Math 202 - Assignment 7

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

Proposition 1 ([DF], p342). Let R be a ring and let M be an R-module. A subset N of M is anR-submodule of M if and only if

1. N is nonempty and

2. x+ ry ∈ N for all r ∈ R and all x, y ∈ N .

Exercise 10.1.4. Let R be a ring with identity, let M be the R-module Rn with component-wiseaddition and multiplication, and let I1, I2, . . . , In be left ideals of R for some n ∈ N. The following aresubmodules of Rn:

a. N1 = {(i1, i2, . . . , in) : ik ∈ Ik for all k ∈ {1, 2, . . . , n}} and

b. N2 = {(x1, x2, . . . , xn) :∑n

k=1 xk = 0}.

Proof. To prove (a), it suffices to show, by Proposition 1, that N1 is nonempty and x + ry ∈ N1 forall r ∈ R and all x, y ∈ N1. For the first condition, (0, 0, . . . , 0) ∈ N1 since Ik is a subgroup of Rcontaining the additive identity 0 for all k ∈ {1, 2, . . . , n}. That is, N1 is nonempty.

For the second condition, let x = (ik)k∈Z+ , y = (yk)k∈Z+ ∈ N1 and let r ∈ R. Then, by definitionof addition and scalar multiplication,

x+ ry = (ik)k∈Z+ + r(yk)k∈Z+

= (ik)k∈Z+ + (ryk)k∈Z+

= (ik + ryk)k∈Z+

∈ N1

since ik + ryk ∈ Ik for all k ∈ {1, 2, . . . , n} by the left ideal axioms. This gives us (a).To establish (b), we apply a similar method as that used in (a). Since

∑nk=1 0 = 0, the element

(0, 0, . . . , 0) ∈ N2. Thus N2 is nonempty. Moreover if x = (ik)k∈Z+ , y = (yk)k∈Z+ ∈ N1 and r ∈ R,then

x+ ry = (ik)k∈Z+ + r(yk)k∈Z+ = (ik + ryk)k∈Z+ .

Therefore, because

n∑k=1

xk + ryk =

n∑k=1

xk + r

(n∑

k=1

yk

)since I is a ring

= 0 + r0 since x, y ∈ N2

= 0,

we have that x+ ry ∈ N2 by definition.

Exercise 10.1.5. Let R be a ring with identity, let I be a left ideal of R and let M be a left R-module.Define

IM :=

∑finite

aimi : ai ∈ I and mi ∈M for all i

.

IM is an R-submodule of M .

Proof. It suffices to show, by Proposition 1, that IM is nonempty and x+ ry ∈ IM for all r ∈ R andall x, y ∈ IM . For the former condition, observe that 0R ∈ I since I is an additive subgroup of R and0M ∈M because M is a group. Hence the finite sum 0R · 0M = 0M satisfies the membership conditionof IM . Therefore IM is nonempty.

For the latter condition, let r ∈ R and let x =∑n

i=1 aimi, y =∑m

i=1 a′im′i ∈ IM such that n,m ∈ N,

ai, a′i ∈ I and mi,m

′i ∈M . Then

x+ ry =

n∑i=1

aimi + r ·

(m∑i=1

a′im′i

)

=

n∑i=1

aimi +

m∑i=1

(ra′i)m′i since M is a left R-module

=

n∑i=1

aimi +

m∑i=1

a′′im′i

for a′′i = ra′i ∈ I. That is, x + ry is a finite sum of elements of the form am such that a ∈ I andm ∈M . Thus x+ ry ∈ IM .

Exercise 10.1.6. Let R be a ring with identity and let M be a left R-module. For any nonemptycollection {Ni}i∈I of R-submodules of M , the intersection

N =⋂i∈I

Ni

is an R-submodule of M .

Proof. Observe that N is a subset of M since, for all n ∈ N , n is an element of some Ni with i ∈ I.Hence n ∈ M . So it suffices to show, by Proposition 1, that N is nonempty and x + ry ∈ N for allr ∈ R and all x, y ∈ N . For the first property, 0M ∈ Ni for all i ∈ I because each Ni is an additivesubgroup of M . Therefore 0M ∈ N = ∩i∈INi and, so, N is nonempty.

For the second property, let r ∈ R and let x, y ∈ N . Then x and y are elements of Ni forall i ∈ I by definition. Thus, by the submodule axioms, x + ry ∈ Ni for each i ∈ I. That is,x+ ry ∈ N = ∩i∈INi.

Exercise 10.1.7. Let R be a ring with identity and let M be a left R-module. If N1 ⊆ N2 ⊆ . . . is anascending chain of R-submodules of M , then

N =

∞⋃i=1

Ni

is an R-submodule of M as well.

Proof. Suppose that N1 ⊆ N2 ⊆ . . . is an ascending chain of R-submodules of M . To prove that Nis also an R-submodule of M , it suffices to show, by Proposition 1, that N is nonempty and thatx+ ry ∈ N for all r ∈ R and all x, y ∈ N .

Since 0 ∈ N1, 0 is an element of the union N . Hence N is nonempty. For the remaining property,let r ∈ R and let x, y ∈ N . Because x and y are elements of N , each must be an element ofa submodule. That is, x ∈ Nj and y ∈ Nk for some j, k ∈ N. By the ascending chain hypothesis,Nmin(j,k) ⊆ Nmax(j,k). Therefore both x and y are members of Nmax(j,k). Moreover, by the submoduleaxioms, x+ ry ∈ Nmax(j,k). Hence, since Nmax(j,k) ⊆ N , we have that x+ ry ∈ N .

Definition. Let R be a ring and let M be a left R-module. A torsion element is an element m ∈ Msuch that rm = 0 for some nonzero r ∈ R.

Definition. Let R be an integral domain and let M be a left R-module. The set

Tor(M) = {m ∈M : m is a torsion element}

is the torsion submodule of M .

Exercise 10.1.8. Let R be a ring with identity and let M be a left R-module.

a. If R is an integral domain, then Tor(M) is an R-submodule of M ,

b. there exists a ring R with identity and a left R-module M such that Tor(M) is not a submodule ofM and

c. if R has zero divisors, then every nonzero left R-module contains nonzero torsion elements.

Proof. To prove (a), we suppose that R is an integral domain. It suffices to show, by Proposition 1,that Tor(M) is nonempty and x+ ry ∈ Tor(M) for all x, y ∈ Tor(M) and all r ∈ R.

For the former condition, 0 ∈ Tor(M) since 1 · 0 = 0. Hence Tor(M) is nonempty. For the finalcondition, let x, y ∈ Tor(M) and let r ∈ R. As torsion elements, there exist nonzero s, t ∈ R such thats · x = 0 and t · y = 0. Thus

(st) · (x+ ry) = (st) · x+ [(st)r] · y by the R-module axioms

= (ts) · x+ [(sr)t] · y by the commutativity of R

= t · (s · x) + (sr) · (t · y) by the R-module axioms

= t · 0 + (sr) · 0 since s · x = 0 and t · y = 0

= 0.

Because R is an integral domain and s, t are nonzero, the product st is nonzero. Therefore, we haveshown that (st) · (x+ ry) = 0 for a nonzero st ∈ R. That is, x+ ry ∈ Tor(M) and (a) is immediate.

To see that (b) holds, consider the ring R = M = Z/6Z and the elements 2, 3 ∈ R. R is a leftR-module with respect to addition and left ring multiplication. Moreover, since

2 · 3 = 6 = 0

and3 · 2 = 6 = 0,

we find that 2 and 3 are elements of Tor(M). However, since 2 + 3 = 5 6∈ Tor(M), M is not closedunder addition. That is, Tor(M) is not a subgroup of M and, hence, it is not a submodule of Meither. Thus there exists a ring R and R-module M with the desired properties.

For (c), suppose that R contains the zero divisors s and r such that sr = 0. Then, for anynonzero left R-module M with nonzero element m, either r ·m = 0 or r ·m 6= 0. In the first case ofr ·m = 0, m ∈ Tor(M) since r is nonzero by hypothesis. In the second case of r ·m 6= 0, we find thatr ·m ∈ Tor(M) because

s · (r ·m) = (sr) ·m by the R-module axioms

= 0 ·m by hypothesis

= 0.

In either case, there exists a nonzero element contained in Tor(M). This is the desired result.

Definition. Let R be a ring with identity and let M be a left R-module. The annihilator of a submoduleN of M is the set

Ann(N) = {r ∈ R : r · n = 0 for all n ∈ N}.

Exercise 10.1.9. Let R be a ring with identity and let M be a left R-module. For any R-submoduleN of M , the annihilator of N in R is a two-sided ideal of R.

Proof. Suppose that N is an R-submodule of M , It suffices to show that Ann(N) is nonempty andthat x+ rys ∈ Ann(N) for all x, y ∈ Ann(N) and all r, s ∈ R. For the former condition, consider theelement 0R. Since

0R · n = 0N

for any n ∈ N , we see that 0R ∈ Ann(N).For the remaining condition, we let x, y ∈ Ann(N) and let r, s ∈ R. For any n ∈ N , we have that

(x+ rys) · n = x · n+ r · (y · (s · n)) by the R-module axioms

= 0 + r · 0 since x, y ∈ Ann(N) and sn ∈ N= 0.

Hence x+ rys ∈ Ann(N).

Definition. Let R be a ring with identity and let M be a left R-module. The annihilator of an idealI of R is the set

Ann(I) = {m ∈M : i ·m = 0 for all i ∈ I}.

Exercise 10.1.10. Let R be a ring with identity and let M be a left R-module. For any ideal I of R,Ann(I) is an R-submodule of M .

Proof. Suppose that I is an ideal of R. To prove the desired result, it suffices to show, by Proposition1, that Ann(I) is nonempty and x+ ry ∈ Ann(I) for all x, y ∈ Ann(I) and all r ∈ R.

To see that Ann(I) is nonempty, observe that

i · 0N = 0N

for all i ∈ I. Thus 0N ∈ Ann(I).For the remaining condition, let x, y ∈ Ann(I) and let r ∈ R. If i ∈ I, then ir ∈ I by the ideal

axioms. Hence

i · (x+ ry) = i · x+ (ir) · y by the R-submodule axioms

= 0 + 0 since x, y ∈ Ann(I) and ir ∈ I= 0.

Therefore x+ ry ∈ Ann(I).

Exercise 10.1.11. Let M be the Z-module Z/24Z× Z/15Z× Z/50Z.

a. Ann(M) = 600Z and

b. Ann(2Z) ∼= G×H such that G = 〈(12 + 24Z, 0 + 15Z, 0 + 50Z)〉 and H = 〈(0 + 24Z, 0 + 15Z, 25 + 50Z)〉.

Proof. For (a), we appeal to the definition of set equality. To see the first inclusion, let a ∈ Ann(M).As it annihilates all elements of M , a must annihilate (1 + 24Z, 1 + 15Z, 1 + 50Z) ∈M . That is,

0 = a · (1 + 24Z, 1 + 15Z, 1 + 50Z) = (a+ 24Z, a+ 15Z, a+ 50Z).

Since 0 = (0 + 24Z, 0 + 15Z, 0 + 50Z), the previous equation implies thata+ 24Z = 0 + 24Z,a+ 15Z = 0 + 15Z and

a+ 50Z = 0 + 50Z.

Hence a ∈ 24Z ∩ 15Z ∩ 50Z = 600Z.For the opposite inclusion, let a ∈ 600Z. Because 24, 15 and 50 are divisors of 600, observe that ab is

an element of each of 24Z, 15Z and 50Z for any b ∈ Z. Therefore, for all (x+24Z, y+15Z, z+50Z) ∈M ,

a · (x+ 24Z, y + 15Z, z + 50Z) = (ax+ 24Z, ay + 15Z, az + 50Z)

= (0 + 24Z, 0 + 15Z, 0 + 50Z)

= 0.

Thus a annihilates all elements of M and, consequently, a ∈ Ann(M) and we have now shown thatAnn(M) = 600Z. This is (a).

For (b), we proceed similarly to the argument given in (a). To demonstrate the first inclusion, leta = (x+ 24Z, y + 15Z, z + 50Z) ∈ Ann(2Z). Then a annihilates all elements of 2Z and, in particular,a annihilates 2. Thus

0 = 2 · a = 2 · (x+ 24Z, y + 15Z, z + 50Z) = (2x+ 24Z, 2y + 15Z, 2z + 50Z).

That is, we have the system of linear congruences2x ≡ 0 (mod 24),

2y ≡ 0 (mod 15) and

2z ≡ 0 (mod 50).

By elementary number theoretic results, this system is equivalent tox ≡ 0 (mod 12),

y ≡ 0 (mod 15) and

z ≡ 0 (mod 25).

Therefore x = 12x′, y = 15y′ and z = 25z′ for some x′, y′, z′ ∈ Z. Moreover,

(x+ 24Z, y + 15Z, z + 50Z) = (12x′ + 24Z, 15y′ + 15Z, 25z′ + 50Z)

= (12x′ + 24Z, 0 + 15Z, 25z′ + 50Z)

= (12x′ + 24Z, 0 + 15Z, 0 + 50Z) + (0 + 24Z, 0 + 15Z, 25z′ + 50Z)

∈ G+H.

Hence Ann(2Z) ⊆ G+H.For the opposite inclusion, let a ∈ G+H. Then

a = n · (12 + 24Z, 0 + 15Z, 0 + 50Z) +m · (0 + 24Z, 0 + 15Z, 25 + 50Z)

= (12n+ 24Z, 0 + 15Z, 0 + 50Z) + (0 + 24Z, 0 + 15Z, 25m+ 50Z)

for some n,m ∈ Z. For any element i in the ideal 2Z, it follows that i = 2j for some j ∈ Z. Thus

i · a = (2j) · [(12n+ 24Z, 0 + 15Z, 0 + 50Z) + (0 + 24Z, 0 + 15Z, 25m+ 50Z)]

= (2j) · (12n+ 24Z, 0 + 15Z, 0 + 50Z) + (2j) · (0 + 24Z, 0 + 15Z, 25m+ 50Z)]

= (24jn+ 24Z, 0 + 15Z, 0 + 50Z) + (0 + 24Z, 0 + 15Z, 50jm+ 50Z)]

= (0 + 24Z, 0 + 15Z, 0 + 50Z) + (0 + 24Z, 0 + 15Z, 0 + 50Z)]

= 0.

Hence a annihilates all elements of 2Z and, therefore, a ∈ Ann(2Z). This establishes the oppositeinclusion of G+H ⊆ Ann(2Z) and, so, Ann(2Z) = G+H. Furthermore, since G∩H = {0}, it followsthat G+H ∼= G×H and, so, Ann(2Z) ∼= G×H.

Exercise 10.1.14. Let R be a ring with identity and let M be a left R-module. For any element z ofthe center of R,

zM = {zm : m ∈M}

is an R-submodule of M .

Proof. Let z be an element of the center of R. It suffices to show, by Proposition 1, that zM isnonempty and x+ ry ∈ zM for all x, y ∈ zM and all r ∈ R. Since 0M ∈M ,

z · 0M = 0M ∈ zM.

That is, zM is nonempty.For the remaining submodule criterion, let x = zm, y = zm′ ∈ zM and let r ∈ R. Then

x+ ry = z ·m+ r · (z ·m′)= z ·m+ (rz) ·m′ by the R-submodule axioms

= z ·m+ (zr) ·m′ since z is in the center of R

= z · (m+ r ·m′) by the R-submodule axioms.

Therefore, because m+ r ·m′ ∈M , we have that x+ ry ∈ zM .

Corollary. Let R = M2(F ) for a field F and let

e =

[1 00 0

]∈M2(F ).

Then e is not in the center of R.

Proof. Consider R as a left R-module in the natural way. The element e = e · 1R is a member of Rand, also,

r =

[1 01 0

]∈ R.

Since

re =

[1 01 0

]·[1 00 0

]=

[1 01 0

]= r

and r 6∈ eR, we find that eR is not closed under scalar multiplication. Therefore eR is not a submoduleof R. This implies, by the contrapositive to Exercise 10.1.14, that e is not in the center of R.

Exercise 10.1.16. Let V = Fn for a field F and let T : V → V be the shift operator defined by

T (v1, v2, . . . , vn) = (v2, . . . , vn, 0).

If we View V as an F [x]-module with x acting by the operator T and if U is an F [x]-submodule of V ,then U = Uk for some Uk = Span({ei : 1 ≤ i ≤ k}) such that k ∈ {0, 1, . . . , n}.

Proof. The abelian group V = Fn is a left F [x]-module when endowed with a scalar multiplication· : F [x]× V → V defined by (

m∑i=0

aixi

)· v =

m∑i=0

aiTi(v)

for all∑m

i=0 aixi ∈ F [x] and all v ∈ V .

With this definition in mind, suppose that U is an F [x]-submodule of V . There exists a maximalelement in the set of integers

S = {m ∈ Z : there exists an element u ∈ U such that u has nonzero mth coordinate}

because there are only n coordinates in any element of V . Let k be this maximal integer of S. Thatis, for all u ∈ U and all l > k, the lth coordinate of u is equal to 0.

We claim that U = Uk for this maximal k. To see this, observe that U is closed under scalarmultiplication by F [x] since it is a submodule. Hence, if f is a nonzero element in the kth coordinateof an element u = (u1, . . . , uk−1, f, 0, . . . , 0) ∈ U , then

(f−1xk−1) · u = f−1T k−1(u)

= f−1 · (f, 0, . . . , 0)

= (1, 0, . . . , 0)

= e1.

Thus e1 ∈ U .Furthermore, since U is also closed under addition,

(f−1xk−2) · u− (f−1uk−1) · e1 = f−1T k−2(u)− (f−1uk−1, 0, . . . , 0)

= f−1(uk−1, f, 0, . . . , 0)− (f−1uk−1, 0, . . . , 0)

= (0, 1, 0, . . . , 0)

= e2.

So e2 ∈ U .Continuing in this way, we find that ei ∈ U for all i ∈ {1, 2, . . . , k}. Therefore Span({ei : 1 ≤ i ≤

k}) = Uk is contained in U by the F [x]-submodule properties. That is, Uk ⊆ U . Moreover, since kis the maximal coordinate for which there is an element with nonzero entry, it follows that U ⊆ Uk.Hence U = Uk, as desired.

Exercise 10.1.18. Let F = R, let V = R2 and let T ∈ EndR(V ) be the map such that

T (v) =

[0 1−1 0

]· v

for all v ∈ V . Then V and 0 are the only F [x]-submodules of V with respect to T .

Proof. Let U be an F [x]-submodule of V . Since F embeds into F [x], we may view V as an F -vectorspace and U as an F -subspace of V . Either U = 0, U = V or the dimension of U as an F -subspace isequal to 1. If dimR(U) = 1, then there exists a nonzero element u ∈ U such that U = spanR(u). ForU to be an F [x]-submodule of V , we must have that T (U) ⊆ U by the discussion in §10.1 of [DF]. Inparticular, T (u) = λu for some λ ∈ R. Therefore, by definition, u is an eignevector of T with a realeigenvalue. However, the eigenvalues of T are ±i and, so, T has no real eigenvalues. This contradictionimplies that such a U cannot occur. That is, U = {0} or U = V .

Exercise 10.1.19. Let F = R, let V = R2 and let T ∈ EndR(V ) be the map defined by T (a, b) = (0, b).Then the set

S = {W ⊆ V : W is an F [x]-submodule of V }

satisfies S = {0,Wx,Wy, V } such that Wx = {(a, 0) ∈ V : a ∈ R} and Wy = {(0, b) ∈ V : b ∈ R}.

Proof. The set of F [x]-submodules is equal to the set

S = {W ⊆ V : W is a subspace of V and W is T -stable}

by an argument given in §10.1 of [DF].For the subspace Wx of V ,

T (a, 0) = (0, 0) ∈Wx

for all (a, 0) ∈Wx. Hence Wx is T -stable and, moreover, Wx ∈ S.

Similarly, for the subspace Wy,T (0, b) = (0, b) ∈Wy

for all (0, b) ∈Wy and, consequently, Wy ∈ S. It follows that, since 0 and V are trivially contained inS, {0,Wx,Wy, V } ⊆ S.

To demonstrate the opposite inclusion, we let W ∈ S. There are two cases of W : either W containsan element (a, b) such that both a and b are nonzero or it does not. In the first case of the existenceof such an element (a, b), we find that, by the T -stability of W ,

T (u) = T (a, b) = (0, b) ∈W.

But (a, b) and (0, b) are R-linearly independent since

det

([a b0 b

])= ab

and a and b are nonzero by hypothesis. Since it is a subspace of the two-dimensional R-vector spaceV containing two linearly independent elements, W must equal V .

In the second case where no elements exist in W of the form (a, b) such that both a and b arenonzero, it follows that W is either equal to 0, Wx or Wy. Thus W ∈ {0,Wx,Wy, V }, as desired.

Lemma 1. Let A be a Z-module, let a ∈ A and let n be a positive integer. The map φa : Z/nZ→ Adefined by φa(k) = ka is a well-defined Z-module homomorphism if and only if na = 0.

Proof. Suppose that φa is a well-defined Z-module homomorphism. In the Z-module Z/nZ, n = 0.Hence, because φa is well-defined,

na = φa(n) = φa(0) = 0a = 0.

That is, na = 0.For the converse statement, suppose that na = 0 and consider the map ψ : Z → A defined by

ψ(k) = ka for all k ∈ Z. This map is a group homomorphism since

ψ(k1 + k2) = (k1 + k2)a

= k1a+ k2a by the Z-module axioms

= ψ(k1) + ψ(k2)

for all k1, k2 ∈ Z. Furthermore, because

ψ(nk) = ψ(kn)

= (kn)a

= k(na) by the Z-module axioms

= k0 since na = 0 by hypothesis

= 0

for all nk ∈ nZ, we find that the subgroup nZ is contained in the kernel of ψ. Therefore ψ descendsto the quotient Z/nZ. That is, there exists a unique group homomorphism φa : Z/nZ→ A such thatφa(k) = ψ(k) = ka. Moreover, since

φa(k1k2) = φa(k1k2)

= (k1k2)a

= k1(k2a) by the Z-module axioms

= k1φa(k2)

for all k1 ∈ Z and all k2 ∈ Z/nZ, it follows that φa is a Z-module homomorphism.

Exercise 10.2.4. Let A be a Z-module and let n be a positive integer. Then HomZ(Z/nZ, A) ∼= An

where An = {a ∈ A : an = 0}.

Proof. Let φ ∈ HomZ(Z/nZ, A) and let a = φ(1). For any k ∈ Z+, that φ is a Z-module homomorphismimplies

φ(k) = φ

(k∑

i=1

1

)=

k∑i=1

φ(1) =

k∑i=1

a =

(k∑

i=1

1

)a = ka.

Thus φ = φa, with φa defined as in Lemma 1.By Lemma 1, φ = φa is a Z-module homomorphism if and only if na = 0. Therefore the func-

tion ψ : HomZ(Z/nZ, A) → An defined by ψ(φa) = a is a bijection. Furthermore ψ is a Z-modulehomomorphism since

ψ(φa + φa′) = ψ(φa+a′)

= a+ a′

= ψ(φa) + ψ(φa′)

and

ψ(kψa) = ψ(φka)

= ka

= kψ(φa)

for all k ∈ Z and all φa, φa′ ∈ HomZ(Z/nZ, A). That is, ψ is an isomorphism and we have thatHomZ(Z/nZ, A) ∼= An.

Exercise 10.2.5. Let a ∈ Z/21Z and let φa : Z/30Z→ Z/21Z denote the map defined by φa(k) = ka.Then

HomZ(Z/30Z,Z/21Z) = {φ0, φ7, φ14}.

Proof. By Exercise 10.2.4, the map ψ : An → HomZ(Z/30Z,Z/21Z) defined by ψ(a) = φa is anisomorphism for An = {a ∈ Z/nZ : 30a = 0}. So it suffices to show that An = {0, 7, 14}.

For the set relation An ⊆ {0, 7, 14}, we let a ∈ An. Then

30a = 30a = 0 = 0 (1)

since a annihilates 30 by the membership condition of An. Thus, because 30a and 0 are elements ofZ/21Z, (1) implies that

30a ≡ 0 (mod 21).

So, by definition, 21 divides 30a and there exists b ∈ Z such that 21b = 30a. Dividing this equationby 3 yields that 7b = 10a and, because 7 is coprime to 10, we find that 7 must divide a. That is,a ∈ {0, 7, 14}.

For the opposite inclusion, we let a ∈ {0, 7, 14}. Then 7 divides a and it is clear that, for someb ∈ Z,

30a = 307b = 30 · 7b = 21 · 10b = 0 = 0.

Hence a ∈ An. Therefore An = {0, 7, 14}, as required.

Exercise 10.2.6. For all positive integers m and n,

HomZ(Z/nZ,Z/mZ) ∼= Z/(n,m)Z.

Proof. Let m and n be positive integers. By Exercise 10.2.4,

HomZ(Z/nZ,Z/mZ) ∼= An

where An = {a ∈ Z/mZ : na = 0}. So it suffices to show that An∼= Z/(n,m)Z.

To see that this is the case, consider an arbitrary element a ∈ An. As an element of An, a annihilatesn and, therefore,

na = na = 0 = 0.

That is, na = 0 in the group Z/mZ. Thus

na ≡ 0 (mod m)

to imply that there exists an integer k such that km = na. Furthermore, we find by dividing by thegreatest common divisor (n,m) that

km

(n,m)=

n

(n,m).

So m(m,n) divides the product n

(n,m)a. It follows, by elementary number theoretic results, that m(n,m) and

n(n,m) are coprime and, hence, n

(n,m) must divide a. Since a is divisible by m(n,m) , a ∈

{km

(n,m) : 1 ≤ k ≤ (n,m)− 1}

.

Therefore, we have shown that

An ⊆{

km

(n,m): 1 ≤ k ≤ (n,m)− 1

}.

The opposite inclusion follows from the observation that each of the implications in the derivation of

the previous set relation are, in fact, equivalences. Or, more pedantically, let a ∈{

km(n,m) : 1 ≤ k ≤ (n,m)− 1

}.

Then m(n,m) divides a and, so, a = k′m

(n,m) for an integer k′. Hence

(n,m)a = (n,m)k′m

(n,m)= k′m ≡ 0 (mod m).

Consequently (n,m)a = 0 and we find that a ∈ An. This gives us that{km

(n,m): 1 ≤ k ≤ (n,m)− 1

}⊆ An.

Each set inclusion holds and we conclude that

An =

{km

(n,m): 1 ≤ k ≤ (n,m)− 1

}.

Moreover An is isomorphic to Z/(n,m)Z by the map φ : An → Z/(n,m)Z defined by φ(

km(n,m)

)= k.

That is, An∼= Z/(n,m)Z and the proof is complete.

Exercise 10.2.9. Let R be a commutative ring. Prove that HomR(R,M) and M are isomorphic asleft R-modules.

Proof. Throughout this problem let a, r, r′ ∈ R and m,m′ ∈ M . We follow the hint. Let ϕ,ψ ∈HomR(R,M). Suppose ϕ(1) = ψ(1) = m. We must have

ϕ(a · 1) = aϕ(1) = am

and

ψ(a · 1) = aψ(1) = am.

Since a was arbitrary, ϕ = ψ. This shows that any ϕ ∈ HomR(R,M) is equal to ϕm for some m.Define f : HomR(R,M) → M ; ϕm 7→ m. The map is obviously well-defined. To see that f is an

R-module homomorphism, first observe

(ϕm + rϕm′)(1) = ϕm(1) + rϕm′(1) = m+ rm′ = ϕm+rm′(1).

Hencef(ϕm + rϕm′) = f(ϕm+rm′) = m+ rm′ = f(ϕm) + rf(ϕm′).

To see that f is injective, suppose f(ϕm) = 0. This implies m = 0. It follows that ϕm(r) = 0 for allr. This shows ϕm = 0 ∈ HomR(R,M), i.e., the kernel is trivial. Finally, f is surjective as well. Tothis end we will first show that the map ϕm : R → M given by 1 7→ m and r 7→ rm is an R-modulehomomorphism:

ϕm(r + ar′) = (r + ar′)m = rm+ ar′m = ϕm(r) + aϕm(r′).

Therefore given m ∈ M , there exists ϕm ∈ HomR(R,M) such that f(ϕm) = m, and the proof iscomplete.

Exercise 10.2.11. Let A1, . . . , An be R-modules and let Bi be a submodule of Ai for each i = 1, . . . , n.Prove that

(A1 × · · · ×An)/(B1 × · · · ×Bn) ∼= (A1/B1)× · · · × (An/Bn).

Proof. Define an R-module homomorphism ϕ : A1 × · · · × An → (A1/B1) × · · · × (An/Bn) byϕ(a1, . . . , an) = (a1, . . . , an), where we let ai denote the image of ai ∈ Ai in the quotient Ai/Bi.It is easy to check that this is indeed an R-module homomorphism. Furthermore, it is clearly surjec-tive. The kernel consists of exactly the tuples (a1, . . . , an) such that ai = 0 for all i, i.e. such thatai ∈ Bi. Thus the kernel is B1 × · · · × Bn, and hence we obtain the desired result from the firstisomorphism theorem.

Exercise 10.2.13. Let I be a nilpotent ideal in a commutative ring R, let M and N be R-modules andlet ϕ : M → N be an R-module homomorphism. Show that if the induced map ϕ : M/IM → N/IN issurjective, then ϕ is surjective.

Proof. Since ϕ is surjective, N/IN = ϕ(M/IM) = (ϕ(M) + IN)/IN . By the lattice isomorphismtheorem for modules, N = ϕ(M) + IN .

We will show N = ϕ(M) + ItN for t ≥ 1 by induction. The base case is already shown. Supposethe equation holds for some t ≥ 1. Then

N = ϕ(M) + ItN = ϕ(M) + It(ϕ(M) + IN) = ϕ(M) + Itϕ(M) + It+1N

= ϕ(M) + ItN.

Since Ik = 0 for appropriate k, the conclusion follows.

References

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