Characterization of the inessential endomorphisms in the category of Abelian groups

Publ. Mat. 47 (2003), 359–372 CHARACTERIZATION OF THE INESSENTIAL ENDOMORPHISMS IN THE CATEGORY OF ABELIAN GROUPS S. Abdelalim and H. Essannouni∗ Abstract An endomorphism fof an Abelian group Ais said to be inessential (in the category of Abelian groups) if it can be extended to an endomorphism of any Abelian group which contains Aas a subgroup. In this paper we show that fis as above if and only if (f−vidA)(A)iscontained in the maximal divisible subgroup of Afor some v∈Z. 1. Introduction Throughout this paper, we will follow the terminology of [2]. Let M be an object of a category Cand f∈End(M), fis called inessential (in C)iffor any monomorphism σ:M→Nthere exists ˜ f∈End(N) such that ˜ fσ =σf,inother words the following diagram Mσ −−−−→N f     ˜ f Mσ −−−−→N commutes. Ines(M) denotes all the inessential endomorphisms of M.Mis called rigid if End(M)=Ines(M). For a concrete category C, the characterization of the inessential endomorphisms is one of the problems raised in [2]. In this paper, we take C=Ab the category of the Abelian groups and we show for an Abelian group A, and an endomorphism fof A, that fis inessential (in Ab)ifand only if there exists v∈Zsuch that 2000 Mathematics Subject Classification. 20K30. Key words. Category, inessential, monomorphism, rigid, divisible, extension. ∗The second author is partially supported by the Minister´ıo de Cienc´ıa y Tecnolog´ıa. Proyecto BEM 2001–2335, and he thanks Professor A. Kaidi for the helpful conversations during his visit to the University of Almer´ıa. 360 S. Abdelalim, H. Essannouni (f−vidA)(A)⊆D, where Dis the maximal divisible subgroup of A.In particular if Ais reduced then Ines(A)=ZidA. The proof of this result uses the properties of the endomorphisms of some extensions of certain direct sums of torsion cyclic groups. From now on, the word group means Abelian group and we adopt the notations of [3]. 2. Some constructions Construction 1. Let (αn)n≥0beasequence of natural numbers such that αn<α n+1 and 2αn+1 −αn+n+3≤αn+2,∀n∈N.Ifweput θn=αn−αn−1−nfor n≥1 then we have θn−θn−1≥n,n≥2. Let p∈N∗and (tn,m)n≥mbeaset of nonzero natural numbers, relatively prime with psuch that ti,jtj,k =ti,k if i≥j≥k. We consider the direct product  n≥1 xnwith o(xn)=pαnand denote by ϕk: n≥1 xn→xkthe canonical projection. For m≥1, we define the element gmof  n≥1 xnby ϕn(gm)=0ifn<m pαn−αmxnif n≥m. We directly check that o(gm)=pαm,xm=gm−pαm+1−αmgm+1 and {gm/m ≥1} = m≥1 gm. Let m∈N∗and ξa function from Ninto {0,1},wedefine the element S(m, ξ)of  n≥1 xnby ϕn(S(m, ξ)) = 0ifn<m ξ(n)tn,mpn−m+αn−1xnif n≥m. We have S(m, ξ)=r  n=m ξ(n)tn,mpn−m+αn−1xn+tr+1,mpr+1−mS(r+1,ξ) if r≥m. Let K1be the subgroup of  n≥1 xngenerated by {gm/m ≥1}∪{S(m, ξ)/m ≥1,ξ∈{0,1}N}. Inessential Endomorphisms in the Abelian Groups 361 Lemma 2.1. The direct sum  n≥1 xnis a subgroup of K1and for all λ∈End(K1)there exist s, N ∈Nand v∈Zsuch that ts,1pαn−nλ(xn)= pαn−nvxn,∀n≥N. Proof: Let λ∈End(K1). Let us show at first that there exists N0≥1 such that if n>m≥N0then ϕn(pαm−mλ(xm)) = 0. If not, we can find a sequence (mk)k≥1such that for all k≥1, there exists nk>m kwith ϕnk(pαmk−mkλ(xmk)) =0and αnk≤mk+1. Let ζ:N→{0,1}be the function defined by ζ(n)=1ifn∈{mk/k ≥1} and ζ(n)=0otherwise. We can write: λ(S(1,ζ)) = a  i=1 cigi+ b  j=1 djS(m, ξj). If we put t=αa, then ptλ(S(1,ζ)) = ptb  j=1 djS(m, ξj). For any k,we have pθmk+1S(1,ζ)=pθmk+1 ×mk+1−1  n=1 ζ(n)tn,1pn−1+αn−1xn+tmk+1,1pmk+1 S(mk+1,ζ)∈pαnkK1 because θmk+1+n−1+αn−1≥αnif mk≥n≥1, ζ(n)=0if mk+1 >n>m kand θmk+1+mk+1 ≥αnk.Ifkis large enough, then ϕnk(pθmk+1λ(S(1,ζ))) = ϕnk pθmk+1 b  j=1 djS(m, ξj) =0 therefore pθnk−θmkdivides v(nk), where v(n)= b  j=1 djξj(n). Since the set {v(n)/n ∈N}is finite and θnk−θmk≥nk, then there exists k1≥1 such that v(nk)=0,∀k≥k1.Onthe other hand pθmk−mk+1S(1,ζ)−tmk,1pαmk−mkxmk∈pαnkK1, therefore ϕnk pθmk−mk+1 b  j=1 djS(m, ξj) =0 for klarge enough. Therefore it exists k2≥1 such that v(nk)=0, ∀k≥k2, which is absurd. Thus there exists N0∈Nsuch that: pαn−nλ(xn)= 362 S. Abdelalim, H. Essannouni pαn−nrnxn,∀n≥N0, where rn∈Z. Since T(K1)=  m≥1 gmand αk≤αn−nfor k<n, therefore pαn−nλ(gn)∈pαn−n( k≥n vgk). Let m≥N0and put for n≥m,un=lif pαn−mλ(gn)=pαn−ml  k=n tkgk with (pαn−mtlgl=0and l>n) and un=0ifpαn−mλ(gn)∈pαn−mgn. Since xn=gn−pαn+1−αngn+1,itiseasy to see that the sequence (un)n≥m is decreasing. Since for un=0wehave un>n, then there exists Mm≥msuch that un=0,∀n≥Mm. Therefore pαn−mλ(gn)∈ pαn−mgn,∀n≥Mm. Let ξ0(n)=1,∀n∈N.Wecan write: pk  λ(S(1,ξ 0)) = pk k  j=1 mjS(s, ξj), where k,k,s ∈N,m1,...,m k∈Z and ξ1,...,ξ k∈{0,1}N. We have pθn−n+1S(1,ξ 0)−tn,1pαn−nxn∈pαnK1thus for nlarge enough tn,1pαn−nϕn(λ(xn)) = pθn−n+1ϕn( k  j=1 mjS(s, ξj)) =⇒pn+s−1 divides ts,1ps−1rn−w(n) where w(n)= k  j=1 mjξj(n). Accordingly, if d∈Zsuch that the set {n∈N/w(n+1)−w(n)=d}is infinite, then pm divides d,∀m≥N0, therefore d=0. Since the set {w(n+1)−w(n)/n ∈ N}is finite, then there exist v0∈Zand N1∈Nsuch that w(n)=v0, ∀n≥N1.Itisclear that ps−1divides v0. Finally if we put v0=ps−1v, we can find N∈Nsuch that ts,1pαn−nλ(xn)=pαn−nvxn,∀n≥N. Construction 2. Let (αn)n≥0be as in Construction 1, and let pand qbe two natural numbers different from zero and relatively prime, we consider the two direct products  n≥1 xnand  n≥1 ynwith o(xn)=pαnand o(yn)=qαn,∀n≥1. The elements hmof  n≥1 ynare defined in the same way as the gm of  n≥1 xn(see Construction 1). The elements S1(m, ξ) (respectively S2(m, ξ)) of  n≥1 xn(respectively  n≥1 yn) are defined like S(m, ξ)of Construction 1 with tn,m =qn−m(respectively tn,m =pn−m). Inessential Endomorphisms in the Abelian Groups 363 We put R(m, ξ)=S1(m, ξ)+S2(m, ξ)∈( n≥1 xn)⊕( n≥1 yn), then we have, R(m, ξ)=r  n=m ξ(n)(pq)n−m(pαn−1xn+qαn−1yn)+(pq)r+1−mR(r+1,ξ) if r≥m. Let K2be the subgroup of (  n≥1 xn)⊕( n≥1 yn) generated by {gm/m ≥1}∪{hm/m ≥1}∪{R(m, ξ)/m ≥1,ξ∈{0,1}N}. Lemma 2.2. The direct sum ( n≥1 xn)⊕( n≥1 yn)is a subgroup of K2and for all λ∈End(K2), there exist v∈Z,N∈Nsuch that pαn−nλ(xn)=pαn−nvxnand qαn−nλ(yn)=qαn−nvyn,∀n≥N. Proof: Let µ:( n≥1 xn)⊕( n≥1 yn)→ n≥1 xnbe the canonical projection. Then µ(K2)isthe group K1of Construction 1 (with tn,m = qn−m). Let λ∈End(K2). There exists λ1∈End(µ(K2)) such that λ1(µ(X)) = µ(λ(X)), ∀X∈K2.ByLemma 2.1 there exist s1,N 1∈N and v1∈Zsuch that qs1pαn−nλ1(xn)=pαn−nv1xn,∀n≥N1, therefore qs1pαn−nλ(xn)=pαn−nv1xn,∀n≥N1.Inthe same way there are s2,N 2∈Nand v2∈Zsuch that ps2qαn−nλ(yn)=qαn−nv2yn,∀n≥N2. We can take s1=s2=sand N1=N2=N. Let ξ0(n)=1,∀n∈N,we can write: (pq)lλ(R(1,ξ 0)) = (pq)lk  j=1 mjR(m, ξj) where l,k,m ∈N∗, m1,...,m k∈Zand ξ1,...,ξ k∈{0,1}N.Wecan take m≥1+s.By applying µto this equality, we obtain: plλ(S(1,ξ 0)) = pl k  j=1 mjS(m, ξj). Then for nlarge enough pn+m−1divides qm−1−spm−1v1−v(n) where v(n)= k  j=1 mjξj(n) (see the proof of Lemma 2.1). Let d∈Zsuch that the set {n∈N/v(n)=d}is infinite, then d=qm−1−spm−1v1in the same way d=pm−1−sqm−1v2.Ifweput v1=qsvand v2=psv, then we can find N∈Nsuch that pαn−nλ(xn)=pαn−nvxnand qαn−nλ(yn)=qαn−nvyn,∀n≥N. 364 S. Abdelalim, H. Essannouni Construction 3. Let (αn)n≥0be as in Construction 1 and (βn)n≥1be a sequence of nonzero natural numbers. Let p, q1,...,q n,... be nonzero relatively prime natural numbers. Let us consider the group (  n≥1 xn)⊕ ( n≥1 zn) with o(xn)=pαnand o(zn)=qβn n,∀n≥1, the elements gm and S(m, ξ)of  n≥1 xnare defined as in Construction 1 with tn,m =             1ifn=m q1···qmif n=m+1 (q1···qm)n−mn−m−1  j=1 qn−m−j m+jif n≥m+2, the element R(m, ξ)of  n≥1 znis defined as follows ϕn(R(m, ξ)) =    0ifn<m ξ(n)pn−mtn,mznif n≥m where ϕk: n≥1 zn→zkis the canonical projection. If we put T(m, ξ)=S(m, ξ)+R(m, ξ)∈  n≥1 xn)⊕( n≥1 zn , we have T(m, ξ)=r  n=m ξ(n)tn,mpn−m(pαn−1xn+zn)+tr+1,mpr+1−mT(r+1,ξ), if r≥m. Let K3be the subgroup of (  n≥1 xn)⊕( n≥1 zn) generated by {gn/n ≥ 1}∪{zn/n ≥1}∪{T(m, , ξ)/m ≥1,ξ∈{0,1}N}. Lemma 2.3. The direct sum ( n≥1 xn)⊕( n≥1 zn)is a subgroup of K3 and for all λ∈End(K3), there exist v∈Zand N,s ∈Nsuch that ts,1pαn−nλ(xn)=pαn−nvxnand ts,1λ(zn)=vzn,∀n≥N. Proof: Let µ:( n≥1 xn)⊕( n≥1 zn)→ n≥1 xnbe the canonical projection. Then µ(K3)=K1is the group of Construction 1. Let λ∈ End(K3), the endomorphism λ1of K1defined by λ1(µ(X)) = µ(λ(X)), Inessential Endomorphisms in the Abelian Groups 365 ∀X∈K3,iswell defined. According to Lemma 2.1 there exist s, N0∈N and v∈Zsuch that ts,1pαn−nλ1(xn)=pαn−nvxn,∀n≥N0.Itis clear that λ(zn)∈zn,∀n≥1. Putting λ(zn)=knzn,∀n≥1, we consider ξ0:N→{0,1}with ξ0(n)=1,∀n∈Nwe can write: plrλ(T(1,ξ 0)) = plr k  j=1 djT(m, ξj) where rand pare relatively prime and m≥s.Byapplying µto this equality, we obtain: plλ1(S(1,ξ 0)) = pl k  j=1 djS(m, ξj). Following the same steps as in Lemmas 2.1 and 2.2, we can find N1∈N such that pn+m−1divides tm,spm−1v−v(n), ∀n≥N1,with v(n)= k  j=1 djξj(n). Then there exists N2such that v(n)=tm,spm−1v,∀n≥N2. If n≥m, then qβn ndivides tm,1pm−1kn−v(n). Finally there exists N∈N such that ts,1pαn−nλ(xn)=vpαn−nxnand ts,1λ(zn)=vzn,∀n≥N. 3. Characterization of the inessential endomorphisms in the category of the Abelian groups In the following, we suppose that Ais a group, and fan endomorphism of Asatisfying the following property. (E): Forany exact sequence 0 →Aσ →Bthere exists ˜ f∈End(B) such that the following diagram 0−−−−→Aσ −−−−→B f     ˜ f 0−−−−→Aσ −−−−→B is commutative. Let (αn)n≥0be a sequence as in Construction 1. Lemma 3.1. For all a∈Aand any q∈N∗, there exists v∈Zsuch that (f(a)−va)∈ n≥0 qnA. Proof: Let us consider the free group L= n≥1 en.Weput G=A⊕L, G0={a−qαnen/n ≥1} and G=G/G0. The homomorphism σ:A→ Gdefined by σ(b)=b+G0is a monomorphism, and if xn=en+σ(A) (en=en+G0) then G/σ(A)=  n≥1 xnand o(xn)=qαn,∀n≥1. Let 366 S. Abdelalim, H. Essannouni K1beasubgroup of  n≥1 xndefined in Construction 1 (with tn,m =1, ∀n≥m). There exists a commutative diagram, whose rows are exact, and which has the following form: 0−−−−→A−−−−→Gπ −−−−→G/σ(A)−−−−→0        0−−−−→Aσ −−−−→Bµ −−−−→K1−−−−→0 (see [3, 24.6]). We can find ˜ f∈End(B) and λ∈End(K1) such that ˜ fσ =σf and λµ =µ˜ f.ByLemma 2.1, there are v∈Zand N∈Nsuch that qαn−nλ(xn)=vqαn−nxn,∀n≥N.Forn≥N,µ[qαn−n(˜ f(en)− ven)] = 0, therefore (f(a)−va)∈qnA,so(f(a)−va)∈ n≥0 qnA. Corollary 3.2. If A1=0, then for all a∈T(A)there exists va∈Z such that f(a)=vaawhere T(A)is the torsion part of A. Proof: Let us put q=o(a) and let v∈Zsuch that (f(a)−va)∈ n≥0 qnA. Let pbeaprime number, if pdivides qthen (f(a)−va)∈ n≥0 pnAand if pand qare relatively prime, we also have (f(a)−va)∈ n≥0 pnA,thus f(a)=va. Lemma 3.3. If A1=0, then there exists v∈Zsuch that f(a)=va, ∀a∈T(A). Proof: We suppose that T(A)isbounded, then there exists x0∈T(A) such that x0is a direct summand of T(A) and o(x0).T(A)=0. If f(x0)=vx0, then ∀a∈T(A), f(a)=va.Wenow suppose that T(A)is not bounded. If pis prime number, we denote by Tpthe p-component of T(A). 1st case: There exists a prime number psuch that Tpis not bounded. Let Sbe a basic subgroup of Tp,wecan write S=  n≥1 an ⊕S0with o(an)=prnand 1 ≤rn<r n+1,∀n≥1. Inessential Endomorphisms in the Abelian Groups 367 For each n≥1, we consider anas an element of the group Xnwith pαnXn=an. There exists a group Gsuch that: A≤G,   n≥1 Xn ≤G, A+  n≥1 Xn =G and A∩  n≥1 Xn = n≥1 an. We put xn=Xn+A, then G/A = n≥1 xnand o(xn)=pαn,∀n≥1. By [3, Proposition 24.6], there exists a commutative diagram, whose rows are exact, and has the following form: 0−−−−→A−−−−→Gπ −−−−→G/A −−−−→0        0−−−−→Aσ −−−−→Bµ −−−−→K1−−−−→0 . K1is the group of Construction 1 (with tn,m =1,∀n≥m). There are ˜ f∈End(B) and λ∈End(K1) such that ˜ fσ =σf and λµ =µ˜ f. There exist v∈Zand N∈Nsuch that pαn−nλ(xn)=vpαn−nxn,∀n≥N (Lemma 2.1). We have for each n≥N,µ[pαn−n(˜ f(Xn)−vXn)] = 0, so that (f(an)−van)∈pnA. Let us put f(an)=knan(Corollary 3.2), then we have pndivides kn−v,∀n≥N.Byusing again Corollary 3.2, we can establish easily that f(an)=van,∀n≥1. Let b∈Tqwith q=p, and put o(b)=qs. Let us consider the free group L= n≥0 en. Let L0be the subgroup of L generated by {qse0}∪{qαnen−e0/n ≥1}.