5.8 Injective and Projective Modules

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Let R be a ring and let P be an R-module. We say that P is projective if

every diagram of the form

M M00 0 (exact)

P

- -

?

_

_

can be embedded in a commutative diagram of the form

M M00 0 (exact)

P

- -

?

_

_

_

_

_ _+

¯_

In an entirely dual sense, the R-module I is said to be injective if every

diagram of the form

0 M0 M (exact)

I

- -

6

μ

_

can be embedded in a commutative diagram of the form

0 M0 M (exact)

I

- -

6

μ

_

Q

Q

Q Qk ¯_

We have the following simple characterization of projective modules.

Theorem 5.8.1 The following conditions are equivalent for the R-module

P.

(i) P is projective.

(ii) Every short exact sequence 0 ! M0 ! M ! P ! 0 splits.

(iii) P is a direct summand of a free R-module.

136 CHAPTER 5. MODULE THEORY

The next result gives a very important class of projective R-modules.

Assume that R is an integral domain with fraction field E. Recall from our

discussions of Dedekind domains that we defined, for any ideal I _ R,

I−1 = {_ 2 E| _I _ R}.

Recall also that the ideal I was called invertible if I−1I = R.

Theorem 5.8.2 Let R be an integral domain and let I _ R be an ideal.

(i) If I is invertible, then I is a projective R-module.

(ii) If I is a finitely generated ideal and is projective, then I is invertible.

Note that the above theorem gives a great number of interesting examples

of projective modules which aren’t free. Indeed, if R is any Dedekind

domain which isn’t a principal ideal domain then there will be non-free ideals

(cf. Exercise 12, Below) of R. However, we have seen that any ideal of

a Dedekind domain is invertible, hence is a projective R-module.

In order to obtain a characterization of injective modules we need a

concept dual to that of a free module. First, however, we need the concept

of a divisible abelian group. An abelian group D is divisible if for every

d 2 D and for every 0 6= n 2 Z, there is some c 2 D such that nc = d.

Example 1. The most obvious example of a divisible group is probably the

additive group (Q, +) of rational numbers.

Example 2. A moment’s thought should reveal that if F is any field of

characteristic 0, then (F, +) is a divisible group.

Example 3. Note that any homomorphic image of a divisible group is

divisible. Of paramount importance is the divisible group Q/Z.

Example 4. If p is a prime, the group Z(p1) is a divisible group. (You

should check this.)

The importance of divisible groups is the following.

Theorem 5.8.3 Let D be an abelian group. Then D is divisible if and only

if D is injective.

5.8. INJECTIVE AND PROJECTIVE MODULES 137

Let R be a ring, and let A be an abelian group. DefineM = HomZ(R,A);

thus M is certainly an abelian group under pointwise operations. Give M

the structure of a (left) R-module via

(a · f)(b) = f(ba), a, b 2 R, f 2 M.

It is easy to check that the above recipe gives HomZ(R,A) the structure of

a left R-module. (See Exercise 10.)

The importance of the above construction is found in the following.

Proposition 5.8.4 Let R be a ring and let D be a divisible abelian group.

Then the R-module HomZ(R,D) is an injective R-module.

Recall that any free R-module is the direct sum of a number of copies

of R, and that any R-module is a homomorphic image of a free module.

We now define a cofree R-module to be the direct product (not sum!) of a

number of copies of the injective module HomZ(R,Q/Z). We then have

Proposition 5.8.5 Let M be an R-module. Then M can be embedded in

a cofree R-module.

Finally we have the analogue of Theorem 41, above.

Theorem 5.8.6 The following conditions are equivalent for the R-module

I.

(i) I is injective.

(ii) Every short exact sequence 0 ! I ! M ! M00 ! 0 splits.

(iii) I is a direct summand of a cofree R-module.

Exercises

1. Prove that the direct sum

M

i2I

Pi is projective if and only if each Pi is.

2. Prove that the direct product

Y

i2I

Ii is injective if and only if each Ii

is.

138 CHAPTER 5. MODULE THEORY

3. Let R be a ring, let A be a fixed R-module, and let _ : M ! N be a

homomorphism of R-modules. Define

__ : HomR(A,M) ! HomR(A,N),

__ : HomR(N,A) ! HomR(M,A),

by setting __(f) = _ _ f, f 2 HomR(A,M), __(f) = f _ _, f 2

HomR(N,A). Prove that __ and __ are both homomorphisms of abelian

groups. (Warning: it need not be the case that either of HomR(A,M),

HomR(M,A) is an R-module.)

4. Let P be an R-module. Prove that P is projective if and only if given

any exact sequence 0 ! M0 μ

! M _! M00 ! 0, the induced sequence

0 ! HomR(P,M0) μ_! HomR(P,M) __! HomR(P,M00) ! 0

is exact.

5. Suppose we have a sequence 0 ! M0 μ

! M _! M00 ! 0 of R-modules.

Prove that this sequence is exact if and only if the sequence

0 ! HomR(P,M0) μ_! HomR(P,M) __! HomR(P,M00) ! 0

is exact for every projective R-module P.

6. Let I be an R-module. Prove that I is injective if and only if given

any exact sequence 0 ! M0 μ

! M _! M00 ! 0, the induced sequence

0 ! HomR(M00, I) __

! HomR(M, I) μ_

! HomR(M0, I) ! 0

is exact.

7. Suppose we have a sequence 0 ! M0 μ

! M _! M00 ! 0 of R-modules.

Prove that this sequence is exact if and only if the sequence

0 ! HomR(M00, I) __

! HomR(M, I) μ_

! HomR(M0, I) ! 0

is exact for every injective R-module I.

8. Let M be an R-module. Prove that

HomR(R,M) _=R M.

5.8. INJECTIVE AND PROJECTIVE MODULES 139

9. Prove that if A_, _ 2 A, is a family of abelian groups, then

HomZ(R,

Y

_2A

A_) _=R

Y

_2A

HomZ(R,A_).

10. Let M be a right R-module, and let A be an abelian group. Prove

that the scalar multiplication (r · f)(m) = f(mr), r 2 R, m 2 M, f 2

HomZ(M,A) gives HomZ(M,A) the structure of a left R-module.

11. Prove that there is a natural isomorphism of abelian groups:

HomR(M,HomZ(R,A)) _= HomZ(M,A),

where M is an R-module and A is an abelian group.

12. Let R be an integral domain in which every ideal is a free R-module.

Prove that R is a principal ideal domain.

13. Let F be a field and let R be the ring

R =

__

a b

0 c

_

| a, b, c 2 F

_

,

with left R-modules

M =

__

a

b

_

| a, b 2 F

_

and

L =

__

a

0

_

| a 2 F

_

as in Exercises 6 and 7 of Section 5.7.

(a) Prove that L is a projective R-module, but that M/L is not.

(b) If we set

I =

__

a b

0 0

_

| a, b 2 F

_

,

show that the ideal I is a projective R-module.

140 CHAPTER 5. MODULE THEORY

14. A ring for which every ideal is projective is called a hereditary ring.

In Section 4.6 we saw that every every ideal of a Dedekind domain

R is invertible. In turn, by Theorem 5.8.2 every invertible ideal is a

projective R-module. Thus, Dedekind domains are hereditary. Prove

that if F is a field, then the ring Mn(F) of n × n matrices over F is

hereditary. The same is true for the ring of lower triangular n × n

matrices over F.

15. Let A be an abelian group and let B _ A be such that A/B is infinite

cyclic. Prove that A _= A/B × B.

16. Let A be an abelian group and assume that A = H × Z1 = K × Z2

where Z1 and Z2 are infinite cyclic. Prove that H _= K. (Hint: First

H/(H \ K) _= HK/K _ A/K _= Z2 so H/(H \ K) is either trivial

or infinite cyclic. Similarly for K/(H \ K). Next A/(H \ K) _=

H/(H \ K) × Z1 and A/(H \ K) _= K/(H \ K) × Z2 so H/(H \ K)

and K/(H \K) are either both trivial (in which case H = K) or both

infinite cyclic. Thus, from Exercise 10 obtain H _= H/(H \ K) × H \

K _= K/(H \ K) × H \ K _= K, done.)

17. Prove Baer’s Criterion: Let I be an R-module and assume that for

any left ideal J _ R and any R-module homomorphism _J : J ! I,

_ extends to an R-module homomorphis _ : R ! I. Show that I is

injective. (Hint: Let M0 _ M be R-modules and assume that there

is an R-module homomorphism _ : M0 ! I. Consider the poset of

pairs (N, _N), where M0 _ N _ M and where _N extends _. Apply

Zorn’s Lemma to obtain a maximal element (N0, _0). If N0 6= M, let

m 2 M −N0 and let J = {r 2 R| rm 2 N0}; note that J is a left ideal

of R. Now what?)

18. Let R be a Dedekind domain with fraction field E. Recall that a

fractional ideal is simply a finitely-generated R-submodule of E. If

J _ E is a fractional ideal, prove that J is a projective R-module.

(Hint: As for ordinary ideals, define J−1 = {_ 2 E| _J _ R}. Using

Lemma 4.5.1 of Section 4.5, argue that J−1J = R. Now argue exactly

as in Theorem 5.8.2, (i) to prove that J is a projective R-module.)

19. Let R be a Dedekind domain with field of fractions E. If I, J _ E are

fractional ideals, and if 0 6= _ 2 HomR(I, J), prove that _ is injective.

(Hint: If J0 = im _, then argue that J0 is a projective R-module.

5.8. INJECTIVE AND PROJECTIVE MODULES 141

Therefore, One obtains I = ker _ _ J0, where J _=R J0. Why is such

a decomposition a contradition?)

20. Let R be a Noetherian domain. Prove that R is a Dedekind domain if

and only if every ideal of R is a projective R-module. (See Exercise 5

of Section 4.6.)

142 CHAPTER 5. MODULE THEORY

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