HOMEWORK 1 SOLUTIONS

MATH 121

Problem (10.1.2). Prove that

R×

and

M

group action of the multiplicative group

satisfy the two axioms in Section 1.7 for a

R×

on the set

M.

Solution. For the rst axiom, we have to check that for

r1 , r2 ∈ R

and

m ∈ M,

we

have

For the second axiom, we have

r1 (r2 m) = (r1 r2 )m. to check that for m ∈ M , 1m = m.

Both of these are

included as part of the denition of a module. Problem (10.1.4). Let

M

Rn described in Example 3 and let I1 , I2 , . . . , In following are submodules of M :

be the module

R. Prove that the {(x1 , x2 , . . . , xn ) | xi ∈ Ii }. {(x1 , x2 , . . . , xn ) | xi ∈ R and x1 + x2 + · · · + xn = 0}.

be left ideals of (a) (b)

for

N . We have to check that N is a subgroup, and that x ∈ N , rx ∈ N . Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ N .

(a) Call the set

Solution.

r ∈R

and

Then

x + y = (x1 + y1 , . . . , xn + yn ) ∈ N since the Ii 's are ideals and hence closed under sums. Now, let

N

and

r ∈ R.

x = (x1 , . . . , xn ) ∈

Then

rx = (rx1 , . . . , rxn ) ∈ N, again because the (b) Call this set

P.

Ii 's

are ideals. Hence

under scalar multiplication.

r ∈ R.

N

is a submodule.

P is a subgroup and is closed x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ P and

As in part (a), we check that Let

Then

x + y = (x1 + y1 , . . . , xn + yn ). From the above, we know that this is in N , so we need to check that (x1 + y1 ) + · · · + (xn + yn ) = 0. This is true because x1 + · · · + xn = y1 + · · · + yn = 0. Similarly,

rx = (rx1 , . . . , rxn ) ∈ N, and

so

Date

(rx1 ) + · · · + (rxn ) = r(x1 + · · · + xn ) = 0, rx ∈ P . Hence P is a submodule.

: 10 January, 2011.

1

2

MATH 121

Problem (10.1.9). If

{r ∈ R | rn = 0 ideal of R. be

Solution. Let

and

n ∈ N.

N

is a submodule of

for all

ann(N )

n ∈ N }.

M,

the annihilator of

N in R is dened to N in R is a 2-sided

Prove that the annihilator of

N in R. Suppose r ∈ ann(N ), a ∈ R, ar, ra ∈ ann(N ). We have

be the annihilator of

We have to show that

(ar)(n) = a(rn) = 0 because

for some

rn = 0,

and

n0 ∈ N .

(ra)(n) = r(an) = rn0 = 0 ann(N ) is a 2-sided ideal of R.

Hence

I in M is dened to be {m ∈ M | am = 0 for all a ∈ I}. Prove that the annihilator of I in M is a submodule of M .

Problem (10.1.10). If

I

is a right ideal of

ann(I) be the annihilator a ∈ I . Then we have

Solution. Let

r ∈ R,

and

R, the annihilator

of

I

in

M.

of

Suppose that

m, m0 ∈ ann(I),

a(m + m0 ) = am + am0 = 0, so

m + m0 ∈ ann(I).

Furthermore,

a(rm) = (ar)m = a0 m = 0 for some

a0 ∈ I .

Hence

Problem (10.1.11). Let

rm ∈ ann(I). M

Thus

ann(I)

is a submodule of

be the abelian group (i.e.,

M.

Z-module) Z/24Z × Z/15Z ×

Z/50Z. (a) Find the annihilator of (b) Let

I = 2Z.

M

in

Z

(i.e., a generator for this principal ideal).

Describe the annihilator of

I

in

M

as a direct product of cyclic

groups.

a ∈ Z annihilates M if and only if am = 0 for all m in a set of generators of M . One choice of generators of M is (1, 0, 0), (0, 1, 0), and (0, 0, 1). Hence, ann(M ) is the intersection of the annihilators

Solution.

(a) An element

of the submodules generated by these elements. Clearly, the annihilators of

(24), (15), and (50), respectively, so their intersection is (lcm(24, 15, 50)) = (600). An element m = (m1 , m2 , m3 ) of M is annihilated by I if and only if these submodules are

(b)

(2m1 , 2m2 , 2m3 ) = 0. This happens if and only if

50Z/50Z, 25Z/50Z.

2m1 ∈ 24Z/24Z, 2m2 ∈ 15Z/15Z, and 2m3 ∈ m1 ∈ 12Z/24Z, m2 ∈ 15Z/15Z, and m3 ∈

or, equivalently, if Hence

ann(I) = 12Z/24Z × 15Z/15Z × 25Z/25Z.

HOMEWORK 1 SOLUTIONS Problem (10.2.5). Exhibit all

Z-module

3

homomorphisms from

Z/30Z

to

Z/21Z.

Solution. A

Z-module homomorphism from a cyclic module to any module is determined by where a generator is sent. (See Problem 10.2.9 below.) Let φ : Z/30Z → Z/21Z be a Z-module homomorphism. Then we must have 30φ(1) = 0. The elements y ∈ Z/21Z so that 30y = 0 are y = 7k (mod 21) for k = 0, 1, 2, so there are three such homomorphisms, given by 1 7→ 7, 1 7→ 14, and 1 7→ 0. HomR (R, M ) and M are isomorphic as left R-modules. [Show that each element of HomR (R, M ) is determined by its value on the identity of R.]

Problem (10.2.9). Let

R

be a commutative ring. Prove that

Solution. We dene a map

ψ : HomR (R, M ) → M given by φ 7→ φ(1). Let's show R-modules. For φ1 , φ2 ∈ HomR (R, M ), we have

that this is a homomorphism of left

ψ(φ1 + φ2 ) = (φ1 + φ2 )(1) = φ1 (1) + φ2 (1) = ψ(φ1 ) + ψ(φ2 ), and for

φ ∈ HomR (R, M )

and

r ∈ R,

we have

ψ(rφ) = (rφ)(1) = rφ(1) = rψ(φ), as desired. Now let's show that that

φ(1) = 0.

Now, let

r ∈ R.

ψ

is injective. Suppose that

ψ(φ) = 0.

This means

We have

φ(r) = rφ(1) = r0 = 0 φ is an R-module homomorphism. Hence φ = 0, as desired. Finally, we show ψ is surjective. Pick m ∈ M . We need to nd φ ∈ HomR (R, M ) so that φ(1) = m. We dene φ(r) = rm, but we need to check that this is actually an R-module homomorphism. Let r1 , r2 ∈ R. We have

since that

φ(r1 + r2 ) = (r1 + r2 )m = r1 m + r2 m = φ(r1 ) + φ(r2 ) and

Hence

φ

is indeed an

φ(r1 r2 ) = (r1 r2 )m = r1 (r2 m) = r1 φ(r2 ). R-module homomorphism, and so ψ is an

Problem (10.2.10). Let

R

be a commutative ring. Prove that

isomorphism.

HomR (R, R)

and

R

are

isomorphic as rings.

HomR (R, R) and R are isomorphic as R-modules, so it suces to check that our map ψ from that problem is in fact a ring map. Hence, we have to check that ψ(φ1 ◦ φ2 ) = ψ(φ1 )ψ(φ2 ) and ψ(1) = 1. For the Solution. We showed in Problem 10.2.9 that

rst one, we have

ψ(φ1 ◦ φ2 ) = (φ1 ◦ φ2 )(1) = φ1 (φ2 (1)) = φ1 (1)φ2 (1) = ψ(φ1 ◦ φ2 ). For the second, we have

ψ(id) = id(1) = 1, as desired. Thus

ψ

is a ring isomorphism.

4

MATH 121

Problem (10.2.11). Let

for each

i = 1, 2, . . . , n.

A1 , A2 , . . . , An

be

R-modules and let Bi

be a submodule of

Ai

Prove that

(A1 × · · · × An )/(B1 × · · · × Bn ) ∼ = (A1 /B1 ) × · · · × (An /Bn ). [Recall Exercise 14 in Section 5.1.] Solution. In Problem 5.1.4, we showed that there is an isomorphism

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

We must now show that it is compatible with the

R-module

structure.

Recall that the map is given by

mod B1 × · · · × Bn ) = (a1

φ((a1 , . . . , an ) Now, let

r∈R

and

(a1 , . . . , an ) ∈ A1 × · · · × An .

mod B1 , . . . , an We have

mod B1 × · · · × Bn ) = φ((ra1 , . . . , ran ) mod B1 × · · · × Bn )

φ(r(a1 , . . . , an )

= (ra1

mod B1 , . . . , ran

= r(a1

mod B1 , . . . , an

= rφ((a1 , . . . , an ) Problem (10.2.13). Let

37, Section 7.3), let homomorphism. then

ϕ

M

I

mod Bn )

mod B1 × · · · × Bn )

R (cf. Exercise ϕ : M → N be an R-module φ : M/IM → N/IN is surjective,

be a nilpotent ideal in a commutative ring

and

N

be

R-modules

and let

Show that if the induced map

n ∈ N.

By hypothesis, we can nd some

ϕ(m) = n + ai ∈ I

and

n0i ∈ N .

For each

i,

X

X

m∈M

so that

ai n0i ,

we can nd some

ϕ(m0i ) = n0i + Hence

mod Bn )

is surjective.

Solution. Let

where

mod Bn ).

m0i

so that

a0ij n00ij .

  X X ϕ m− ai m0i = n − ai a0ij n00ij ∈ N + I 2 N. i,j

M → N/I 2 N

is surjective. Continuing in

the same manner, we can see that all the induced maps r Since some I = 0, this shows that the original map ϕ :

M → N/I k N are surjective. M → N is surjective.

Hence we've shown that the induced map

Note that we didn't use the fact that

I

is nilpotent until the last step. Consider the

R = Z, M = N = Z, I = (3), and suppose that ϕ : M → N is multiplication by 5. Now, the induced map M/IM → N/IN is surjective, which means that every integer is a multiple of 5 up to a multiple of 3. For example, if n = 4, then n is a multiple of 5 up to a multiple of 3 because 4 + 2 × 3 = 10 is a multiple of 5. The argument above shows that we

more down-to-earth example that sets aside that hypothesis. Let

HOMEWORK 1 SOLUTIONS

5

can improve the situation to saying that 4 is a multiple of 5 up to a multiple of 9,

4 + 4 × 9 = 40 is a multiple of 27, because 4 + 3 × 27 = 85 is

which is true because

5. Similarly, 4 is a multiple of 5

up to a multiple of

a multiple of 5, and so on. The

only dierence here is that we cannot conclude that no power of

I

ϕ

is actually surjective because

is 0.

R-module M is called a torsion module if for each m ∈ M element r ∈ R such that rm = 0, where r may depend on m (i.e.,

Problem (10.3.4). An

there is a nonzero

M = Tor(M )

in the notation of Exercise 8 of Section 1).

abelian group is a torsion that is a torsion

Z-module. Z-module.

Solution. Suppose that

A

Give an example of an innite abelian group

is a nite abelian group (or

by Lagrange's Theorem, for each

Prove that every nite

a ∈ A, na = 0.

Hence

example of an innite abelian group that is a torsion

Z-module) of order n. Then A is a torsion Z-module. An

Z-module

is

∞ Y

(Z/2Z).

i=1

R-module M is called irreducible if M 6= 0 and if 0 and M are the only submodules of M . Show that M is irreducible if and only if M 6= 0 and M is

Problem (10.3.9). An

a cyclic module with any nonzero element as its generator. Determine all irreducible

Z-modules. Solution. Suppose that

of

M.

Then

Hence,

M

Rm

M

is an irreducible

is a submodule of

M,

R-module.

Let

m

be a nonzero element

so by hypothesis, we must have

is cyclic, and any nonzero element of

M

si a generator.

M = Rm.

For the other

M is a cyclic module, and that any nonzero element is a generator. N is a nonzero submodule of M . Then for any nonzero n ∈ N , Rn of M , because n is also a nonzero element of M . As any submodule of some Rn, M must be irreducible. The irreducible Z-modules, then, are

direction, suppose Now suppose that must be all

M

contains

exactly the cyclic groups of prime order. Problem (10.3.11). Show that if

nonzero

M

M1

and

M2

are irreducible

R-modules,

then any

R-module homomorphism from M1 to M2 is an isomorphism. Deduce that if EndR (M ) is a division ring (this result is called Schur's Lemma.)

is irreducible then

[Consider the kernel and image.] Solution. Let

φ : M1 → M2

R-module

homomorphism between two

Then ker φ and im φ are submodules of M1 and M2 , respecker φ = 0 and im φ = M2 . Hence φ is an isomorphism. Now, let φ ∈ EndR (M ) be nonzero. By the previous argument, φ is an automorphism, so it has an inverse. Since φ was arbitrary, every nonzero element of EndR (M ) is invertible, so EndR (M ) is a division ring.

irreducible

R-modules.

be a nonzero

tively, so we must have

6

MATH 121

Problem (10.3.20). Let

R-module.

I

be a nonempty index set and for each

Mi

The direct product of the modules

i∈I

let

Mi

be an

is dened to be their direct product

R componentwise Mi is dened to be the restricted direct

as abelian groups (cf. Exercise 15 in Section 5.1) with the action of multiplication. The direct sum of the modules product of the abelian groups

R

Mi

(cf. Exercise 17 in Section 5.1) with the action of

QMi 's is the subset of the direct product, prodi∈I Mi , which consists of the elements i∈I mi such that only nitely many of the components mi are nonzero; the action of R on the Q Q direct product or direct sum is given by r i∈I mi = i∈i rmi (cf. Appendix I for componentwise multiplication. In other words, the direct sum of the

the denition of Cartesian products of innitely many sets). The direct sum will be denoted by

L

i∈i

Mi .

(a) Prove that the direct product of the of the

Mi 's

Mi 's

is an

R-module

is a submodule of their direct product. R = Z, I = Z+ and Mi is the cyclic group of order

(b) Show that if

i,

and the direct sum

then the direct sum of the

Mi 's

i

for each

is not isomorphic to their direct product.

[Look at torsion.] Solution.

(a) We have

(mi )i∈I + (m0i )i∈I = (mi + m0i )i∈I ∈

Y

Mi

and

Y r(mi )i∈I = (rmi )i∈I ∈ Mi , Q L Q so Mi is an R-module. To show that Mi is a submodule of Mi , we 0 must show that if all but nitely many of the mi and mi are zero, then this 0 is also true of the mi + m + i and rmi . For the sum, there are only nitely 0 many indices i so that at least one of mi and mi is nonzero, so there can only be nitely many indices in which the sum is nonzero. Similarly, for the scalar product, there are only nitely many the only positions in which

rmi

i

in which

mi

can be nonzero.

is nonzero, and these are Hence the direct sum is

indeed a submodule of the direct product.

Q M is a torsion group, whereas Mi is i L not. To see that Mi is torsion, let (mi ) be any element of Mi , and suppose that n is large enough so that mi = 0 for i > n. Then n!(mi ) = 0, so (mi ) is L a torsion element. Since (mi ) was arbitrary, Mi is a torsion group. Q Now, Q let's illustrate a non-torsion element of Mi . Take x = (1, 1, 1, . . .) ∈= Mi . th place, n Suppose that nx = 0. Then in order to annihilate the 1 in the i + must be a multiple of i. Hence, n must be a multiple of i for each i ∈ Z . The only integer with this property is 0, so x is torsion-free. L Mi is countable (as a set), whereas Q Another way to do this is to note that Mi is uncountable.

(b) One way to do this is to observe that

L

HOMEWORK 1 SOLUTIONS

7

R be a Principal Ideal Domain, let M be a torsion R-module p be a prime in R (do not assume M is nitely generated, hence it need not have a nonzero annihilator  cf. Exercise 5). The p-primary component of M is the set of all elements of M that are annihilated by some positive power of p. (a) Prove that the p-primary component is a submodule. [See Exercise 13 in

Problem (10.3.22). Let

(cf. Exercise 4) and let

Section 1.] (b) Prove that this denition of in Exercise 18 when (c) Prove that as

p

M

M

p-primary

component agrees with the one given

has a nonzero annihilator.

is the (possibly innite) direct sum of its

runs over all primes of

p-primary components,

R.

M [p] denote the p-primary component of M . Suppose that m, n ∈ M [p] and r ∈ R. Suppose that m and n are annihilated by pa and pb ,

Solution.

(a) Let

respectively. Then

pa+b (m + n) = pa+b m + pa+b n = 0, so

m + n ∈ M [p].

Also,

pa (rm) = r(pa m) = r0 = 0, so

rm ∈ M [p].

Hence

M [p]

is a submodule of

(b) In this case, any element of

M [p]

M.

is annihilated by

p αp ,

so the denitions

coincide.

isomorphism.

φ : M →

L

M [p], as follows. Let m ∈ M . Then m is contained in a nitely generated R-submodule of M , say N = Rm. Then N has an annihilator ann(N ), so by Exercise 10.3.18, N is the direct sum ∼ L of its p-primary components, θ : N → N [p]. Now, we embed each N [p] inside the corresponding M [p] in the natural way: ιp : N [p] ,→ M [p]. Now, L we dene φ(m) = ( ιp ◦ θ)(m). One can check that φ is in fact an R-module

(c) We dene a map