A ring is said to be strongly right bounded if every nonzero right ideal contains a nonzero ideal. In this paper strongly right bounded rings are characterized, conditions are determined which ensure that the split-null (or trivial) extension of a ring is strongly right bounded, and we characterize strongly right bounded right quasi-continuous split-null extensions of a left faithful ideal over a semiprime ring . This last result partially generalizes a result of C. Faith concerning split-null extensions of commutative FPF rings

Birkenmeier, Gary F.

G. F. Birkenmeier

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Publicacions Matemàtiques

Split-null extensions of strongly right bounded rings

Birkenmeier, G. F.

Revistes Catalanes amb Accés Obert

Publicacions Matemátiques, Vol 33 (1989), 37-44 .
Abstract
SPLIT-NULL EXTENSIONS OF STRONGLY
RIGHT BOUNDED RINGS
GARY F . BIRKENMEIER
A ring is said to be strongly right bounded if every nonzero right ideal
contains a nonzero ideal . In this paper strongly right bounded rings are
characterized, conditions are determined which ensure that the split-null
(or trivial) extension of a ring is strongly right bounded, and we characte-
rize strongly right bounded right quasi-continuous split-null extensions of
a left faithful ideal over a semiprime ring . This last result partially genera-
lizes a result of C . Faith concerning split-null extensions of commutative
FPF rings .
Examples of strongly right bounded rings are : right duo rings (e.g ., commu-
tative rings and strongly regular rings) [8], [18] and [26] ; right subdirectly
irreducible rings [9] and [10] ; right valuation rings which are not subdirectly
irreducible [24, p . 216] ; and bounded principal ideal domains [20, p . 41] . In
[13, p . 364] an example of a strongly left bounded right primitive ring is given .
In [16, p . 5 .3] an example of a strongly right bounded right self-injective ring
which is not left selfinjective is presented . Strongly right bounded rings play
a fundamental role in the theory of FPF rings (e.g ., a strongly right bounded
right selfinjective ring is right FPF and the basic ring of a semiperfect right
FPF ring is strongly right bounded [16]) . In fact, according to [17, p . 310], C .
Faith has conjectured that a right FPF ring is Morita equivalent to a strongly
right bounded ring .
All rings are associative, R denotes a ring with unity andM will always be a
unital (R, R)-bimodule . The split-null (or trivial extension S(R, M) ofM by
R is the ring formed from the Cartesian product R x M with component--wise
addition and with multiplication given by (a, m)(b, k) = (ab, ak+mb) (cf., [12],
[15], and [22]) . Annihilators will be symbolized as [A(X) = {a E AjaX = 0}
and rA(X) = {a E AIXa = 0} . A (ring) direct summand of R will mean a
right ideal generated by a (central) idempotent . From [16], R is right FPF if
every finitely generated faithful right R-module generates the category mod-
R . From [3], R is right quasi-FPF if, whenever a faithful right R-module is a
direct sum of finitely many cyclic modules, then it is a generator for mod-R.
A ring R is (quasi-) Baer (cf., [7] a.nd [23]) if the right annihilator of every
38
	
G.F . BIRKENMEIER
(ideal) nonempty subset ofR is a direct summand of R. Semiprime right FPF
rings are quasi-Baer [11, p . 168] . From [6] a ring is right CS if every right
ideal is essential in a direct summand . From [21], R is right quasi-continuous
(also known as 7r-injective [19]) if it is right CS and if P and Q are direct
summans of R such that P f1 Q = 0, then P ® Q is a direct summand of
R. Note that if R is right CS and every idempotent is central, then R is right
quasi- continuous . Thus in [14, p . 83] Faith has shown that every commutative
FPF ring is quasi-continuous . R satisfies the fntersection left annihilator sum
property, ILAS, if whenever X and Y are right ideals such that X f1 Y = 0,
then CR(X)R + CR(Y)R = R (e.g ., right uniform rings, right selfinjective rings
[25, p . 275],and right quasi-FPF rings [3, Lemma 1]) .
Proposition 1 . The following conditions are equivaleni :
(i) R is a strongly right bounded ring .
(ii) If xR is a faithful cyclic module, then rR(x) = 0 .
R is directly ftnite and every faithful cyclic module is isomorphic to R.
Proof.
(i) --> (ii) . If rR(x) qÉ 0, then there exists a nonzero ideal Y C_ rR(x). Hence
xRY = 0. Contradiction!
(ii) --> (iii) . Assume R= Jr' ®S where X and S are right ideals and S is iso-
morphic to R. Hence R/X is faithful . Therefore, X = 0 . Consequently,
R is directly finite . Clearly every faithful cyclic module is isomorphic to
R.
(iii) ---> (i) . Let X be a right ideal containing no nonzero ideals . Then R/X
is isomorphic to R. Hence R=X® S where S is a right ideal . Since R
is directly finite, !I' = 0 . Consequently, R is strongly right bounded .
Lemma 2. Let R be a strongly right bounded ring .
(i) Every nonzero right ideal is an essential extension of an ideal of R.
(ii) R is right nonsingular if and only if R is semiprime if and only if R is
reduced (i . e., R has no nonzero nilpotent elements).
Proof. Part (i) is in [16, Note 1.3D] . Part (ii) is in [4, Proposition 1] .
Proposition 3 . Le¡ R be a strongly right bounded ring . Then the following
conditions are equivalent :
(i) R is quasi-Baer .
(ii) R is semiprime right quasi-continuous .
(iii) R is semiprime right quasi-FPF.
Proof.: This result follows from [2, Proposition 1.2], [3, Propositions 4 and
6], and Lemma 2 .
The following notation will be used : if V C_ S(R, M), then Vl and VZ are the
sets of first and second components of V, respectively.
Lemma 4 .
(i) If V is a right ideal of S(R, M), then Vi is a right ideal of R, V2 is a
right R-submodule of M, and {0} x ViM is a right S(R, M)-submodule
of V.
(ii) IfW is a right ideal of R and K is a right R-submodule of M such that
WM C K, then W x K is a right ideal of S(R, M) .
(iii) Let V C S(R, M) . Then [IR(V1) n IR(V2)] x IM(V,) C IS(R,M)(V) .
(iv)
	
The right ideal {0} x M is right essential in S(R, M) if and only if M
is left faithful (i.e ., IR(M) = 0)-
(v) If V and W are right ideals of S(R, M) such that V n W = 0, then
VlMnWlM=0.
(vi) Let S(R, M)) be strongly right bounded where M is an ideal of R. Then
R is strongly right bounded and if IR(M) 7~ 0, then IR(M)n rR(M) 0-
(vi¡) Let M be a module such that whenever An B = 0, then AM nBM = 0
where A and B are right ideals of R (e.g ., M is an ideal) . If S(R, M)
satisfies the ILAS condition, then R satisfies ¡he ILAS condition.
(viii) Le¡ M be and ideal of R. Then S(R, M) is right uniform if and only if
R is right uniform and M is left faithful .
Proof.
SPLIT-NULL EXTENSIONS 39
(i) Clearly Vi is a right ideal of R and V2 is a right R-submodule of M.
Let w E Vl and m E M. There exists k E V2 such that (w, k) E V.
Then (w, k)(0, m) = (0, wm) E V. Thus {0} x ViM is a right S(R, M)-
submodule of V.
(ii) and (iii) are straightforward .
(iv) Suppose {0} x M is right essential in S(R, M) and 0 7~ t E IR(M) .
There exists (w, m) E S(R, M) such that 0 :~ (t, 0)(w,m) E {0} x
M. Contradiction! Hence All is left faithful . Conversely, let (w, m) E
S(R, M). If w == 0, we are finished. So assume w :~ 0. There exists
k E Msuch that 0 74 (w, ?n)(0, k) = (0, wk) E {0} x 1VI . Hence {0} x M
is right essential in S(R, M) .
(v) Assume vm = wk E VlMn 1471M where v E Vi, w E W1 , and m, k E M.
There exists x E V2 and y E W2 such that (v, x) E V and (w, y) E W.
Consider (v, x)(0, m) = (0, vm) = (0, wk) = (w, y)(0, k) E V n W = 0.
Therefore, VlMnWlM = 0.
(vi) Let Y be a nonzero right ideal of R. There exists an ideal J of S(R, M)
such that J is essential in Y x YM. Since Jl and J2 cannot both be
zero, Y contains a nonzero ideal. Hence R is strongly right bounded.
If IR(M) jÉ 0, then there exists a nonzero ideal H C_ IR(M) x {0} .
Hence Hl is a nonzero ideal of R and ({0} x M)H = {0} x MHl C_ H.
Therefore, 0 7É Hl C_ IR(M) n rR(M).
(vii) Let A and B be right ideals of R such that AnB = 0. Let A* = A xAM
and B* = B x BM. Hence A* n B* = 0. Now IS(R,M)(A*) = IR(A) x
[m(A) and IS(R,M)(B*) = IR(B) x Im(B) . Consequently, IR(A)R +
40
	
G.F . BIRKENMEIER
CR(B)R = R.
(vi¡¡) Assume S(R, M) is right uniform and let Y be a nonzero right ideal
of R . By part (iv) M is left faithful . Let 0 qÉ w E R. There exists
(t, m) E S(R, M) such that 0 :~ (w, 0)(t, m) = (wt, wm) E Y x YM.
Therefore, R is right uniform. Conversely, let V be a nonzero right ideal
of S(R, M) and 0 qÉ (t, m) E S(R, M). By part (iv) 0 q¿ V fl ({0} x M) =
{0} x V2 is essential in V . If t qÉ 0, there exists y E R such that
0 ty E V2 . Since M is left faithful, there exists k E M such that
0 tyk E V2 . Thus 0 7É (t, m)(0, yk) = (0, tyk) E {0} x V2 . If t = 0,
then m 7É 0 and there exists q E R such that 0 , mq E V2 . Thus
0 :~ (t, m)(q, 0) = (0, mq) E {0} x V2 . Consequently, in all cases {0} x V2
is right essential in S,(R, M). Therefore, S(R, AJ) is right uniform .
We note that if R is commutative and M is and ideal of R, then S(R, M) is
commutative . However, in Example 9 we shall provide a strongly right bounded
ring Ti and an ideal (T, 0) such that S(TI , (T, 0)) is not strongly right bounded .
Also in [9, Example 2 .2] the ring R is a strongly right bounded ring ; however,
from Lemma 4 (vi), S(R, R(x i , 0)R) is not strongly right bounded . Thus it
is natural to investigate conditions on R and M which insure that S(R, M) is
strongly right bounded . We say M is a strongly right bounded module if every
nonzero right R -submodule contains a nonzero (R, R)-bisubmodule of M .
Theorem 5. LeíR be a strongly right bounded ring . If either of the following
conditions is satisfied, then S(R, M) is a strongly right bounded ring .
(i) M is a strongly right bounded module such that CR(M) contains no non-
zero nilpotent ideals ofR and IR(M) C rR(M) .
(ii) M is an ideal of R such that CR(M) fl M = 0 .
Proof- Let V be a nonzero right ideal of S(R, M). If Vi = 0 or V f1 ({0} x
M) :jÉ 0, then there exists a nonzero (R, R)-bisubmodule K C_ V2 such that
{0} x K C_ V is an ideal of S(R, M). So assume Vi :~ 0 and V f1({0} x M) = 0 .
Let D be a nonzero ideal of R such that D C_ VI . Note that with either condition
(i) or (ii), Vi M = 0 = MVI . If condition (i) is satisfied, then V2 = V2 x {0} q¿ 0 .
Hence D2 x {0} C V is a nonzero ideal of S(R, M). Now assume condition (ii)
is satisfied . If V2 = 0, then D x {0} C_ V is a nonzero ideal of S(R, M). If
V2 7~ 0, then V2M qÉ 0 . But V(M x {0}) = {0} x V2M C V f1 ({0} x M) = 0 .
Contradiction! Therefore, in all cases V contains a nonzero ideal of S(R, M).
Consequently, S(R, M) is strongly right bounded .
We note that when M is an ideal of R, then S(R, M) is isomorphic to a
subring of T2(R) (i .e ., the 2 x 2 lower triangular matrix ring over R) . However,
from [4, Proposition 10], T (R) is never strongly right bounded for n > 1 .
Corollary 6 . Leí M be an ideal of `R . Then S(R, M) is strongly right
bounded right uniform if and only if R is strongly right bounded right uniform
and M is left faithful .
Proof. This result follows from Theorem 5 and Lemma 4 (vi¡¡) .
Thus, if R is a strongly right bounded domain and M is any ideal of R, then
S(R, M) is a strongly right bounded right uniform ring . The ring H[x] where
H denotes the real quaternions provides an example of a strongly bounded
domain which is neither left nor right duo .
Proposition 7. Let M be a left faithful ideal of R.
	
Then the following
equivalentes are true :
(i) Every ideal of R is right essential in a (ring) direct summand of R if
and only if every ideal of S(R, M) i,s right essential in a (ring) direct
summand of S(R, M) .
(ii) Every right ideal is right essential ira a ring direct summand of R if and
only if the same is true for S(R, M).
Proof.
SPLIT-NULL EXTENSIONS 41
(i) Let S denote S(R, M) and assume every ideal of R is right essential in a
direct summand of R. Let Y be an ideal of S and V = Y fl {0} xM. By
Lemma 4 (iv), V is right essential in Y, V = {0} x V2, and V2 is an ideal of
R. Hence there exists a (central) idempotent e E R such that V2 is right
essential in eR . Consider (e, 0)S . Let (x, m) E S; then (e, 0)(x, m) =
(ex, em). Suppose 0 :~ (ex, em) . If ex :~ 0, then there exists t E R
such that 0 :~ ext E V2 . Hence 0 7É (ex, em)(0, t) = (0, ext) E V . If
ex = 0, then there exists w E R such that 0 :~ emw E V2 . Hence
0 :~ (ex, em)(w, 0) = (0, emw) E V. Therefore, in all cases, V is right
essential in (e, 0)S . Hence Y is right essential in (e, 0)S . Consequently,
every ideal of S(R, M) is right essential in a (ring) direct summand of
S(R, M) .
Conversely, suppose every ideal of S is right essential in a (ring) direct
summand of S . Let K be an ideal of R. Then there exists a (central)
idempotent (e, m) E S such that {0} x KM is right essential in (e, m)S .
Note that eme = 0 . Hence (e, m) is central in S if and only if e is central
in R and M = 0. Now {0} x KM C_ (e, m)({0} x M) C (e, m)S . Hence
KM is right essential in eM and eM is right essential in eR because M
is left faithful in R. Since K is an ideal and KM is right essential in K,
then K is right essential in eR.
(ii) This part is proved in a manner similar to that of part (i) .
In [15] Faith characterizes when S(R, M) is FPF where R is commutative
and M is faithful . He poses this characterization as an open problem when R
is noncommutative . The following result partially generalizes Faith's result .
Corollary 8. Le¡ R be a semiprime or a right nonsingular ring and M be
a left faithful ideal of R. Then ¡he following conditions are equivalent :
(i) R is strongly right bounded and right quasi-continuous .
42
	
G.F . BIRKENMEIER
(ii) S(R, M) is strongly right bounded and right quasi-continuous .
(iii) S(R, M) is strongly right bounded and right quasi-FPF .
Proof.
(i) -> (ii) By Lemma 2, R is reduced . Hence every idempotent of R is
central . Thus every idempotent of S(R, M) is central . By Theorem 5
and Proposition 7, S(R, M) is strongly right bounded and right quasi-
continuous .
(ii) + (iii) By Lemma 4 (vi) and Lemma 2, R is reduced . Hence every
idempotent of S(R, M) is central . By [3, Proposition 6], S(R, M) is
right quasi-FPF .
(iii) -> (i) By Lemma 4 (vi) and Lemma 2, R is reduced strongly right
bounded ring . By Lemma 4 (vi¡), R satisfies the ILAS condition . From
[1, Lemma 2.2] and Proposition 3, R is right quasi-continuous .
When R is quasi-Baer strongly right bounded and M is a,left faithful
ideal of R, the sequence of embeddings
R -> S(R, M) -> T2 (R)
is interesting in that S(R, M) is strongly right bounded (and right quasi-
continuous) but not quasi-Baer (cf ., Proposition 3) and T2(R) is quasi-
Baer [23] but not strongly right bounded .
The following example is a special case of a general procedure indicated in
Example 9 . Let I denote the ring of integers and T the semigroup ring
of A over IZ (Le ., integers modulo 2) where A is the semigroup on the set
{a, b} satisfying the relation xy = y for x, y E A . Thus T = {0, a, b, a -f- b} .
Let TI denote the Dorroh extension of T (i .e ., the ring with unity formed
from T x I with componentwise addition and with multiplication given by
(x, k)(y, n) = (xy + nx + ky, kn)) . T1 has the following properties :
(i) The set of nilpotent elements of Ti, N(T1 ) = {(0, 0), (a + b, 0)}, is the
Jacobson radical and equals the right socle of TI .
(ii) Every nonzero right ideal of T1 contains either N(T1 ) or a nonzero ideal
of the form (0,2kI) _ «0, 2k%*) E Ti ¡ k is a fixed integer and i E I} .
Therefore, TI is strongly right bounded .
(iii) Ti is not right duo since (a, 1)T1 is not an ideal .
(iv) Ti is not strongly left bounded .
(v) Ti does not satisfy the ILAS condition since IT,(N(T1 )) -{- ITl ((a +
b, 2)T1 )T1 Ti . However if {X2} is a nonempty set of ideals of Ti such
that nX; = 0 then R = EIT,(X;) . Thus TI satisfies the ILAS condition
defined in [1] .
(vi) Ti is not right CS, since (a+ b, 2)Tl is not essential in a direct summand .
However, every ideal is right essential in a direct summand of Tl .
(vi¡) S(I, N(Tj » (i .e ., split-null extension) is ring isomorphic to the subring
(0, I) + N(T1 ) of Ti . S(I, N(Tj )) provides an example for Theorem 5
SPLIT-NULL EXTENSIONS
	
43
(vi¡¡) S(T1 , (0, k2I» provides an example for Theorem 5 (ii) .
(ix) S(TI , (T, 0)) is an example of a split-null extension of a strongly right
bounded ring which is not strongly right bounded (cf. Theorem 5) .
To see this observe ((a, l), (0, 0»S(Ti, (T, 0)) = {((ka, k), (0, 0))¡k E I}
contains no nonzero ideals since ((b, 0), (0, 0»((ka, k), (0, 0)) = ((k(a +
b), 0), (0, 0)) .
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1980 Mathematics subject classifications : 16A15
Department of Mathematics
University of Southwestern Louisiana
Lafayette, LA 70504
U .S.A .
Rebut el 12 d'Octubre de 1988


Diposit Digital de Documents de la UAB

Publicacions Matemátiques, Vol 33 (1989), 37-44 .AbstractSPLIT-NULL EXTENSIONS OF STRONGLYRIGHT BOUNDED RINGSGARY F . BIRKENMEIERA ring is said to be strongly right bounded if every nonzero right idealcontains a nonzero ideal . In this paper strongly right bounded rings arecharacterized, conditions are determined which ensure that the split-null(or trivial) extension of a ring is strongly right bounded, and we characte-rize strongly right bounded right quasi-continuous split-null extensions ofa left faithful ideal over a semiprime ring . This last result partially genera-lizes a result of C . Faith concerning split-null extensions of commutativeFPF rings .Examples of strongly right bounded rings are : right duo rings (e.g ., commu-tative rings and strongly regular rings) [8], [18] and [26] ; right subdirectlyirreducible rings [9] and [10] ; right valuation rings which are not subdirectlyirreducible [24, p . 216] ; and bounded principal ideal domains [20, p . 41] . In[13, p . 364] an example of a strongly left bounded right primitive ring is given .In [16, p . 5 .3] an example of a strongly right bounded right self-injective ringwhich is not left selfinjective is presented . Strongly right bounded rings playa fundamental role in the theory of FPF rings (e.g ., a strongly right boundedright selfinjective ring is right FPF and the basic ring of a semiperfect rightFPF ring is strongly right bounded [16]) . In fact, according to [17, p . 310], C .Faith has conjectured that a right FPF ring is Morita equivalent to a stronglyright bounded ring .All rings are associative, R denotes a ring with unity andM will always be aunital (R, R)-bimodule . The split-null (or trivial extension S(R, M) ofM byR is the ring formed from the Cartesian product R x M with component--wiseaddition and with multiplication given by (a, m)(b, k) = (ab, ak+mb) (cf., [12],[15], and [22]) . Annihilators will be symbolized as [A(X) = {a E AjaX = 0}and rA(X) = {a E AIXa = 0} . A (ring) direct summand of R will mean aright ideal generated by a (central) idempotent . From [16], R is right FPF ifevery finitely generated faithful right R-module generates the category mod-R . From [3], R is right quasi-FPF if, whenever a faithful right R-module is adirect sum of finitely many cyclic modules, then it is a generator for mod-R.A ring R is (quasi-) Baer (cf., [7] a.nd [23]) if the right annihilator of every38	G.F . BIRKENMEIER(ideal) nonempty subset ofR is a direct summand of R. Semiprime right FPFrings are quasi-Baer [11, p . 168] . From [6] a ring is right CS if every rightideal is essential in a direct summand . From [21], R is right quasi-continuous(also known as 7r-injective [19]) if it is right CS and if P and Q are directsummans of R such that P f1 Q = 0, then P ® Q is a direct summand ofR. Note that if R is right CS and every idempotent is central, then R is rightquasi- continuous . Thus in [14, p . 83] Faith has shown that every commutativeFPF ring is quasi-continuous . R satisfies the fntersection left annihilator sumproperty, ILAS, if whenever X and Y are right ideals such that X f1 Y = 0,then CR(X)R + CR(Y)R = R (e.g ., right uniform rings, right selfinjective rings[25, p . 275],and right quasi-FPF rings [3, Lemma 1]) .Proposition 1 . The following conditions are equivaleni :(i) R is a strongly right bounded ring .(ii) If xR is a faithful cyclic module, then rR(x) = 0 .R is directly ftnite and every faithful cyclic module is isomorphic to R.Proof.(i) --> (ii) . If rR(x) qÉ 0, then there exists a nonzero ideal Y C_ rR(x). HencexRY = 0. Contradiction!(ii) --> (iii) . Assume R= Jr' ®S where X and S are right ideals and S is iso-morphic to R. Hence R/X is faithful . Therefore, X = 0 . Consequently,R is directly finite . Clearly every faithful cyclic module is isomorphic toR.(iii) ---> (i) . Let X be a right ideal containing no nonzero ideals . Then R/Xis isomorphic to R. Hence R=X® S where S is a right ideal . Since Ris directly finite, !I' = 0 . Consequently, R is strongly right bounded .Lemma 2. Let R be a strongly right bounded ring .(i) Every nonzero right ideal is an essential extension of an ideal of R.(ii) R is right nonsingular if and only if R is semiprime if and only if R isreduced (i . e., R has no nonzero nilpotent elements).Proof. Part (i) is in [16, Note 1.3D] . Part (ii) is in [4, Proposition 1] .Proposition 3 . Le¡ R be a strongly right bounded ring . Then the followingconditions are equivalent :(i) R is quasi-Baer .(ii) R is semiprime right quasi-continuous .(iii) R is semiprime right quasi-FPF.Proof.: This result follows from [2, Proposition 1.2], [3, Propositions 4 and6], and Lemma 2 .The following notation will be used : if V C_ S(R, M), then Vl and VZ are thesets of first and second components of V, respectively.Lemma 4 .(i) If V is a right ideal of S(R, M), then Vi is a right ideal of R, V2 is aright R-submodule of M, and {0} x ViM is a right S(R, M)-submoduleof V.(ii) IfW is a right ideal of R and K is a right R-submodule of M such thatWM C K, then W x K is a right ideal of S(R, M) .(iii) Let V C S(R, M) . Then [IR(V1) n IR(V2)] x IM(V,) C IS(R,M)(V) .(iv)	The right ideal {0} x M is right essential in S(R, M) if and only if Mis left faithful (i.e ., IR(M) = 0)-(v) If V and W are right ideals of S(R, M) such that V n W = 0, thenVlMnWlM=0.(vi) Let S(R, M)) be strongly right bounded where M is an ideal of R. ThenR is strongly right bounded and if IR(M) 7~ 0, then IR(M)n rR(M) 0-(vi¡) Let M be a module such that whenever An B = 0, then AM nBM = 0where A and B are right ideals of R (e.g ., M is an ideal) . If S(R, M)satisfies the ILAS condition, then R satisfies ¡he ILAS condition.(viii) Le¡ M be and ideal of R. Then S(R, M) is right uniform if and only ifR is right uniform and M is left faithful .Proof.SPLIT-NULL EXTENSIONS 39(i) Clearly Vi is a right ideal of R and V2 is a right R-submodule of M.Let w E Vl and m E M. There exists k E V2 such that (w, k) E V.Then (w, k)(0, m) = (0, wm) E V. Thus {0} x ViM is a right S(R, M)-submodule of V.(ii) and (iii) are straightforward .(iv) Suppose {0} x M is right essential in S(R, M) and 0 7~ t E IR(M) .There exists (w, m) E S(R, M) such that 0 :~ (t, 0)(w,m) E {0} xM. Contradiction! Hence All is left faithful . Conversely, let (w, m) ES(R, M). If w == 0, we are finished. So assume w :~ 0. There existsk E Msuch that 0 74 (w, ?n)(0, k) = (0, wk) E {0} x 1VI . Hence {0} x Mis right essential in S(R, M) .(v) Assume vm = wk E VlMn 1471M where v E Vi, w E W1 , and m, k E M.There exists x E V2 and y E W2 such that (v, x) E V and (w, y) E W.Consider (v, x)(0, m) = (0, vm) = (0, wk) = (w, y)(0, k) E V n W = 0.Therefore, VlMnWlM = 0.(vi) Let Y be a nonzero right ideal of R. There exists an ideal J of S(R, M)such that J is essential in Y x YM. Since Jl and J2 cannot both bezero, Y contains a nonzero ideal. Hence R is strongly right bounded.If IR(M) jÉ 0, then there exists a nonzero ideal H C_ IR(M) x {0} .Hence Hl is a nonzero ideal of R and ({0} x M)H = {0} x MHl C_ H.Therefore, 0 7É Hl C_ IR(M) n rR(M).(vii) Let A and B be right ideals of R such that AnB = 0. Let A* = A xAMand B* = B x BM. Hence A* n B* = 0. Now IS(R,M)(A*) = IR(A) x[m(A) and IS(R,M)(B*) = IR(B) x Im(B) . Consequently, IR(A)R +40	G.F . BIRKENMEIERCR(B)R = R.(vi¡¡) Assume S(R, M) is right uniform and let Y be a nonzero right idealof R . By part (iv) M is left faithful . Let 0 qÉ w E R. There exists(t, m) E S(R, M) such that 0 :~ (w, 0)(t, m) = (wt, wm) E Y x YM.Therefore, R is right uniform. Conversely, let V be a nonzero right idealof S(R, M) and 0 qÉ (t, m) E S(R, M). By part (iv) 0 q¿ V fl ({0} x M) ={0} x V2 is essential in V . If t qÉ 0, there exists y E R such that0 ty E V2 . Since M is left faithful, there exists k E M such that0 tyk E V2 . Thus 0 7É (t, m)(0, yk) = (0, tyk) E {0} x V2 . If t = 0,then m 7É 0 and there exists q E R such that 0 , mq E V2 . Thus0 :~ (t, m)(q, 0) = (0, mq) E {0} x V2 . Consequently, in all cases {0} x V2is right essential in S,(R, M). Therefore, S(R, AJ) is right uniform .We note that if R is commutative and M is and ideal of R, then S(R, M) iscommutative . However, in Example 9 we shall provide a strongly right boundedring Ti and an ideal (T, 0) such that S(TI , (T, 0)) is not strongly right bounded .Also in [9, Example 2 .2] the ring R is a strongly right bounded ring ; however,from Lemma 4 (vi), S(R, R(x i , 0)R) is not strongly right bounded . Thus itis natural to investigate conditions on R and M which insure that S(R, M) isstrongly right bounded . We say M is a strongly right bounded module if everynonzero right R -submodule contains a nonzero (R, R)-bisubmodule of M .Theorem 5. LeíR be a strongly right bounded ring . If either of the followingconditions is satisfied, then S(R, M) is a strongly right bounded ring .(i) M is a strongly right bounded module such that CR(M) contains no non-zero nilpotent ideals ofR and IR(M) C rR(M) .(ii) M is an ideal of R such that CR(M) fl M = 0 .Proof- Let V be a nonzero right ideal of S(R, M). If Vi = 0 or V f1 ({0} xM) :jÉ 0, then there exists a nonzero (R, R)-bisubmodule K C_ V2 such that{0} x K C_ V is an ideal of S(R, M). So assume Vi :~ 0 and V f1({0} x M) = 0 .Let D be a nonzero ideal of R such that D C_ VI . Note that with either condition(i) or (ii), Vi M = 0 = MVI . If condition (i) is satisfied, then V2 = V2 x {0} q¿ 0 .Hence D2 x {0} C V is a nonzero ideal of S(R, M). Now assume condition (ii)is satisfied . If V2 = 0, then D x {0} C_ V is a nonzero ideal of S(R, M). IfV2 7~ 0, then V2M qÉ 0 . But V(M x {0}) = {0} x V2M C V f1 ({0} x M) = 0 .Contradiction! Therefore, in all cases V contains a nonzero ideal of S(R, M).Consequently, S(R, M) is strongly right bounded .We note that when M is an ideal of R, then S(R, M) is isomorphic to asubring of T2(R) (i .e ., the 2 x 2 lower triangular matrix ring over R) . However,from [4, Proposition 10], T (R) is never strongly right bounded for n > 1 .Corollary 6 . Leí M be an ideal of `R . Then S(R, M) is strongly rightbounded right uniform if and only if R is strongly right bounded right uniformand M is left faithful .Proof. This result follows from Theorem 5 and Lemma 4 (vi¡¡) .Thus, if R is a strongly right bounded domain and M is any ideal of R, thenS(R, M) is a strongly right bounded right uniform ring . The ring H[x] whereH denotes the real quaternions provides an example of a strongly boundeddomain which is neither left nor right duo .Proposition 7. Let M be a left faithful ideal of R.	Then the followingequivalentes are true :(i) Every ideal of R is right essential in a (ring) direct summand of R ifand only if every ideal of S(R, M) i,s right essential in a (ring) directsummand of S(R, M) .(ii) Every right ideal is right essential ira a ring direct summand of R if andonly if the same is true for S(R, M).Proof.SPLIT-NULL EXTENSIONS 41(i) Let S denote S(R, M) and assume every ideal of R is right essential in adirect summand of R. Let Y be an ideal of S and V = Y fl {0} xM. ByLemma 4 (iv), V is right essential in Y, V = {0} x V2, and V2 is an ideal ofR. Hence there exists a (central) idempotent e E R such that V2 is rightessential in eR . Consider (e, 0)S . Let (x, m) E S; then (e, 0)(x, m) =(ex, em). Suppose 0 :~ (ex, em) . If ex :~ 0, then there exists t E Rsuch that 0 :~ ext E V2 . Hence 0 7É (ex, em)(0, t) = (0, ext) E V . Ifex = 0, then there exists w E R such that 0 :~ emw E V2 . Hence0 :~ (ex, em)(w, 0) = (0, emw) E V. Therefore, in all cases, V is rightessential in (e, 0)S . Hence Y is right essential in (e, 0)S . Consequently,every ideal of S(R, M) is right essential in a (ring) direct summand ofS(R, M) .Conversely, suppose every ideal of S is right essential in a (ring) directsummand of S . Let K be an ideal of R. Then there exists a (central)idempotent (e, m) E S such that {0} x KM is right essential in (e, m)S .Note that eme = 0 . Hence (e, m) is central in S if and only if e is centralin R and M = 0. Now {0} x KM C_ (e, m)({0} x M) C (e, m)S . HenceKM is right essential in eM and eM is right essential in eR because Mis left faithful in R. Since K is an ideal and KM is right essential in K,then K is right essential in eR.(ii) This part is proved in a manner similar to that of part (i) .In [15] Faith characterizes when S(R, M) is FPF where R is commutativeand M is faithful . He poses this characterization as an open problem when Ris noncommutative . The following result partially generalizes Faith's result .Corollary 8. Le¡ R be a semiprime or a right nonsingular ring and M bea left faithful ideal of R. Then ¡he following conditions are equivalent :(i) R is strongly right bounded and right quasi-continuous .42	G.F . BIRKENMEIER(ii) S(R, M) is strongly right bounded and right quasi-continuous .(iii) S(R, M) is strongly right bounded and right quasi-FPF .Proof.(i) -> (ii) By Lemma 2, R is reduced . Hence every idempotent of R iscentral . Thus every idempotent of S(R, M) is central . By Theorem 5and Proposition 7, S(R, M) is strongly right bounded and right quasi-continuous .(ii) + (iii) By Lemma 4 (vi) and Lemma 2, R is reduced . Hence everyidempotent of S(R, M) is central . By [3, Proposition 6], S(R, M) isright quasi-FPF .(iii) -> (i) By Lemma 4 (vi) and Lemma 2, R is reduced strongly rightbounded ring . By Lemma 4 (vi¡), R satisfies the ILAS condition . From[1, Lemma 2.2] and Proposition 3, R is right quasi-continuous .When R is quasi-Baer strongly right bounded and M is a,left faithfulideal of R, the sequence of embeddingsR -> S(R, M) -> T2 (R)is interesting in that S(R, M) is strongly right bounded (and right quasi-continuous) but not quasi-Baer (cf ., Proposition 3) and T2(R) is quasi-Baer [23] but not strongly right bounded .The following example is a special case of a general procedure indicated inExample 9 . Let I denote the ring of integers and T the semigroup ringof A over IZ (Le ., integers modulo 2) where A is the semigroup on the set{a, b} satisfying the relation xy = y for x, y E A . Thus T = {0, a, b, a -f- b} .Let TI denote the Dorroh extension of T (i .e ., the ring with unity formedfrom T x I with componentwise addition and with multiplication given by(x, k)(y, n) = (xy + nx + ky, kn)) . T1 has the following properties :(i) The set of nilpotent elements of Ti, N(T1 ) = {(0, 0), (a + b, 0)}, is theJacobson radical and equals the right socle of TI .(ii) Every nonzero right ideal of T1 contains either N(T1 ) or a nonzero idealof the form (0,2kI) _ «0, 2k%*) E Ti ¡ k is a fixed integer and i E I} .Therefore, TI is strongly right bounded .(iii) Ti is not right duo since (a, 1)T1 is not an ideal .(iv) Ti is not strongly left bounded .(v) Ti does not satisfy the ILAS condition since IT,(N(T1 )) -{- ITl ((a +b, 2)T1 )T1 Ti . However if {X2} is a nonempty set of ideals of Ti suchthat nX; = 0 then R = EIT,(X;) . Thus TI satisfies the ILAS conditiondefined in [1] .(vi) Ti is not right CS, since (a+ b, 2)Tl is not essential in a direct summand .However, every ideal is right essential in a direct summand of Tl .(vi¡) S(I, N(Tj » (i .e ., split-null extension) is ring isomorphic to the subring(0, I) + N(T1 ) of Ti . S(I, N(Tj )) provides an example for Theorem 5SPLIT-NULL EXTENSIONS	43(vi¡¡) S(T1 , (0, k2I» provides an example for Theorem 5 (ii) .(ix) S(TI , (T, 0)) is an example of a split-null extension of a strongly rightbounded ring which is not strongly right bounded (cf. Theorem 5) .To see this observe ((a, l), (0, 0»S(Ti, (T, 0)) = {((ka, k), (0, 0))¡k E I}contains no nonzero ideals since ((b, 0), (0, 0»((ka, k), (0, 0)) = ((k(a +b), 0), (0, 0)) .References1. G . F . BIRKENMEIER, A generalization of FPF rings, Comm. Algebra 17(1989),855-884 .2 . G . F . BIRKENMEIER, A decomposition of rings generated by faithful cyclicmodules, Canad. Math . Bull . (to appear) .3 . G . F . BIRKENMEIER, Rings generated by faithful direct sums, Tam- kangJ. Math . (to appear) .4 . G .F . BIRKENMEIER, R.P . TUCCI, Homomorphic images and the sin-gular ideal of a strongly right bounded ring, Comm. Algebra 16 (1988),1099-1112 .5 . G . F . BIRKENMEIER, H . E . H EATHERLY, Embeddings of strongly rightbounded rings and algebras, Comm. Algebra 17 (1989), 573-586 .6 . A.W . CHATTERS, C .R . HAJARNAVIS, Rings in which every comple-rrient right ideal is a direct summand, Quart. J. Math . Oxford 28 (1977),61-80 .7 . W . E . CLARK, Twisted matrix units semigroup algebras, Duke Math . J.34 (1967), 417-424 .8 . R. C . COURTER, Maximal duo algebras of matrices, J. Algebra 63 (1980),444-458 .9 . M . G . DESHPANDE, Structure of right subdirectly irreducible rings I, J.Algebra 17 (1971), 317-325 .10 . M . G . DESHPANDE, V. K . DESHPANDE, Rings whose proper homomor-phic images are right subdirectly irreducible, Pac. J. Math . 52 (1974),45---51 .11 . C . FAITH, "Injective quotient rings of commutative rings, in ModuleTheory," Springer Lecture Notes 700, 151-203, Springer-Verlag, Berlin,1979 .12 . C . FAITH, Self-injective rings, Proc . Amer. Math . Soc. 77 (1979), 157-164 .44	G.F . BIRKENMEIER13 . C . FAITH, "Algebra I. Rings, Modules and Categories," Grundlehrender Mathematischen Wissenschaften 190, Berlin and New York : Sprin-ger-Verlag, 1981 .14 . C . FAITH, "Injective Modules and Injective Quotient Rings," Lecture No-tes in Pure and Applied Mathematics 72, New York : Marcel Dekker, 1982 .15 . C . FAITH, Commutative FPF rings arising as split-null extensions, Proc .Amer . Math . Soc. 90 (1984), 181-185 .16 . C . FAITH, S . PAGE, "FPF Ring Theory : Faithful Modules and Genera-tors of Mod-R," London Math. Soc . Lecture Notes Series 88, Cambridge :Cambridge University Press, 1984 .17 . T .G . FATICONI, Semi-perfect FPF rings and applications, J. Algebra107 (1987), 297-315 .18 . E . H . FELLER, Properties of primary noncommutative rings, Trans. Amer .Math . .Soc . 89 (1958), 79-91 .19 . V .K . GOEL, S .K . JAIN, 7r--injective modules and rings whose cyclicsare ir-injective, Comm. Algebra 6 (1978), 59-73 .20 . N . JACOBSON, "The Theory of Rings," Amer. Math. Soc . MathematyicalSurveys II . Providence : American Mathematical Society, 1943 .21 . L . JEREMY, Modules et anneaux quasi-continuous, Canad. Math . Bull.17 (1974), 217-228 .22 . Y . KITAMURA, On quotient rings of trivial extensions, Proc . Amer. Math .Soc. 88 (1983), 391-396 .23 . A . POLLINGHER, A . ZAKS, On Baer and quasi-Baer rings, Duke Math .J . 37 (1970), 127-138 .24 . M . SATYANARAYANA, M . G . DESHPANDE, Rings with unique maximalideals, Math . Nachr. 87 (1979), 213-219 .25 . B . STENSTROM, "Rings of Quotients," New York, Springer -Verlag 1975 .26 . G . T HIERRIN, On duo rings, Can. Math . Bull . 3 (1960), 167---172 .1980 Mathematics subject classifications : 16A15Department of MathematicsUniversity of Southwestern LouisianaLafayette, LA 70504U .S.A .Rebut el 12 d'Octubre de 1988

Split-null extensions of strongly right bounded rings

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