theorem 3.2.1, we have the following theorem.
Theorem 3.2.2 If K[G] is continuous, then G is a locally finite group.
Since K[G] is principally self-injective if and only ifG is locally finite (Theorem1.3.2), we have the following corollary.
Corollary 3.2.2 If K[G] is continuous, then K[G] is principally self-injective.
The following example shows that the above result is not true for arbitrary rings.
Letf be the ring homomorphism f(a) =a for alla∈Q and f(x i ) =x 2 i
Then S is a subring ofA The only nontrivial right ideal of S is
which is principal Thus S is right continuous If S is right self-injective then S is quasi-Frobenius and hence left artinian which is not true Therefore, S is not right self-injective If S is right principally injective then S is right self-injective, a contradiction.
For a CS group algebra, we have the following theorem.
Theorem 3.2.3 For a CS group algebra K[G], the following are equivalent.
(iii) K[G] is principally self-injective.
Proof (ii)⇐⇒ (iii): by Theorem 1.3.2.
(iii) =⇒ (i): Suppose K[G] is principally self-injective, then K[G] has C2 ([32]
Theorem 1.2 (1)) SinceK[G] is CS by assumption, K[G] is continuous.
Corollary 3.2.3 Every group algebraK[G]of locally finite group with no nontrivial idempotents is continuous.
Proof By Theorem 3.2.3 it is enough to prove thatK[G] is uniform as a rightK[G]- module Let αand β be two nonzero elements in K[G] andH =hSupp α, Supp βi. Then H is finite by assumption So, K[H] is self-injective with no nontrivial idem- potent, and hence it is right uniform Therefore, αx = βy 6= 0 for some nonzero elements x and y in K[H] Thus, K[G] is right uniform The result follows by
We now give a number of examples of countable locally finite groups such that
Example 3.2.2 LetG =S ∞ = ∞ ∪ n=1 S n and K be any field Then G is a countable locally finite group We show that ∆(G) = 1 Let α∈S ∞ and α6= 1 Sinceα 6= 1, α(k)6=k for some k ∈N Thenβαβ −1 (β(k))6=β(k) Therefore, βαβ −1 6= 1 Since
S ∞ is a transitive group, there are infinitly many conjugates of α Thus ∆(G) = 1 and hence the group algebra K[G] is prime (Theorem 1.3.3) By ([36], Theorem
7.4.8) J(K[G]) = 0 By ([36], Theorem 9.2.5) K[G] is primitive Since a primitive continuous ring is simple (Lemma 1.3.7 and [40], Theorem 4.6), it follows thatK[G] is not continuous.
Example 3.2.3 Let p be a prime, P = Z p ∞ , the Pr¨ufer p-group, and H be any finite group Let G=P ×H and R =Q[G] ThenR is not continuous For ifR is continuous then by ([11], Corollary 10.11)R is semiperfect SinceR is regular ([36], Theorem 3.1.5), R is semisimple artinian, a contradiction because G is an infinite group (Theorem 1.3.8) In particular,Q[Z p ∞] is not continuous.
Remark 3.2.2 It can be similarly shown that if p and q are distinct primes,
P = Z p ∞ , H a finite group of order p n m where m > 1, q does not divide m, and G =P ×H then R =Z q [G] is not continuous For if R is continuous then by
([11], Corollary 10.11) R is semiperfect To show R is regular it is enough to show that for each element (g, h) ∈ P ×H, q does not divide o((g, h)) Let G 1 = hgi. Then o(G 1) = p k for some k ∈ N Clearly, G 1 ×H is a subgroup of P ×H and (g, h)∈G 1×H If q divideso((g, h)), then q divideso(G 1×H) =p k+n m, a contra- diction.
We now give an example of a prime local continuous group algebra.
Example 3.2.4 Let p be a prime and G = P ∞ = ∞ ∪ n=1 P n where for each n, P n is a Sylow p-subgroup of S p n and P n ⊂ P n+1 Then G is a locally finite p-group and
∆(G) = 1 Let K be a field of characteristic p Then K[G] is a prime local group algebra (see [36], p.314-315) K[G] is continuous by Corollary 3.2.3.
Since a prime regular continuous ring is simple (Lemma 1.3.7), we note that there does not exist any nontrivial prime regular CS (equivalently, continuous) group algebra K[G].
Although, the concept of almost injectivity has been studied for more than a decade, we find that a number of interesting and useful properties have remained unnoticed In this chapter the properties of endomorphism rings of indecomposable almost self-injective modules have been investigated Also we study the relationship of almost injectivity with other injectivity like conditions that include generlized injectivity (ojectivity), essentially injectivity, and CS property Theorem 4.1.1 shows that the endomorphism ring of an indecomposable almost self-injective module is local Moreover, the endomorphism ring of a uniserial almost self-injective right module is left uniserial (Corollary 4.1.2) It is shown for a domain D, D is right almost self-injective if and only ifD is left almost self-injective if and only if D is a two sided valuation domain (Corollary 4.1.3) Also, it is shown that a finite direct
59 sum of indecomposable almost self-injective modules is almost self-injective if the indecomposable modules are relatively almost injective (Remark 4.2.2).
Endomorphism Rings of Almost Self-Injective Modules
Recall that for two rightR-modulesM andN, M is called almost N-injective if for each submoduleX of N and each homomorphismf :X →M, either there existsg such that diagram (1) commutes or there existshsuch that diagram (2) commutes, where
N 1 is a nonzero direct summand of N, and π : N → N 1 is a projection onto N 1.Henceforth, these diagrams will be referred to as diagram (1) and diagram (2), re- spectively M is called almost self-injective ifM is almost M-injective A ringR is called right almost self-injective if it is almost self-injective as a right module over itself Left almost self-injective rings are defined, similarly.
We begin with a simple fact.
Lemma 4.1.1 An indecomposable almost self-injective module is quasi-continuous, hence uniform.
For two uniform modules M andN we give below a characterization as to when
M is almost N-injective in terms of their injective hulls.
Proposition 4.1.1 Let M and N be uniform modules Then M is almost N- injective if and only if for every f ∈ Hom(E(N), E(M)) either f(N)⊆ M or f is an isomorphism and f −1 (M)⊆N.
Proof Assume M is almost N-injective Let f ∈ Hom(E(N), E(M)) and X {n ∈ N| f(n) ∈ M} Then f| X : X → M Since M is almost N-injective, either the diagram (1) or the diagram (2) holds If (1) holds, then there existsg :N →M such that g| X = f| X We claim M ∩ (g −f)(N) = 0 Let m ∈ M such that m = (g−f)(n), for some n ∈ N Then f(n) = g(n)−m ∈ M Hence n∈ X So m=g(n)−f(n) = 0 ButM ⊆ e E(M) Hence (g−f)(N) = 0 That isf(N)⊆M.
If (2) holds, then there existsh :M →N such that h◦f = 1 X Hencef is one to one Sof is an isomorphism sinceE(N) is injective andE(M) is an indecomposable module Clearly,h| f (X) =f −1 | f(X ) We claimN∩(f −1 −h)(M) = 0.Letn 0 ∈N such that n 0 = (f −1 −h)(m 0 ) for some m 0 ∈M Thenf −1 (m 0 ) =h(m 0 ) +n 0 ∈N Apply f to both sides, we getm 0 =f f −1 (m 0 ) =f(h(m 0 ) +n 0 ) which impliesm 0 ∈f(X) So n 0 = (f −1 −h)(m 0 ) = 0 because h| f (X) =f −1 | f(X ) and m 0 ∈f(X) Hence our claim is true Since N ⊆ e E(N), (f −1 −h)(M) = 0 That meansf −1 (M) =h(M) ⊆N. The converse is clear.
Proposition 4.1.2 Let R be a ring with no nontrivial idempotent Then R is right almost self-injective if and only if for every q ∈E(R R ), either q∈ R or there exists r∈R such that qr= 1.
Proof Assume first R is right almost self-injective Then R R is uniform by Lemma 4.1.1 Let q ∈ E(R R ) and l q : R −→ E(R R ) be the left multiplication homo- morphism Then there exists f : E(R R ) −→ E(R R ) such that l q | R = f| R By Proposition 4.1.1 either f(R) ⊆ R or f is an isomorphism and f −1 (R) ⊆ R If f(R) ⊆ R, then q ∈R If f is an isomorphism and f −1 (R) ⊆ R, then there exists r∈R such that f(r) = 1 Soqr=l q (r) =f(r) = 1.
Conversely, suppose for every q ∈ E(R R ), either q ∈ R or there exists r ∈ R such that qr = 1 We claim that E(R R ) is uniform For if e ∈ End(E(R R )) is an idempotent, then either e(1) ∈ R or there exists r ∈ R such that e(1)r = 1 If e(1)∈R, thene(1) is an idempotent inR and by assumption e(1) = 0 or e(1) = 1.
Hencee= 0 or e= 1 E (R R ) because R ⊆ e E(R R ) If e(1)r = 1 for some r∈R, then e(r) = 1 Soe(1) =e(e(r)) =e 2 (r) =e(r) = 1 Soe| R
R = 1 R R We proceed to show that e= 1 E(R R ) Else, there existsx∈E(R R ) such that e(x)6=x, then ex−x 6= 0. SinceR ⊆ e E(R R ), there exists r 0 ∈R such that (ex−x)r 0 6= 0 and (ex−x)r 0 ∈R.
Because (ex−x)r 0 ∈ R, (ex−x)r 0 =e(ex−x)r 0 = 0, a contradiction to the fact that (ex−x)r 0 6= 0 Therefore, e = 1 E(R R ) This proves E(R R ) is indecomposable and hence uniform Thus, R R is uniform Now let f ∈ End(E(R R )) Then by assumption eitherf(1) ∈Rorf(r) = 1 for somer∈R f(1) ∈Rimpliesf(R)⊆R.
Iff(r) = 1 for somer∈R, thenf| rR :rR−→Ris an isomorphism BecauseE(R R ) is uniform and injective,f is an isomorphism onE(R R ) andf −1 (R) =rR⊆R By
Proposition 4.1.1 R R is almost self-injective.
Corollary 4.1.1 Let D be a domain and Q its maximal right ring of quotient. Then D is right almost self-injective if and only if for every q ∈ Q, either q or q −1 ∈D.
It is known that the endomorphism ring of an indecomposable quasi-injective (more generally continuous) module is local We prove an analogous result for indecomposable almost self-injective module The following result is a fundamental theorem in this chapter.
Theorem 4.1.1 If M is an indecomposable almost self-injective module, thenEnd(M) is local.
For a proof of this theorem, we need first to prove the following two lemmas.
Lemma 4.1.2 Let M be an indecomposable almost self-injective module Then for every f, g ∈ S = End(M), (i) if ker(f) ( ker(g), then Sg ( Sf, (ii) if ker(f) =ker(g), then Sf ⊆Sg or Sg ⊆Sf.
Proof Defineϕ:f(M) −→g(M) byϕ(f(m)) =g(m) Clearly,ϕ is a well defined
R-homomorphism (i) We have ker(f) ( ker(g) Then ϕ is not a one to one map By assumption ϕ can be extended to M Hence there existsψ∈S such that ψ(f(m)) =ϕ(f(m)) for every m∈ M Thus g(m) = (ψ◦f)(m) for every m ∈M. Consequently Sg ( Sf (ii) Let ker(f) = ker(g) In this case ϕis one to one So either ϕ can be extended to an endomorphism ψ ∈ S or there exists η ∈ S such that η ◦ ϕ= 1 f(M ) If ϕ=ψ on f(M), then as aboveSg ⊆Sf If η ◦ ϕ= 1 f(M ), thenf(m) = (η◦ϕ)(f(m)) =η(ϕ(f(m))) =η(g(m)) = (η◦g)(m) for everym∈M. ThusSf ⊆Sg.
Corollary 4.1.2 Let M be a uniserial almost self-injective right R-module Then End(M) is left uniserial.
Lemma 4.1.3 LetM be an indecomposable almost self-injective module and letS End(M) Then the left ideal H of S generated by non-isomorphic monomorphisms in S is a two-sided ideal.
Proof It is enough to show that f g ∈ H for each g ∈ S and for each non- isomorphismf ∈S with ker(f) = 0 Ifker(f g)6= 0, then by Lemma 4.1.2 f g∈H. Now assume that f g is 1-1 If f g were an isomorphism f would be onto, a contra- diction Thus f g∈H.
P roof of Theorem 4.1.1 Let S =End(M) Then S has no idempotents other than 1 and 0 Recall that since M is indecomposable almost self-injective, it is uniform (Lemma 4.1.1) Let F be the set of all non-isomorphic monomorphisms in S If F is empty, then ϕ ∈ S is an isomorphism if and only if Ker(ϕ) = 0.
Let h+g ∈ U(S), where U(S) is the group of units of S Since M is uniform, either ker(h) = 0 or ker(g) = 0 This means either h or g is an isomorphism.
Hence S is local Suppose F is not empty Let H = P f∈F
S\U(S)⊂H Now leth∈H We show that h is not invertible Writeh P n i=1 g i f i , where f i ∈F and g i ∈S By Lemma 4.1.2, Sf 1, Sf 2, ã ã ã, Sf n are linearly ordered.
So, Sf 1 ⊆ Sf 2 ⊆ ã ã ã ⊆Sf n , after reordering if necessary Hence h =gf n for some g ∈ S Now if h is invertible, then f n is left invertible Since S has no nontrivial idempotents,f n is invertible, a contradiction because f n ∈F Thus H =S\U(S).
Since H is a two-sided ideal of S (Lemma 4.1.3), it follows that S is local This completes the proof of the theorem.
Let M = ⊕ α∈I M α A submodule N of M is said to be finitely contained (written as f.c.) in the direct sum, with respect to the decomposition M = ⊕ α∈I
M α , if there exists{α 1 , α 2 ,ã ã ã, α n } ⊆I such thatN ⊆ ⊕ n i=1 M α i [18] A module M is said to have the CS property of uniform (resp f.c uniform) module if every uniform (resp f.c. uniform) submodule is essential in a direct summand of M It is well-known that if
M has a finite uniform dimension and the CS property of uniform module thenM is CS (see [11], Corollary 7.8).
Theorem 4.1.2 ([18], Theorem 10) Assume {M α } α∈I is a set of completely inde- composable R-modules, each M α is uniform, and M = ⊕ α∈I
M α Then the following conditions are equivalent:
(i) M has the CS property of f.c uniform module.
(ii) For any pairα, β in I, any homomorphismf of a submodule A α in M α toM β is extended to an element in Hom R (M α , M β )orf −1 is extended to an element in Hom R (M β , M α ), provided ker(f) = 0.
Since the direct sum of CS modules is not necessarily CS, it has been a subject of active research to find conditions as to when the direct sum of a family of CS indecomposable modules is CS The following remark gives one such condition in terms of almost injectivity.
Remark 4.1.1 For a module M which can be expressed as a finite direct sum of indecomposable modules {M i } n i=1 it follows from Theorem 4.1.1 and Theorem 4.1.2 that the following are equivalent: (i) M ⊕M is CS and End(M i ) is local for each i; (ii) M is finitely P
-CS and End(M i ) is local for each i; (iii) M i is almost M j - injective for everyi and j.
Almost Injectivity and Other Injectivity Conditions
In [17], Hanada, Kuratomi, and Oshiro introduced a generalization of relative injectivity For two modulesM and N, they calledM to be generalizedN-injective (or N-ojective as in [31]) , if for any submodule A of N and any homomorphism f :A→M, there exist decompositionsN =N⊕N,M =M⊕M, a homomorphism f : N → M, and a monomorphism g :M →N satisfying the following properties (∗), (∗∗)
(∗∗) For a∈A, we expressa inN =N⊕N as a=a+a, where a∈N and a∈N. Thenf(a) =f(a) +f(a), where f =g −1
M is called generalized self-injective if M is generalized M-injective A decom- position M = ⊕ i∈I M i is called exchangeable if for any direct summand N of M,
The following two theorems due to Mohamed and M¨uller [31], which are essen- tially the same results of Hanada et al but with quite different proofs (see [17],
Theorem 4.2.1 ([31], Theorem 10) Let M = M 1 ⊕M 2 Then M is CS and the decomposition is exchangeable if and only if M i is CS and generalized M j -injective for all i6=j.
Theorem 4.2.2 ([31], Theorem 13) Let M = M 1 ⊕ ã ã ã ⊕M n where the M i are uniform Then M is CS and the decomposition is exchangeable if and only if M i is generalized M j -injective for all i6=j.
Proposition 4.2.1 If M is generalized N-injective, then M is almost N-injective.
Proof LetAbe a submodule ofN andf :A→M be a homomorphism Then there exists decompositions N = N ⊕N, M = M ⊕M, a homomorphism f : N → M, and a monomorphismg :M →N satisfying the properties (∗), (∗∗) Iff can not be extended toN, then N 6=N This meansN 6= 0 Defineh:M →N byh=g◦π M , where π M :M → M is the canonical projection of M onto M with respect to the decompositionM =M⊕M For everya∈A, expressainN =N⊕N asa =a+a, where a∈N and a∈N Then by (∗∗) h◦f(a) = h(f(a) +f(a)), where f =g −1
Remark 4.2.1 Clearly, ifM andN are indecomposable modules, thenM is almost
N-injective if and only ifM is generalizedN-injective.
For two modules M and N, M is said to be essentially N-injective if for every submodule A of N, any homomorphism f : A → M with ker(f) ⊆ e A, f can be extended to a homomorphism g :N →M (see [11], p 16-17).
Observe that if M is almost N-injective, then M is essentially N-injective For if A is a submodule of N and f : A → M a homomorphism with ker(f) ⊆ e A, then it follows from [4], Lemma B, that f can be extended to a homomorphism g : N → M provided ker(f) ⊆ e N Let B be a complement of A in N and de- fine h : A⊕ B → M by h(a +b) = f(a) for every a ∈ A and b ∈ B Then ker(h) = ker(f)⊕B ⊆ e A⊕B ⊆ e N So, ker(h) ⊆ e N and hence h can be ex- tended to a homomorphism g :N →M Clearly,g is an extention of f.
We may record the following implications just for reference only:
From the above discussion and Proposition 4.2.1 we have the following Proposi- tion of K Hanada et al [17]:
Proposition 4.2.2 ([17], Proposition 1.4 (2)) IfM is generalized N-injective, then
We close this section by a remark that enable us to produce examples of almost self-injective modules.
Remark 4.2.2 Let {M i } n i=1 be a finite set of indecomposable almost self-injective modules If M i is almost M j -injective for any pair i and j in {1,2, , n}, then
Proof By Lemma 4.1.1 eachM i is uniform By assumption and Remark 4.2.1M i is generalizedM j -injective for everyiandj LetM =⊕ n i=1 M i andX =M⊕M Then
X is CS and the decomposition X = ⊕ n i=1 (M i ⊕M i ) is exchangeable by Theorem 4.2.2 (also see [17], Corollary 2.10) This implies X is CS and the decomposition
X =M⊕M is exchangeable HenceM is generalized self-injective by Theorem 4.2.1 (c.f [17], Theorem 2.1) and so M =⊕ n i=1 M i is almost self-injective by Proposition 4.2.1.
As a consequence of Remark 4.2.2 it follows that for a prime p, Z/p 2 Z ⊕Z/pZ is almost self-injective but not quasi-injective If M is any indecomposable almost self-injective module, then M n is also almost self-injective for all positive integers n In particular, since any two sided valuation domain D is right and left almost self-injective, we obtain thatD n is right and left almost self-injective as aD-module.
It was shown by Jain-Kanwar-Malik-Srivastava that K[D ∞] is CS if and only if char(K)6= 2 ([24], Theorem 3.6).
Question (1): Find conditions on a ringR that force the group ring R[D ∞] to be CS?
Recall that every almost self-injective group algebra K[G] with no nontrivial idempotent is self-injective (Corollary 3.1.1).
Question (2): Study almost self-injective group algebra, in general?
There does not exist any nontrivial indecomposable regular continuous (equiv- alently, CS) group algebra K[G] since an indecomposable contiuous regular ring is simple (Lemma 1.3.7).
Question (3): Is it true that a regular continuous group ring (equivalently, CS group ring) self-injective?
In Proposition 4.2.1, we show that generalized injectivity condition implies al- most injectivity condition.
Question (4): Is the converse true? In particular, is it true that a CS almost self-injective module is generalized self-injective?
Question (5): Can we replace the generalized injectivity condition by almost injectivity condition in Theorem 4.2.1? More general, suppose M and N are CS and M is almost N-injective IsM generalized N-injective?
[1] Adel N Alahmadi, S K Jain, and J.B.Srivastava,Semiprime CS Group Algebra of Polycyclic-By-Finite Group Without Domains as Summands is Hereditary, to appear in J Algebra.
[2] Adel N Alahmadi, S K Jain, P Kanwar, and J.B.Srivastava, Group Algebras in which Complements are Direct Summands, (Russian) Fundam Prikl Mat.,
[3] Adel N Alahmadi, and S K Jain, Note On Almost Injectivity, submitted.
[4] Yoshitomo Baba, Note on Almost M-injective, Osaka J Math., 26, 687-698
[5] Antonio Behn, Polycyclic Group Rings Whose Principal Ideals Are Projective,
[6] K I Beidar, S K Jain, P Kanwar, and J B Srivastava, CS matrix rings over local rings,J Algebra, 264(1), 251-261 (2003).
[7] K I Beidar, S K Jain, and P Kanwar, Nonsingular CS-rings coincide with tight PP rings,J Algebra, 282 (1), 626-637 (2004).
[8] Gary F Birkenmeier, Jin Y Kim, and Jae K Park, A counterexample for CS rings, Glasg Math J., 42(2), 263-269 (2000).
[9] A W Chatters and C R Hajarnavis, Rings in which every complement right ideal is a direct summand, Quart J Mathematics,28, 61-80 (1977).
[10] A W Chatters and C R Hajarnavis, Rings with Chain Conditions, Pitman, London, 1980.
[11] N V Dung, D V Huynh, P F Smith, and R Wisbauer, Extending Modules,
[12] C Faith,Algebra: Rings, Modules and Categories I, Springer-Verlag, New York
[13] Daniel R Farkas, A note on locally finite group algebras, Proc Amer Math. Soc.,48(1), 26-28 (1975).
[14] V K Goel, and S K Jain, π-injective modules and rings whose cyclics are π-injective, Comm Alg.6(1), 59-73 (1978).
[15] K R Goodearl, and R B Warfield, An Introduction to Noncommutative Noetherian Rings, Cambridge University Press, 2004.
[16] K R Goodearl, Von Neumann Regular Rings, Second Edition, Krieger Publ.
[17] K Hanada, Y Kuratomi, and K Oshiro,On Direct Sums of Extending Modules and Internal Exchange Property,J Algebra, 250, 115-133 (2002).
[18] M Harada and K Oshiro,On Extending Property On Direct Sums of Uniform Modules, Osaka J Math.,18, 767-785 (1981).
[19] M Harada and A Tozaki, Almost M-projectives and Nakayama Rings, J Al- gebra, 122(2), 447-474 (1989).
[20] M Harada,Direct Sums of Almost Relative Injective Modules,Osaka J Math.,
[21] M Harada,Note On Almost Relative Projectives and Almost Relative Injectives,
[22] D V Huynh, S K Jain, S R L´opez-Permouth, On the Symmetry of the Goldie and CS Conditions for Prime Rings, Proc Amer Math Soc.,128(11),
[23] S K Jain, Husain S Al-Hazmi and Adel N Alahmadi,Right-Left Symmetry of Right Nonsingular Right Max-Min CS Prime Rings, to appear in Comm Alg.
[24] S K Jain, P Kanwar, S Malik and J B Srivastava, K[D ∞]is a CS-Algebra, Proc Amer Math Soc.,128(2), 397-400 (2000).
[25] S K Jain, P Kanwar, and Sergio R L´opez-Permouth,Nonsingular semiperfect
CS rings II, Bull London Math Soc.,32(4), 421-431 (2000).
[26] S K Jain, P Kanwar, and J B Srivastava, Survey of some recent results on
CS group algebras and open questions, Advances in Algebra, Proceedings ICM
Satellite Conference in Algebra and Related Topics, World Scientific, 2002. [27] T Y Lam,A First Course in Noncommutative Rings, Springer-Verlag, 2001.
[28] T Y Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics,
[29] J Lambek,Lectures on Rings and Modules,Chelsea Publishing Company, New York, 1986.
[30] S H Mohamed and B J M¨uller,Continuous and Discrete Modules,Cambridge University Press, 1990
[31] S H Mohamed and B J M¨uller, Ojective Modules, Comm Alg., 30(4),