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Math 202 - Assignment 7 Solutions

Authors: Yusuf Goren, Miguel-Angel Manrique and Rory Laster

Exercise 10.3.2. Let R be a commutative ring with identity. For al l positive integers n and m,Rn =Rm if and only ifn = m.

Proof. Let : Rn Rm be an isomorphism ofR-modules and let I R be a maximal ideal. Thenthe map : Rn Rm/IRm given by () = () is a morphism ofR-modules. Moreover

ker= { Rn | () = 0} = { Rn | () IRm} =1(IRm) = I Rn.

Therefore by the first isomorphism theoremRn/IRn =Rm/IRm. We already showed that Rm/IRm=(R/I)m. Since I is maximal, F := R/I is a field and we have an isomorphism of F-vector spacesR/I)n =(R/I)m. Hencen = m.

Exercise 10.3.6. Let R be a ring with identity and let M be a left R-module. If M is a finitelygeneratedR-module that is generated byn elements, then every quotient ofM may be generated bynor fewer elements. In particular, quotients of cyclic modules are cyclic.

Proof. Assume that M is generated by A = {a1, . . . , an} and let N M1. Then M/N is generated

by A= {a1+ N , . . . , an+ N}since for any m M, m = r1a1+ + rnan for somer1, . . . , rn Rand

m+N=r1a1+ +rnan+N= (r1a1+N) + + (rnan+N) = r1(a1+N) + +rn(an+N).

Exercise 10.3.7. LetR be a ring with identity and letN be a leftR-submodule ofM. If bothM/NandNare finitely generated, thenM is also.

Proof. Assume thatN Mwith generatorsb1, . . . , bk and thatM/Nis generated by a1 + N , . . . , an +N. Letm Mbe arbitrary. Then there exist r1, . . . , rn R such that

m+N=r1(a1+N) + +rn(an+N) = r1a1+ +rnan+N = m (r1a1+ +rnan) N.

Hence there existrn+1, . . . , rn+k R such that

m (r1a1+ + rnan) = rn+1b1+ + rn+kbk = m= r1a1+ + rnan+ rn+1b1+ + rn+kbk.

ThereforeM is generated by a1, . . . , an, b1, . . . , bk.

Definition. LetR be a ring. AnR-moduleM is irreducible ifM = 0 and if0 andM are the onlyR-submodules ofM.

Exercise 10.3.9. LetR be a ring with identity and letMbe a leftR-module. M is irreducible if andonly ifMis a nonzero cyclic module such that any nonzero element ofM is a generator.

Proof. Assume M= 0 is irreducible and let 0 =m Mbe arbitrary. Then the submodule generatedbym is, as the name asserts, a non-trivial submodule ofMhence it must be equal to M. ThereforeM is cyclic and generated by any non-zero element.

Conversely, assume that M is cyclic module generated by any non-zero element. Assume thatN M is a non-trivial submodule. Since it is non-trivial, it contains a non-zero element, say n.Therefore the submodule generated byn is a submodule ofN and henceRn = M Nwhich impliesthatN=M. So M is irreducible.

1I tend to denote subgroups and subrings by ; on the other hand normal subgroups, ideals and submodules are

denoted by . The idea is to distinguish the substructures that you can quotient out and the ones that you cant.

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Exercise 10.3.11. LetR be a ring with identity and let M1 andM2 be leftR-modules. IfM1 andM2 are irreducible, then any nonzero R-module homomorphism from M1 to M2 is an isomorphism.Moreover, for every irreducible leftR-moduleM, EndR(M) is a division ring.

Proof. Let : M1 M2 be a non-zero morphism of irreducible modules. Then M1 = ker M1 and

hence ker = 0. On the other hand 0 = (M1)

M2 and hence (M1) = M2. Therefore is anisomorphism of irreducible modules. IfM1 = M2 = Mis irreducible, every endomorphism is invertibleand therefore EndR(M) is a division ring.

Exercise 10.3.16. LetA1, . . . , Ak be any ideals in the ringR. Prove that the map

f :M M/A1M M/AkM; m (m+A1M , . . . , m+AkM)

is anR-module homomorphism with kernelA1M A2M AkM.

Proof. Letr R and m, m M.To see that f is an R-module homomorphism, write

(m+rm) = ((m+rm) +A1M, (m+rm) +A2M , . . . , (m+rm

) +AkM)

= (m+A1M, m+A2M , . . . , m+AkM) +r(m +A1M, m

+A2M , . . . , m +AkM)

=(m) +r(m).

Now it is obvious that A1M A2M AkM ker(f). Suppose m ker(f). We have

f(m) = (0 +A1M, 0 +A2M , . . . , 0 +AkM),

whence m AiM, for i {1, 2, . . . , k}.

Exercise 10.3.17. In the notation of the previous exercise, assume further that R is commutativeand that the idealsA1, A2, . . . , Ak are pairwise comaximal. Then

M/(A1. . . Ak)M=M/A1M M/AkM.

Proof. We will first show that for any two comaximal idealsAiandAj , we haveAiMAjM=AiAjM.

The inclusionAiAjM AiM AjM is clear. For the reverse, supposex AiM AjM, and writeai+aj = 1 with ai Ai andaj Aj . Then

x= 1 x= (ai+aj)x= aix+ajx

implies aix, ajx AiAjM since x AiM AjM. To go from n = 2 to higher n, we will use aninduction, needing to show at each step that A1 Ak1 andAk are comaximal. By assumption,Ak andAk are comaximal fori {1, . . . , k 1}, so there existxi Ai andyi Ak so thatxi + yi = 1.Then

1 =

k1i=1

(xi+yi)

=x1. . . xk1+ (terms involvingy s).The sum of terms involving ys lies in Ak, and x1. . . xk1 A1 Ak1, whence 1 A1 Ak1+Ak, as desired.

Now we will use the first isomorphism theorem for modules to complete the problem. To do so itremains to show the map f is surjective. Let (m1 + A1, . . . , mk + Ak) M/A1M M/AkM.be arbitrary. Since the A1, . . . , Ak are pairwise coprime, for i, j {1, . . . , k} and i = j, there exist

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aij Ai and bij Aj so that aij + bij = 1. For l {1, . . . , k}, define cl = a1la2l all akl. Alsodefinex = c1m1+ +ckmk. We have

f(x) = ((c1m1+ +ckmk) +A1M , . . . , (c1m1+ +ckmk) +AkM)

= ((c1m1+A1M) + + (ckmk+A1M), . . . , (c1m1+AkM) + + (ckmk+AkM))

Noticecj = (a1ja2j ai1,jai+1,j ajj akj )aij . Sinceaij Ai, we havecj Ai ifi =j . Also,forn {1, . . . , k}, we have

cl = (1 b1lAl

)(1 b2lAl

) (1 bll) (1 bklAl

)

= 1 +

(products with at least one term from Al)

This impliescl 1 +AlM, and thus cl 1 modAlM, l {1, . . . , k}. Thus we have

f(x) = (c1m1+A1M , . . . , ckmk+AkM)

= (m1+A1M , . . . , mk+AkM),

as desired.

Exercise 10.3.24. LetMi be the freeZ-moduleZ for all positive integers i and letM be the directproduct

iZ+Mi. ThenM is not a freeZ-module.

Proof. Here N and Mare the direct sum and the direct product of {Mk}kZ+ respectively -whereMk = Z. We view Nas a submodule ofM.

(a) For any = (k)kZ+ N, let (n) Z+ be the index such that k = 0 for any k

> (k).Then : N Z+ has countable fibers since1(k) =Zk as sets. Therefore, as it is the preimage of acountable set under a countable-to-one map, N is countable2.

(b) Let F(I) denote the set of finite subsets of I and let B= {ei}iIbe a basis for M. Similarto part (a), one can talk about a set-theoretic map B : M F(I) defined as follows: since for anyelement m Mwe have m =

iIiei where i Z, let B(m) ={i I| i = 0}. So we can write

the sum as m = iB(m)

iei. The direct sumN has a canonical basis {fk}kZ+ where (fk)j = 1 if

k= j and (fk)j = 0 otherwise. LetI1 = kZ+

B(fk) andB1= {ei}iI1 . Notice that since eachB(fk)

is finite, I1 and, hence, B1 is countable. Let N1 be the submodule generated by B1. By the sameargument in part (a), N1 is also countable.

(c) ClearlyB B1 = {ei | i I I1} is a basis for Mand, hence, it is free. So for 0 = x M wehave x =

iB(x)

iei and therefore x=

iB(x)I1

iei. This last sum makes sense since x = 0 implies

that J := B(x) I1 is non-empty. Let = {k Z+ | k dividesj for all j J}, the set ofk for

which the equation x = km has a solution with m =

jJjk

.

(d)An elements Scan be written ass = (k(k!))kZ+ wherek = 1. So : S {1}Z+

given

by (s) = (k) is clearly a bijection of sets. Therefore, since {1}Z+

is uncountable,Sis uncountable.SinceN1 is countable, Scannot be contained in N1 and there exists an s S N1.

(e) Let s S N1 and for k Z+ let mk Msuch that (mk)j = 0 for j < k and (mk)j =sj/k

forj k. So (s kmk)j =sj forj < k and (s kmk)j = 0 for j k. Hence s kmk N N1 andtherefore s= kmk. This contradicts (c) and therefore Mis not free.

Exercise 10.3.27. LetM be theZ-module

iZ+Z and letR be its endomorphism ring, EndZ(M).Define1, 2 R by

1(a1, a2, a3, . . .) = (a1, a3, a5, . . .) and

2(a1, a2, a3, . . .) = (a2, a4, a6, . . .).

2The proof of this fact is an application of axiom of choice.

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a. {1, 2} is a free basis of the leftR-moduleR and

b. R =Rn for all positive integersn.

Proof. (a) Define 1, 2 R by 1(a1, a2, a3, . . .) = (a1, 0, a2, 0, a3, 0, . . .) and 2(a1, a2, a3, . . .) =(0, a1, 0, a2, 0, a3, 0, . . .). It follows that11+22 is the identity map on Mand, so, for any r R

we have that r = r11+ r22. That is, R = R1+ R2. To conclude that the sum is direct, itremains to show that R1 R2 = {0}. It follows that 11 = 22 = id and 21 = 12 = 0.Therefore, if r11 = r22, then multiplying on the right by 1 gives r1 = 0 and multiplying by 2givesr2 = 0, as desired.

(b) By part (a), the map A : R2 R given by (r1, r2)r11+r22 is an isomorphism. Hence,Rn =Rn1 due to the isomorphism (id, . . . , id, A) where there are exactly n 1 identity maps. ThusR =Rn by induction.

References

[DF] Dummit, David and Foote, Richard. Abstract Algebra, 3rd edition.