Locally Split Homomorphisms Relative To A Sub module

In this article,for a positive integer n ,the concept of n-locally split homomorphism relative to a sub module has been introduced and studied,which will turn out to be most useful in the studying and providing characterizations of local projectivity, local-regularity in the sense of (Zelmanowitz, Field house and Ware) relative to a sub module.They present generalization of projectivity and the three types of regularity which have been mentioned respectivily.

A right R-module M is Zelmanowitz-regular, if for each x M, there exists an R-homomorphism M* = HomR(M,R) satisfies x = x (x).The Field house-regular module Was defined as one whose sub modules are pure, while Ware-module was defined as a projective module in which every principal sub module is direct summand.The concept of locally split homomorphisms was introduced in [3].Let M and N be R-modules, and : N M an Rhomomorphism. is called locally split, if for each xo (N), there is an R-homomorphism : M N such that (xo)) = xo.This concept had been utilized to characterize Zelmanowitz-regula modules, and modules in which every sub module is locally split, that is, the inclusion mapping i : N M is locally split for each sub modules N of M.
Many algebraic structures had been restudied relative to a class of sub modules, as semi-regular modules relative to a fully invariant sub module [2], uniform extending modules [4], quasi-injective modules relative to the closed sub modules class [7], pseudo-injective modules relative to a principal sub modules class [10].Recently, projective module relative to a sub module has been studied in [1].Let P be an R-module and T a sub module of P. P is called (T)-projective , if for every R-epimorphismf : A B and R-homomorphism g : P B, there exists an R-homomorphism h : P A such that foh(x) g(x) g(T) for all x P.
In section two of this work we introduce the concept of n-locally split Rhomomorphism relative to a sub module.Several properties have been given and considered modules in which the inclusion mapping of every sub module is locally split with respect to a sub module.In section three, we utilize locally split homomorphism relative to a sub module to characterize locally(T)-projective modules which homomorphism s : M N such that s(xi) xi T for each i = 1, 2, ..., n. the notion of n-locally (0)-split sub moduleswas introduced by Ramamurthi and Rangaswany [6] by the name of strongly pure sub modules.
Recall that a sub module N of an R-module M is fully invariant if (N) N for each R-endomorphism of M. It is known that an R-homomorphism : A M is n-locally (0)-split if and only if it is 1-locally(0)-split [3].Relative to a non-zero sub module we have the following: Proposition 2.2: Let A and B be R-modules and T a fully invariant sub moduleof B.Then an R-homomorphism : A B is n-locally (T)-split if and only if it is1-locally(T)-split.
Proof: The "only if" part is clear for any arbitrary sub module T of B. We shall use induction to prove the "if" part.Suppose that our statement is true Furthermore, for i = 1, 2, ..., n, we have xi .This shows that,xi (xi))_T for each i = 1, 2, ..., n and hence is n-locally(T)-split.
According to the above proposition, all results follow will doing either in thesense of (1 locally) or (n locally) concept for arbitrary sub module,these results will be true in the sense of (n locally) or (1 locally) concept respectively for fully invariant sub module, unless otherwise mentioned.B be an R-homomorphism, h be the Repimorphismfrom A onto h(A) and T be a fully invariant sub module of h(A).Then the following are equivalent: (1) h is n-locally (T)-split, (2) h is n-locally (T)-split and h(A) is n-locally (T)-split sub module in B.
The following corollary follows directly from proposition (2.3) and proposition (2.2).
Let M be an R-module and T a sub module of M. A sub module N of M is called (T)-pure, if MA N = NA + T (MA N) for each right ideal A of R.This is equivalent to saying that, for every finite sets {mi} M,{nj} N and {rij} R with nj = m i r ij n i=1 ,j = N T for each j = 1, 2, ...,m [1].In the Z-module Z,2Z is (6Z)-pure sub module in Z which is not pure.
Proposition 2.5: Let M be an R-module and T a sub module of M. Then (1) Every 1 locally ( hence n locally) (T)-split sub module in M is (T)pure. ( This shows that N is (T)-pure.For the second statement, let h : M M be an R-homomorphism and nj = x i r ij m i=1 Ker(h).Then h(x i )r ij m i=1 = 0.
Since h(xi) h(M) = M there is an R-homomorphism q : M M such that h(q(h(xi))) h(xi) = h(ti) for some ti T and each i = 1, 2, ...,m, and hence ti+xi q(h(xi)) T .This shows that Ker(h) is (T)-split in M.
Corollary 2.6:Let h : A B be an R-homomorphism and T be a fully invariant sub module of h(A).If h is 1 locally(T)-split homomorphism, then h(A) is a (T)-pure sub module of B.
We call an R-module M, n (T) regular, if each sub module of M is locally(T)-split, where T is a sub module of M.
It is clear that, if M is n (T) regularR module, then it is k (T) regularfor each k n, in particular, every n-(T)-regular R-module is 1-(T)-regular.Recall that a sub module N of an R-module M is (T)-direct summand in M, ifthere exists a sub module K of M such that M = N + K and K N T, where T is a sub module of M [1].Let Q be the group of rational numbers and p be a prime number.Consider the two subgroups of Q, Qp ={a/b Q : b is relatively prime to p } and Qp={a/pn Q: n is non-negative integer } .

Locally Split Homomorphisms Relative To A Sub module
] 28 [ Proposition 2.8: Let M be an R-module and T a sub module of M.Then This shows that M = N + s 1(T N), and it is easy to check that N 1(T N) T. Thus N is(T)-direct summand.
(2) Let N be m-generated sub module of M. Without loss of generality, we canassume that m > n.As T fully invariant, then M is m (T) regular andhence by (1), N is (T)-direct summand.
Recall that a sub module N of an R-module M is (T)-direct summand in M, ifthere exists a sub module K of M such that M = N + K and K N T, whereT is a sub module of M [1].Let Q be the group of rational numbers and p be aprime number.Consider the two subgroups of Q, Qp ={a/b Q : b is relatively prime to p } and Qp={a/pn Q: n is non-negative integer } .

T h e n , i t i s k n o w n t h a
Proposition 2.9: Let M be an R-module and T a sub module of M.Then

N-Locally Projective Modules Relative To A Sub module Locally Split Homomorphisms Relative To A Sub module
] 30 [ Projectivity relative to a sub module had been studied in [1].Let M be anRmodule and T a sub module of M. M is called (T)-projective, if for each Repimorphism : A B and R-homomorphism : M B, there exists an R-homomorphism : M A such that (x) (x) (T) for each x in M.
In this section, we consider the local property of (T)-projective modules.
We characterize these modules by means of locally homomorphisms relative to a sub module.First we introduce the following: .This shows that M is n-locally (T)-projective.
We call the third statement of the above theorem, the dual basis lemma forn-locally(T)-projectiive modules.the original statements in the theorem are equivalent to the respective statements for fully invariant sub modules, more precisely, every n-locally(T)-projective module is k-locally(T)projective for each k n.Also,if M is 1-locally(T)-projective module and T is fully invariant in M, then M is n-locally(T)-projective.
Corollary 3.3: The following statements are equivalent for an R-module M and a sub module T of M: (1) M is n-locally(resp.1-locally)(T)-projective, (2) For each free R-module F, each R-epimorphism : F M is nlocally(resp.1-locally)(T)-split.
Let{a i } I be a generated set of A and let F be a free R-module with basis (2) As an application of theorem(3.2),it is easy to see the following: If R is a commutative ring and Mi is 1-locally(Ti)-projective R-module,i = 1, 2.
In particular, tensor product of projective module M with 1-locally(T)projective module is 1-locally(M T)-projective.
(3) the following statement is obvious.It follows directly by the dual basis lemma for 1-locally(T)-projective modules.Let R be a commutative ring.

Locally Split Homomorphisms Relative To A Sub module
] 34 [ subset of R, then S 1 M i s 1 -locally(S 1T)-projective S 1R-module, inparticular MP is 1-locally(TP )projective RP-module for each prime ideal P of R.
(4) The converse of ( 3) is not true in general, as we see in the following example.Let R be a Von Neumann regular ring which has no finitely generated maximal ideal and hence has no maximal ideal which is a direct summand.Let{P } be the family of maximal ideals of R. Consider the R-module M = (R/P ).Then R P is a field and hence M P is 1locally(T)-projective R -module for each sub module (T) of M. We claim t h a t M * = 0 .F o r , i f f : R / P R , t h e n e i t h e r f = 0 o r f i s Rmonomorphism.If f is R-monomorphism, then R/P is isomorphic to an ideal W in R. As R regular, then W is a pure in R, and P = annR(W ).Proof: Let x M/T .then there exist a pair of dual basis x j j J on M such that j (x) 0 for only finitely many j J and x -x j J j (x) T .For each j J , define j : M/T R/t(T) by j (m ) = j (m) + t(T) form M where t(T) is the trace ideal of T. It is clear that j is well-defined R/t(T)homomorphism and x = x j j (x ) J . This shows that M/T is locally projective R/t(T)-module.As t(T) is two-sided ideal of R, then M/T is locally projective R-module.Conversely, let x M. Then there a pair of dual basis x j , j J on M/T such that j (x ) 0 for only finitely many j J and x = x j j (x) J .Thus x x j j J (x) T where is the natural epimorphism.This shows that M is 1-locally(T)-projective.
It is well-known that, if M is a projective R-module, then M = Mt(T), and (M) =and (t(M)) and t(M) is a pure ideal of R , provided that R is a commutative ring, where t(M) = (M), the sum runs over all M*.For 1-locally(T)-projective modules, we have the following.(2) t(M) is (t(T))-pure ideal of R. Theorem 3.7: The following statements are equivalent for an R-module M and a sub module T of M: (1) M is n-locally(T)-projective, (2) For each k-generated sub module M0 of M where k n, there exist a finitely generated free R-module F and R-homomorphisms f : M F and g : F M such that g(f(x)) -x T for each x M0.
Proof: (1) (2) : Let Qbe a free R-module having an R-epimorphism h : Q M. Then h is n-locally(T)-split.Thus we can find an R-homomorphism q : M Q such that h(q(x)) x T for all x M0.As q(M0) is a finitely generated sub module of Q, there exists a finite subset {u1, u2, ..., uk} of the free basis of Q such that q(M0) is contained in a finitely generated free sub module F = u1R + u2 + ... + ukR of Q.Since F is a direct summand of Q, then let : Q F be the natural projection of Q onto F. Put f = q : M F and g = h|F : F M. Then clearly g(f(x)) -x T for each x M0.
( It is well-known that, every projective module is isomorphic to a direct summanda free module.In [11], it was proved that, every locally projective R-module is a pure sub module of a direct product of copies of R.

for n-1 where n > 1 .
Then there exists an R-homomorphism 1 : B A such that xi 1(xi)) T f o r e a c h i = 1 , 2 , . . ., n 1 .A s x n 1(xn)) (A),there exists an Rhomomorphism 2 The following result gives a good motivaton for considering relativity in moduletheory.Proposition 2.7: Let M be an R-module and T a sub module of M. If M is 1locally ( n-locally )-(T)-regular, then M/T is regular.The converse is true if M/T is locally projective.Proof: Let N/T be a sub module of M/T and x N/T .Then there exists an Rhomomorphism : M N such that x (x) T .Hence induces a mapping : M/T N/T .This implies that (x ) = x which means that M/T is regular.Conversly, let N be a sub module of M and x1, x2, ..., xn be a finite number of elements of N. Then there is an R-homomorphism s : M/T N + T/T such that s(x i ) = x for each i = 1, 2, ..., n.Local projectivity of M/T implies that there is an R-homomorphisms : M/T N such that s (x ) = s(x i ) for each i = 1, 2, ..., n, where is the natural R-epimorphism of N onto N +T/T .Put = s : M N. Then (xi) xi T for each i = 1, 2, ..., n and hence M is n (T) regular.If M is n-(T)-regular R-module where T is a fully invariant sub module of M,then M is m-(T)-regular for each m n.Also, if M is n-(0)-regular Rmodule,then it is n-(T)-regular for each sub module T of M.

( 1 )
If M is n-(T)-regular, then every k-generated sub module of M is (T)direct summand where k n. (2) If further, T is fully invariant in M, then every finitely generated sub moduleof M is (T)-direct summand.Proof: (1) Let N = x i k i=1 R be k-generated sub module of M, for each k n.By hypothesis, there exists an R-homomorphism s : M N such thats(xi) xi_T for each i = 1, 2, ..., k, and hence s(x) x T for each x in N.For each m M, we have s(m) N and s(s(m) m) = s(s(m)) s(m) N T.

( 1 )
If M is n-(T)-regular, then every k-generated sub module of M is (T)direct summand where k n.The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 29 [ (2) If further, T is fully invariant in M, then every finitely generated sub moduleof M is (T)-direct summand.Proof: (1) Let N = x i k i=1 R be k-generated sub module of M, for each k n.By hypothesis, there exists an R-homomorphism s : M N such that s(xi) xi T for each i = 1, 2, ..., k, and hence s(x) x T for each x in N.For each m M, we have s(m) N and s(s(m) m) = s(s(m)) s(m) N T.This shows that M = N + s 1(T N), and it is easy to check that N s 1(T N) T. Thus N is(T)-direct summand.(2) Let N be m-generated sub module of M. Without loss of generality, we can assume that m > n.As T fully invariant, then M is m (T) regular and hence by (1), N is (T)-direct summand.Corollary 2.10: Let M be an R-module and T asub module of M. If M is n-(T)-regular, then J(M) T. Proof: Let x J(M).Then xR is small and (T)-direct summand of M. This implies that J(M) T. Proposition 2.11: Let M be an R-module and T a fully invariant sub module of M. If M is n-(T)-regular and S is the endomorphism ring of M, then, as an S-module, M is n-(T)-regular.Proof: We consider M a left S-module and hence (S R)-bimodule.Let N be an S-sub module of M and x0 N. Then there exists an R-homomorphism s : M N such that s(x0) -x0 T .We consider s is an element of S. Define s : M N bys (y) = s•y.It is clear that s 0 is an S-homomorphism and hence s (x0) x0 T .This shows that N is 1locally ( and hence n-locally) (T)-split.Thus M is n -(T)-regular S-module.

Definition 3 . 1 :(
Let M be an R-module, T a sub module of M and n a positive integer.M is called n-locally projective relative to T (or simply nlocally (T)-projective ), if for each R-epimorphism : A B and Rhomomorphism : M B, then for any finite number of x1, x2, ..., xn_M, there exists an R-homomorphism : M A such that (xi) (xi) (T) for each i = 1, 2, ..., n.It is clear that, every (T)-projective module is n-locally(T)-projective for each positive integer n and each sub module T. In particular, every projective module is locally projective module which introduced byZimmermann in[11].Also, it is clear that n-locally (T)-projective module is (T)-projective if it is finitely generated by n elements.The Z-module Q is not 1-locally (Z)-projective, if not, let x_Q which is not in Z.Assume F is a free Z-module having a Z-epimorphism : F Q, then there exists a Z-homomorphism f : Q F such that f(x) -x Z.But HomZ(Q,Z) = 0and hence HomZ(Q, F) = 0.This implies that x Z which contradicts the choiceof x.More generally, KR is not 1-locally (R)projective where R is a domain and K is the field of quotients of R as Rmodule.The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 31 [ In the following , we give characterizations of n-locally (T)-projective modules in terms of n-locally (T)-split homeomorphisms.Theorem 3.2: The following are equivalent for an R-module M and a sub module T of M (1) M is n-locally (resp.1-locally) (T)-projective, (2) Every R-epimorphism into M (from any R-module) is n-locally(resp.1locally)(T)-split, (3) For any finite number of x1, x2, .., xn (resp.x)M, there exist families m j J M and j J M* such that for each i = 1, 2, ..., n (a) j (xi)(resp.(x)) 0 for only finitely many j J : Let A be any R-module and : A M be an R-epimorphism .Then there exists an R-homomorphism : M A such that (xi) -xi T for each i = 1, 2, ..., n.This shows that is n-locally (T)-split.= R for every j J, by f((rj)) = m j J rj .Clearly, f isan R-homomorphism.By (2), there is an R-homomorphism : M Rj such that f( (xi))-xi T for each i = 1, 2, ..., n.Then for each j J, there is j : M Rj such that (m) = ( j (m)) for each m M, in particular, (xi) = ( j(xi)) for each i = 1, 2, ..., n.Thus xi m j J j (xi) = xi f( j (xi)) = xi f( j (xi)) T .Let J0 = {j J : j (xi) 0}.Then J0 is a finite subset of J and xi m j J j (xi) Tfor each i = 1, 2, ..., n. (3) (1) : Let : A B be an R-epimorphism and : M B be an Rhomomorphism.By (3), for each finite number of elements x1, x2, ..., xn of Locally Split Homomorphisms Relative To A Sub module ] 32 [M, there are families m j J M and j J M* such that j (xi) 0 for finitely many j J and xi m j (xi) T for each i = 1, 2, ..., n.Since (mj) B, there exists aj A such that (aj) = (mj) for each j J {z i } I .Define : F A by (zi) = ai.Clearly is an R-epimorphism.By (2), is n-locally (T)-split, that is for any finite number of x1, x2, .., xn M, there exists an R homomorphism : M F such that ( (xi)) xi T for each i = 1, 2, ..., n.Put = , then : M A and satisfies The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 33 [(xi)) xi T for each i = 1, 2, ..., n.This shows that is n-locally(T)-split and hence by theorem(3.2),M is n-locally(T)-projective.Remark 3.4: (1) Let K be an R-module which is not (T)-projective for some sub module T of K ( Q is not (Z)-projective Z-module), and M = K H where H = R is a direct sum of countable number of copies of R.Since every module is projective relative to itself, then H is (H)propositon(3.7))impliesthat K is (T)-proective.We claim that M is 1 locally(T H) projective .Let {x } be a basis for H. Then T {x i } N is a generated set of M. For each j N , d e f i n e f j : H R b y f (j )( x i ) extended ( by linearity ) to all H, therefore, ifx = x in i=1 r i , then fj(x) = rj .Again fj can be extended to an R-homomorphism gj : M R by putting gj(t) = 0 for all t T .Then {gj} M*.Let m M. It is clear that gj(m) 0 for only finitely many j N and m = t + x where t T and x H. Then we have m x gj(m) = m x fj(m) = m -x T .By dual basis lemma for 1-locally(T)-projective modules we have M is 1-locally(T)projective and hence 1-locally( T H )-projective.

( 5 )
Maximality of P_ implies that P +W = R and hence P is a direct summand of R, which contradicts the choice of R. Thus M* = 0. Let K be a proper sub module of M and m M/K.By the above, M P i s 1 locally(KP ) projective R P module.If M is 1-locally(K)projective R-module, then by the dual basis lemma for 1-locally(K)projective modules we have m K which is a contradiction.IfM is 1-locally (and hence n-locally)(T)-projective R-module, then J(M) MJ(R)+T(direct application of dual basis lemma for 1-locally(T)projective modules).Further, if T a small sub module of M, then J(M) = MJ(R)+T (6) The following result gives a motivation for studying n-locally(T)projective modules: Let M be an R-module and T a sub module of M. Then M is 1-locally(resp.n-locally)(T)-projective R-module if and only if M/T is locally projective R-module.The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 35 [

) ( 1 )
: Consider a finite number of x1, x2, ..., xn_M and let N be the The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 37 [sub module of M generated by these elements.By the hypothesis, there exist a finitely generated free R-module F and R-homomorphisms f : M F, g : F M such that g(f(x)) -x T for each x_N,in particular g(f(xi))xi T for each i = 1, 2, ..., n.Let {u1, u2, .., uk} be a free basis for F. For each j = 1, 2, .., k we define an R-homomorphism : for each m M. Let yj = g(uj) for each j = 1, 2, ..., k.Then for each i = 1, 2, ..., n we have y ) xi = g(f(xi)) xi T .Theorem(3.2) implies that M is nlocally(T)-projective._ Corollary 3.8: If M is n-locally(T)projective R-module.then for every kgenerated sub module N of M where k n, there exists s EndR(M)suchthat s(x) -x T for each x N. The last corollary suggests a weak concept of n-locally(T)-split sub modules.LetN be a sub module ofM.N is called weak n-locally(T)-split if for each finite number of x1, x2, ..., xn N,there exists an R-endomorphism s of M such that s(xi) xi T for each i=1,2,...,n.It is clear that nlocally(T)-split sub modules are weak n-locally(T)-split.The converse is not true.Thus, in n-(T)-regular modules, every sub module is weak n-locally(T)-split, while in n-locally(T)-projective modules, every kgenerated sub module is weak n-locally(T)-split where k n.We have mentioned before that every k-generated n-locally(T)-projective module where k n is (T)-projective, for countably generated modules we have the following: Theorem 3.9: Let M be an R-module and T a fully invariant sub module of M. If M is countably generated n-locally( T )-projective module, then it is (T)-projective.Locally Split Homomorphisms Relative To A Sub module ] 38 [ Proof:Let {x1, x2, x3, ...} be a countably generated set of M. Let M1 = x1R.Then theorem (3.7) implies that there are a finitely generated free R-module F1 and R-homomorphisms f1 : M F1, g1 : F1 M such that g1(f1(x)) -x T for each x M1.Let M2 = g1(F1) + x2R.Since M2 is finitely generated, again by theorem (3.7), there exist a finitely generated free R-module F2 and R-homomorphismsf2 : M F2, g2 : F2 M such that g2(f2(x)) -x T for each x M2.Observethat g1(F1) g2(F2) + T and x2 g2(F2) + T. In this manner, for each n > 1, wecan find a finitely generated free R-module Fn and R-homomorphismsfn : M Fn, gn : Fn M such that gn(fn(x)) -x T for each x Mn = gn 1(Fn 1) + xnR.This is equivalent to saying that gn(fn(gn 1(y))) gn 1(y) T for each y Fn 1 and gn(fn(xn)) xn T and hence gn 1(Fn 1) gn(Fn) + T and xn gn(Fn) + T. Thus we have an ascending chain g1(F1) g2(F2) + T g3(F3) + T ... of sub modules of M whose union is equal to M. Let sn = gn fn : M gn(Fn) for each n .Then sn EndR(M) satisfying that sn _ gn 1(m) gn 1(m) T and hence sn sn 1(m) sn 1(m) T for each m_M and n > 1.Thus sn gr(m) -m T and sn sr(m) sr(m) T for each m M, whenever r < n, because gr(Fr) + T gn 1(Fn 1) + T and so sn(gr(y)) gr(y) T for every y Fr.We shall convene that for any two Rhomomorphisms and , = modulo T means (x) (x) T for each x in their common domain.Let F = Fn.Then F is a countably generated free module.Define g : F M by g((wn)n) = g n (wn), n=1,2,3,... .Thus g(F) = g n ( F n ) + T = M a n d h e n c e g i s a n Repimorphism.We claim that g is (T)-split R-homomorphism, that is, there exists an R-homomorphism f : M F such that g f(x) x T for all x M. The 6 th Scientific Conference of the College of Computer Sciences & Mathematics ] 39 [ Now, let qn : Fn F be the canonical injection for each n.Then g qn = gn.We shall construct an R-homomorphism hn : Fn F such that g hn = gn modulo T and hn fn gn 1 = hn+1 fn+1 gn 1 modulo T if n > 1 by induction on n.Let h1 = q1.Then g h1 = g1.Suppose n > 1 and thereis given an R-homomorphism hn : Fn F such that g hn = gn modulo T andhn fn gn 1 = hn+1 fn+1 gn 1 modulo T. We define hn+1 = (hn fn+gn+2 fn+2 (1 sn)) gn+1.Then we have g h n + 1 = ( g hn fn+gn+2 fn+2 (1 sn)) gn+1= (gn fn +gn+2 fn+2 (1 sn)) gn+1 modulo T = (sn +sn+2 (1 sn)) gn+1modulo T = (sn + sn+1 sn+2 sn) gn+1 modulo T = sn+2 gn+1 modulo T = gn+1 modulo T. On the other hand, we have hn+1 fn+1 gn 1 = (hn fn + gn+2 fn+2 (1 sn)) gn+1 fn+1 gn 1 = (hn fn + gn+2 fn+2 (1 sn)) gn modulo T = hn fn gn 1 + gn+2 fn+1 gn 1 gn+2 fn+2 gn 1 modulo T= hn fn gn 1modulo T. Thus we get a desired sequence of R-homomorphismshn.Let x M. Then there exists n > 0 such that x gn+1(Fn 1) + T, that is, x = gn 1(y)+t for some t T a n d y Fn 1 .T h e n w e have hn(fn(x)) = hn(fn(gn 1(y)+ t)) = hn+1(fn+1(gn 1(y) + t)) + t1 = hn+1(fn+1(x)) + t1 for some t1 T .Moreover, since x gn(Fn) + T, in this case, by replacing n by n+1 we should have hn+1(fn+1(x)) = hn+2(fn+2(x)) + t1.Continuing in this way, we confirm that hn(fn(x)) = hm(fm(x)) + t1 for every m >n.This shows that hn(fn(x)) is independent of the choice of n so for as x in gn 1(Fn 1).Define f(x) = hn(fn(x)) for each x M , w e h av e a n Rhomomorphism f : M F, which satisfies g(f(x)) x = gn(fn(x)) -x T (since x gn 1(Fn 1)).Finally, the R-module F is projective and hence F is (f(T)) projective.Then f(m) F .By dual basis lemma for (f(T))projective modules ([1],theorem(3.8))there exist two families w j J F Locally Split Homomorphisms Relative To A Sub module ] 40 [ and j J F* such that (f(m)) 0 for only finitely many j J andf(m) w j j j (f(m)) f(T).But m = g(f(m)) + v for some v T. Thus m (m) +t1 for some t, t1 T .Thus the two families g(w j ) J and j f J satisfy the dual basis lemma for (T)-projective modules and hence M is (T)-projective.
For nlocally (T)-projective modules we have the following.Proposition 3.10: Every n-locally (T)-projective R-module is isomorphic to a (t(T)I )-pure sub module of RI Proof: Let M be n-locally (T)-projective R-module and denote M* as a family(f i ) I .Define : M RI by (m) = (f i (m)) I for m M. Clearly is an R-homomorphism and M is isomorphic to (M).We claim that (M) is (t(T)I ) -pure sub module of RI .Consider a system of equations (mk) =(f i (m k )) I = r l L s lk where rl = (r li ) I RI , slk R, L is a finite set and k K (finite set).Theorem (3.2) implies that there exist a finite subset J I and a family{x j } J M such that mk = x j J fj(mk) + tkwhere tk T for all k K. (mk) =(f i ( x j J f j (m k )+t k ) (f i (t k )) I = f i ( x j r lj )s lk j l + (f i (t k )) I for all i_I, k_K.But (f i (t k )) I t(T)I .